1.4: Set Operations with Two Sets
After completing this section, you should be able to:
- Determine the intersection of two sets.
- Determine the union of two sets.
- Determine the cardinality of the union of two sets.
- Apply the concepts of AND and OR to set operations.
- Draw conclusions from Venn diagrams with two sets.
The movie Yours, Mine, and Ours was originally released in 1968 and starred Lucille Ball and Henry Fonda. This movie, which is loosely based on a true story, is about the marriage of Helen, a widow with eight children, and Frank, a widower with ten children, who then have an additional child together. The movie is a comedy that plays on the interpersonal and organizational struggles of feeding, bathing, and clothing twenty people in one household.
If we consider the set of Helen's children and the set of Frank's children, then the child they had together is the intersection of these two sets, and the collection of all their children combined is the union of these two sets. In this section, we will explore the operations of union and intersection as it relates to two sets.
The Intersection of Two Sets
The members that the two sets share in common are included in the intersection of two sets . To be in the intersection of two sets, an element must be in both the first set and the second set. In this way, the intersection of two sets is a logical AND statement. Symbolically, intersection is written as: . intersection is written in set builder notation as: .
Let us look at Helen's and Frank's children from the movie Yours, Mine, and Ours. Helen's children consist of the set \(H=\{\) Colleen, Nick, Janette, Tommy, Jean, Phillip, Gerald, Theresa, Joseph\} and Frank's children are included in the set \(F=\{\) Mike, Rusty, Greg, Rosemary, Loise, Susan, Veronica, Mary, Germaine, Joan, Joseph \}. H intersection \(F\) is the set of children they had together. \(H \cap F=\{\) Joseph \(\}\), because Joseph is in both set \(H\) and set \(F\).
Example 1.23
Finding the Intersection of Set and Set
Set and Find intersection
- Answer
-
The intersection of sets and include the elements that set and have in common: 3, 5, and 7.
Your Turn 1.23
Notice that if sets and are disjoint sets, then they do not share any elements in common, and intersection is the empty set, as shown in the Venn diagram below.
Figure 1.21Example 1.24
Determining the Intersection of Disjoint Sets
Set and set Find
- Answer
-
Because sets and are disjoint, they do not share any elements in common. So, the intersection of set and set is the empty set.
Your Turn 1.24
Notice that if set is a subset of set , then intersection is equal to set , as shown in the Venn diagram below.
Figure 1.22Example 1.25
Finding the Intersection of a Set and a Subset
Set and set Find
- Answer
-
Because set is a subset of set , intersection is equal to set . the set of odd natural numbers.
Your Turn 1.25
The Union of Two Sets
Like the union of two families in marriage, the union of two sets includes all the members of the first set and all the members of the second set. To be in the union of two sets, an element must be in the first set, the second set, or both. In this way, the union of two sets is a logical inclusive OR statement. Symbolically, union is written as: union is written in set builder notation as:
Let us consider the sets of Helen's and Frank's children from the movie Yours, Mine, and Ours again. Helen's children is set \(H=\{\) Colleen, Nick, Janette, Tommy, Jean, Phillip, Gerald, Theresa, Joseph\} and Frank's children is set \(F=\{\) Mike, Rusty, Greg, Rosemary, Loise, Susan, Veronica, Mary, Germaine, Joan, Joseph \(\}\). The union of these two sets is the collection of all nineteen of their children, \(H \cup F=\{\) Colleen, Nick, Janette, Tommy, Jean, Phillip, Gerald, Theresa, Joseph, Mike, Rusty, Greg, Rosemary, Loise, Susan, Veronica, Mary, Germaine, Joan\}. Notice, Joseph is in both set \(H\) and set \(F\), but he is only one child, so, he is only listed once in the union.
Example 1.26
Finding the Union of Sets and When and Overlap
Set and set . Find union .
- Answer
-
union is the set formed by including all the unique elements in set , set , or both sets and : The first five elements of the union are the five unique elements in set . Even though 3, 5, and 7 are also members of set , these elements are only listed one time. Lastly, set includes the unique element 2, so 2 is also included as part of the union of sets and .
Your Turn 1.26
When observing the union of sets and , notice that both set and set are subsets of union . Graphically, union can be represented in several different ways depending on the members that they have in common. If and are disjoint sets, then union would be represented with two disjoint circles within the universal set, as shown in the Venn diagram below.
Figure 1.23If sets and share some, but not all, members in common, then the Venn diagram is drawn as two separate circles that overlap.
Figure 1.24If every member of set is also a member of set , then is a subset of set , and union would be equal to set . To draw the Venn diagram, the circle representing set should be completely enclosed in the circle containing set .
Figure 1.25Example 1.27
Finding the Union of Sets and When and Are Disjoint
Set and set Find
- Answer
-
Because sets and are disjoint, the union is simply the set containing all the elements in both set and set .
Your Turn 1.27
Set \(A=\{\) red, yellow, blue \(\}\) and set \(B=\) \{orange, green, purple \(\}\). Find \(A \cup B\).
Example 1.28
Finding the Union of Sets and When One Set is a Subset of the Other
Set and set Find
- Answer
-
Because set is a subset of set , union is equal to set .
Your Turn 1.28
1. Set \(A=\{a, b, c, \ldots, z\}\) and set \(B=\{a, e, i, o, u\}\). Find \(A \cup B\).
Video
The Basics of Intersection of Sets, Union of Sets and Venn Diagrams
Tech Check
Set Operation Practice
Sets Challenge is an application available on both Android and iPhone smartphones that allows you to practice and gain familiarity with the operations of set union, intersection, complement, and difference.
Figure 1.26The Sets Challenge application/game uses some notation that differs from the notation covered in the text.
- The complement of set in this text is written symbolically as but the Sets Challenge game uses to represent the complement operation.
- In the text we do not cover set difference between two sets and , represented in the game as In the game this operation removes from set all the elements in For example, if set and set are subsets of the universal set then and There is a project at the end of the chapter to research the set difference operation.
Determining the Cardinality of Two Sets
The cardinality of the union of two sets is the total number of elements in the set. Symbolically the cardinality of union is written, . If two sets and are disjoint, the cardinality of union is the sum of the cardinality of set and the cardinality of set . If the two sets intersect, then intersection is a subset of both set and set . This means that if we add the cardinality of set and set , we will have added the number of elements in intersection twice, so we must then subtract it once as shown in the formula that follows.
FORMULA
The cardinality of union is found by adding the number of elements in set to the number of elements in set , then subtracting the number of elements in the intersection of set and set . or
Checkpoint
If sets and are disjoint, then and the formula is still valid, but simplifies to
Example 1.29
Determining the Cardinality of the Union of Two Sets
The number of elements in set is 10, the number of elements in set is 20, and the number of elements in intersection is 4. Find the number of elements in union .
- Answer
-
Using the formula for determining the cardinality of the union of two sets, we can say
Your Turn 1.29
1. If \(n(A)=23, n(B)=17\), and \(n(A \cap B)=7\), then find \(n(A \cup B)\).
Example 1.30
Determining the Cardinality of the Union of Two Disjoint Sets
If and are disjoint sets and the cardinality of set is 37 and the cardinality of set is 43, find the cardinality of union .
- Answer
-
To find the cardinality of union , apply the formula, Because sets and are disjoint, is the empty set, therefore and
Your Turn 1.30
Applying Concepts of “AND” and “OR” to Set Operations
To become a licensed driver, you must pass some form of written test and a road test, along with several other requirements depending on your age. To keep this example simple, let us focus on the road test and the written test. If you pass the written test but fail the road test, you will not receive your license. If you fail the written test, you will not be allowed to take the road test and you will not receive a license to drive. To receive a driver's license, you must pass the written test AND the road test. For an “AND” statement to be true, both conditions that make up the statement must be true. Similarly, the intersection of two sets and is the set of elements that are in both set and set . To be a member of intersection , an element must be in set and also must be in set . The intersection of two sets corresponds to a logical "AND" statement.
The union of two sets is a logical inclusive "OR" statement. Say you are at a birthday party and the host offers Leah, Lenny, Maya, and you some cake or ice cream for dessert. Leah asks for cake, Lenny accepts both cake and ice cream, Maya turns down both, and you choose only ice cream. Leah, Lenny, and you are all having dessert. The “OR” statement is true if at least one of the components is true. Maya is the only one who did not have cake or ice cream; therefore, she did not have dessert and the “OR” statement is false. To be in the union of two sets and , an element must be in set or set or both set and set .
Example 1.31
Applying the "AND" or "OR" Operation
and
Find the set consisting of elements in:
- Answer
-
- because only the elements 0 and 12 are members of both set and set .
- because the set or is the collection of all elements in set or set , or both.
- because the set or is the collection of all elements in set or set , or both.
- Parentheses are evaluated first: because the only member that both set and set share in common is 8. So, now we need to find Because the word translates to the union operation, the problem becomes which is equal to
Your Turn 1.31
Example 1.32
Determine and Apply the Appropriate Set Operations to Solve the Problem
Don Woods is serving cake and ice cream at his Juneteenth celebration. The party has a total of 54 guests in attendance. Suppose 30 guests requested cake, 20 guests asked for ice cream, and 12 guests did not have either cake or ice cream.
- How many guests had cake or ice cream?
- How many guests had cake and ice cream?
- Answer
-
- The total number of people at the party is 54, and 12 people did not have cake or ice cream. Recall that the total number of elements in the universal set is always equal to the number of elements in a subset plus the number of elements in the complement of the set, That means , or equivalently, A total of 42 people at the party had cake or ice cream.
- To determine the number of people who had both cake and ice cream, we need to find the intersection of the set of people who had cake and the set of people who had ice cream. From Question 1, the number of people who had cake or ice cream is 42. This is the union of the two sets. The formula for the union of two sets is Use the information given in the problem and substitute the known values into the formula to solve for the number of people in the intersection: Adding 30 and 20, the equation simplifies to Which means
Your Turn 1.32
Who Knew?
The Real Inventor of the Venn Diagram
John Venn, in his writings, references works by both John Boole and Augustus De Morgan, who referred to the circle diagrams commonly used to present logical relationships as Euler's circles. Leonhard Euler's works were published over 100 years prior to Venn's, and Euler may have been influenced by the works of Gottfried Leibniz.
So, why does John Venn get all the credit for these graphical depictions? Venn was the first to formalize the use of these diagrams in his book Symbolic Logic, published in 1881. Further, he made significant improvements in their design, including shading to highlight the region of interest. The mathematician C.L. Dodgson, also known as Lewis Carroll, built upon Venn’s work by adding an enclosing universal set.
Invention is not necessarily coming up with an initial idea. It is about seeing the potential of an idea and applying it to a new situation.
References:
Margaret E. Baron. "A Note on the Historical Development of Logic Diagrams: Leibniz, Euler and Venn." The Mathematical Gazette , vol. 53, no. 384, 1969, pp. 113-125. JSTOR , www.jstor.org/stable/3614533. Accessed 15 July 2021.
Deborah Bennett. "Drawing Logical Conclusions." Math Horizons , vol. 22, no. 3, 2015, pp. 12-15. JSTOR , www.jstor.org/stable/10.4169/mathhorizons.22.3.12. Accessed 15 July 2021.
Drawing Conclusions from a Venn Diagram with Two Sets
All Venn diagrams will display the relationships between the sets, such as subset, intersecting, and/or disjoint. In addition to displaying the relationship between the two sets, there are two main additional details that Venn diagrams can include: the individual members of the sets or the cardinality of each disjoint subset of the universal set.
A Venn diagram with two subsets will partition the universal set into 3 or 4 sections depending on whether they are disjoint or intersecting sets. Recall that the complement of set , written is the set of all elements in the universal set that are not in set
Figure 1.27Example 1.33
Using a Venn Diagram to Draw Conclusions about Set Membership
- Find
- Find
- Find .
- Find
- Answer
-
- because union is the collection of all elements in set or set or both.
- Because and are disjoint sets, there are no elements that are in both and . Therefore, intersection is the empty set,
- The complement of set is the set of all elements in the universal set that are not in set :
-
The cardinality, or number of elements in set
is n ( B ′ ) = 6.
Your Turn 1.33
Example 1.34
Using a Venn Diagram to Draw Conclusions about Set Cardinality
- Find
- Find
- Find
- Answer
-
- The number of elements in or is the number of elements in union :
- The number of elements in and is the number of elements in intersection :
- The number of elements in set is the sum of all the numbers enclosed in the circle representing set :


16.20: Chains Subordinate to the Poisson Process
Basic Theory
Introduction
Recall that the standard Poisson process with rate parameter \( r \in (0, \infty) \) involves three interrelated stochastic processes. First the sequence of interarrival times \(\bs{T} = (T_1, T_2, \ldots) \) is independent, and each variable has the exponential distribution with parameter \( r \). Next, the sequence of arrival times \( \bs{\tau} = (\tau_0, \tau_1, \ldots) \) is the partial sum sequence associated with the interrival sequence \( \bs{T} \): \[ \tau_n = \sum_{i=1}^n T_i, \quad n \in \N \] For \( n \in \N_+ \), the arrival time \( \tau_n \) has the gamma distribution with parameters \( n \) and \( r \). Finally, the Poisson counting process \( \bs{N} = \{N_t: t \in [0, \infty)\} \) is defined by \[ N_t = \max\{n \in \N: \tau_n \le t\}, \quad t \in [0, \infty) \] so that \( N_t \) is the number of arrivals in \( (0, t] \) for \( t \in [0, \infty) \). The counting variable \( N_t \) has the Poisson distribution with parameter \( r t \) for \( t \in [0, \infty) \). The counting process \( \bs{N} \) and the arrival time process \( \bs{\tau} \) are inverses in the sense that \( \tau_n \le t \) if and only if \( N_t \ge n \) for \( t \in [0, \infty) \) and \( n \in \N \). The Poisson counting process can be viewed as a continuous-time Markov chain.
Suppose that \( X_0 \) takes values in \( \N \) and is independent of \( \bs{N} \). Define \( X_t = X_0 + N_t \) for \( t \in [0, \infty) \). Then \( \bs{X} = \{X_t: t \in [0, \infty)\} \) is a continuous-time Markov chain on \( \N \) with exponential parameter function given by \( \lambda(x) = r \) for \( x \in \N \) and jump transition matrix \( Q \) given by \( Q(x, x + 1) = 1 \) for \( x \in S \).
Proof
This follows directly from the basic structure of a continuous-time Markov chain. Given \( X_t = x \), the holding time in state \( x \in \N \) is exponential with parameter \( r \), and the next state is deterministically \( x + 1 \). Note that the addition of the variable \( X_0 \) is just to allow us the freedom of arbitrary initial distributions on the state space, as is routine with Markov processes.
Note that the Poisson process, viewed as a Markov chain is a pure birth chain. Clearly we can generalize this continuous-time Markov chain in a simple way by allowing a general embedded jump chain.
Suppose that \( \bs{X} = \{X_t: t \in [0, \infty)\} \) is a Markov chain with (countable) state space \( S \), and with constant exponential parameter \( \lambda(x) = r \in (0, \infty) \) for \( x \in S \), and jump transition matrix \( Q \). Then \( \bs{X} \) is said to be subordinate to the Poisson process with rate parameter \( r \).
- The transition times \( (\tau_1, \tau_2, \ldots) \) are the arrival times of the Poisson process with rate \( r \).
- The inter-transition times \( (\tau_1, \tau_2 - \tau_1, \ldots) \) are the inter-arrival times of the Poisson process with rate \( r \) (independent, and each with the exponential distribution with rate \( r \)).
- \( \bs{N} = \{N_t: t \in [0, \infty)\} \) is the Poisson counting process, where \( N_t \) is the number of transitions in (0, t] for \( t \in [0, \infty) \).
- The Poisson process and the jump chain \( \bs{Y} = (Y_0, Y_1, \ldots) \) are independent, and \( X_t = Y_{N_t} \) for \( t \in [0, \infty) \).
Proof
These results all follow from the basic structure of a continuous-time Markov chain.
Since all states are stable, note that we must have \( Q(x, x) = 0 \) for \( x \in S \). Note also that for \( x, \, y \in S \) with \( x \ne y \), the exponential rate parameter for the transition from \( x \) to \( y \) is \( \mu(x, y) = r Q(x, y) \). Conversely suppose that \( \mu: S^2 \to (0, \infty) \) satisfies \( \mu(x, x) = 0 \) and \( \sum_{y \in S} \mu(x, y) = r \) for every \( x \in S \). Then the Markov chain with transition rates given by \( \mu \) is subordinate to the Poisson process with rate \( r \). It's easy to construct a Markov chain subordinate to the Poisson process.
Suppose that \( \bs N = \{N_t: t \in [0, \infty)\} \) is a Poisson counting process with rate \( r \in (0, \infty) \) and that \( \bs Y = \{Y_n: n \in \N\} \) is a discrete-time Markov chain on \( S \), independent of \( \bs N \), whose transition matrix satisfies \( Q(x, x) = 0 \) for every \( x \in S \). Let \( X_t = Y_{N_t} \) for \( t \in [0, \infty) \). Then \( \bs X = \{X_t: t \in [0, \infty)\} \) is a continuous-time Markov chain subordinate to the Poisson process.
Generator and Transition Matrices
Next let's find the generator matrix and the transition semigroup. Suppose again that \( \bs{X} = \{X_t: t \in [0, \infty)\} \) is a continuous-time Markov chain on \( S \) subordinate to the Poisson process with rate \( r \in (0, \infty) \) and with jump transition matrix \( Q \). As usual, let \( \bs P = \{P_t: t \in [0, \infty)\} \) denote the transition semigroup and \( G \) the infinitesimal generator.
The generator matrix \( G \) of \( \bs{X} \) is \( G = r (Q - I) \). Hence for \( t \in [0, \infty) \)
- The Kolmogorov backward equation is \( P^\prime_t = r (Q - I) P_t \)
- The Kolmogorov forward equation is \( P^\prime_t = r P_t (Q - I) \)
Proof
This follows directly from the general theory since \( G(x, x) = -\lambda(x) = -r \) for \( x \in S \) and \( G(x, y) = \lambda(x) Q(x, y) = r Q(x, y) \) for distinct \( x, \, y \in S \).
There are several ways to find the transition semigroup \( \bs{P} = \{P_t: t \in [0, \infty)\} \). The best way is a probabilistic argument using the underlying Poisson process.
For \( t \in [0, \infty) \), the transition matrix \( P_t \) is given by \[ P_t = \sum_{n=0}^\infty e^{-r t} \frac{(r t)^n}{n!} Q^n \]
Proof from the underlying Poisson process
Let \( N_t \) denote the number of transitions in \( (0, t] \) for \( t \in [0, \infty) \), so that \( \bs{N} = \{N_t: t \in [0, \infty)\} \) is the Poisson counting process. Let \( \bs{Y} = (Y_0, Y_1, \ldots) \) denote the jump chain, with transition matrix \( Q \). Then \( \bs{N} \) and \( \bs{Y} \) are independent, and \( X_t = Y_{N_t} \) for \( t \in [0, \infty) \). Conditioning we have \begin{align*} P_t(x, y) & = \P(X_t = y \mid X_0 = x) = \P\left(Y_{N_t} = y \mid Y_0 = x\right) \\ & = \sum_{n=0}^\infty \P\left(Y_{N_t} = y \mid N_t = n, Y_0 = y\right) \P(N_t = n \mid Y_0 = y) \\ & = \sum_{n=0}^\infty \P(Y_n = y \mid Y_0 = x) \P(N_t = n) = \sum_{n=0}^\infty e^{-r t} \frac{(r t)^n}{n!} Q^n(x, y) \end{align*}
Proof using the generator matrix
Note first that for \( n \in \N \), \[ G^n = [r (Q - I)]^n = r^n \sum_{k = 0}^n \binom{n}{k}(-1)^{n-k} Q^k \] Hence \begin{align*} P_t & = e^{t G} = \sum_{n=0}^\infty \frac{t^n}{n!} G^n = \sum_{n=0}^\infty \frac{t^n}{n!} r^n \sum_{k=0}^\infty \binom{n}{k} (-1)^{n-k}Q^k \\ & = \sum_{n=0}^\infty \sum_{k=0}^n \frac{(r t)^n}{k! (n - k)!} (-1)^{n-k} Q^k = \sum_{k=0}^\infty \sum_{n=k}^\infty \frac{(r t)^n}{k! (n - k)!} (-1)^{n-k} Q^k \\ & = \sum_{k=0}^\infty \frac{(r t)^k}{k!} Q^k \sum_{n=k}^\infty \frac{1}{(n - k)!}(- r t)^{n-k} = \sum_{k=0}^\infty e^{-r t} \frac{(r t)^k}{k!} Q^k \end{align*}
Potential Matrices
Next let's find the potential matrices. As with the transition matrices, we can do this in (at least) two different ways.
Suppose again that \( \bs{X} = \{X_t: t \in [0, \infty)\} \) is a continuous-time Markov chain on \( S \) subordinate to the Poisson process with rate \( r \in (0, \infty) \) and with jump transition matrix \( Q \). For \( \alpha \in (0, \infty) \), the potential matrix \( U_\alpha \) of \( \bs{X} \) is \[ U_\alpha = \frac{1}{\alpha + r} \sum_{n=0}^\infty \left(\frac{r}{\alpha + r}\right)^n Q^n \]
Proof from the definition
Using the previous result, \begin{align*} U_\alpha(x, y) & = \int_0^\infty e^{-\alpha t} P_t(x, y) \, dt = \int_0^\infty e^{-\alpha t} \sum_{n=0}^\infty e^{-r t} \frac{(r t)^n}{n!} Q^n(x, y) \, dt \\ & = \sum_{n=0}^\infty Q^n(x, y) \frac{r^n}{n!} \int_0^\infty e^{-(r + \alpha) t} t^n dt \end{align*} The interchange of sum and integral is justified since the terms are nonnegative. Using the change of variables \( s = (r + \alpha) t \) gives \[ U_\alpha(x, y) = \frac{1}{\alpha + r} \sum_{n=0}^\infty \left(\frac{r}{\alpha + r}\right)^n \frac{1}{n!} Q^n(x, y) \int_0^\infty e^{-s t} s^n \, ds \] The last integral is \( n! \).
Proof using the generator
From the result above, \[ \alpha I - G = \alpha I - r (Q - I) = (\alpha + r) I - r Q = (\alpha + r)\left(I - \frac{r}{\alpha + r} Q\right) \] Since \( \left\| \frac{r}{\alpha + r} Q \right\| = \frac{r}{\alpha + r} \lt 1 \) we have \[ (\alpha I - G)^{-1} = \frac{1}{\alpha + r}\left(I - \frac{r}{\alpha + r} Q\right)^{-1} = \frac{1}{\alpha + r} \sum_{n=0}^\infty \left(\frac{r}{\alpha + r}\right)^n Q^n \]
Recall that for \( p \in (0, 1) \), the \( p \)-potential matrix of the jump chain \( \bs{Y} \) is \( R_p = \sum_{n=0}^\infty p^n Q^n \). Hence we have the following nice relationship between the potential matrix of \( \bs{X} \) and the potential matrix of \( \bs{Y} \): \[ U_\alpha = \frac{1}{\alpha + r} R_{r / (\alpha + r)} \] Next recall that \( \alpha U_\alpha(x, \cdot) \) is the probability density function of \( X_T \) given \( X_0 = x \), where \( T \) has the exponential distribution with parameter \( \alpha \) and is independent of \( \bs{X} \). On the other hand, \( \alpha U_\alpha(x, \cdot) = (1 - p) R_p(x, \cdot) \) where \( p = r \big/ (\alpha + r) \). We know from our study of discrete potentials that \( (1 - p) R_p(x, \cdot) \) is the probability density function of \( Y_M \) where \( M \) has the geometric distribution on \( \N \) with parameter \( 1 - p \) and is independent of \( \bs{Y} \). But also \( X_T = Y_{N_T} \). So it follows that if \( T \) has the exponential distribution with parameter \( \alpha \), \( \bs{N} = \{N_t: t \in [0, \infty)\} \) is a Poisson process with rate \( r \), and is independent of \( T \), then \( N_T \) has the geometric distribution on \( \N \) with parameter \( \alpha \big/ (\alpha + r) \). Of course, we could easily verify this directly, but it's still fun to see such connections.
Limiting Behavior and Stationary Distributions
Once again, suppose that \( \bs{X} = \{X_t: t \in [0, \infty)\} \) is a continuous-time Markov chain on \( S \) subordinate to the Poisson process with rate \( r \in (0, \infty) \) and with jump transition matrix \( Q \). Let \( \bs{Y} = \{Y_n: n \in \N\} \) denote the jump process. The limiting behavior and stationary distributions of \( \bs{X} \) are closely related to those of \( \bs{Y} \).
Suppose that \( \bs{X} \) (and hence \( \bs{Y} \)) are irreducible and positive recurrent
- \( g: S \to (0, \infty) \) is invariant for \( \bs{X} \) if and only if \( g \) is invariant for \( \bs{Y} \).
- \( f \) is an invariant probability density function for \( \bs{X} \) if and only if \( f \) is an invariant probability density function for \( \bs{Y} \).
- \( \bs{X} \) is null recurrent if and only if \( \bs{Y} \) is null recurrent, and in this case, \( \lim_{n \to \infty} Q^n(x, y) = \lim_{t \to \infty} P_t(x, y) = 0 \) for \( (x, y) \in S^2 \).
- \( \bs{X} \) is positive recurrent if and only if \( \bs{Y} \) is positive recurrent. If \( \bs{Y} \) is aperiodic, then \( \lim_{n \to \infty} Q^n(x, y) = \lim_{t \to \infty} P_t(x, y) = f(y) \) for \( (x, y) \in S^2 \), where \( f \) is the invariant probability density function.
Proof
All of these results follow from the basic theory of stationary and limiting distributions for continuous-time chains, and the fact that the exponential parameter function \( \lambda \) is constant.
Time Reversal
Once again, suppose that \( \bs{X} = \{X_t: t \in [0, \infty)\} \) is a continuous-time Markov chain on \( S \) subordinate to the Poisson process with rate \( r \in (0, \infty) \) and with jump transition matrix \( Q \). Let \( \bs{Y} = \{Y_n: n \in \N\} \) denote the jump process. We assume that \( \bs X \) (and hence \( \bs Y \)) are irreducible. The time reversal of \( \bs X \) is closely related to that of \( \bs Y \).
Suppose that \( g: S \to (0, \infty) \) is invariant for \( \bs X \). The time reversal \( \hat{\bs X} \) with respect to \( g \) is also subordinate to the Poisson process with rate \( r \). The jump chain \( \hat{\bs Y} \) of \( \hat{\bs X} \) is the (discrete) time reversal of \( \bs Y \) with respect to \( g \).
Proof
From the previous result, \( g \) is also invariant for \( \bs Y \). From the general theory of time reversal, \( \hat{\bs X} \) has the same exponential parameter function as \( \bs X \) (namely the constant function \( r \)) and so is also subordinate to the Poisson process with rate \( r \). Finally, the jump chain \( \hat{\bs Y} \) of \( \hat{\bs X} \) is the reversal of \( \bs Y \) with respect to \( r g \) and hence also with respect to \( g \).
In particular, \( \bs X \) is reversible with respect to \( g \) if and only if \( \bs Y \) is reversible with respect to \( g \). As noted earlier, \( \bs X \) and \( \bs Y \) are of the same type: both transient or both null recurrent or both positive recurrent. In the recurrent case, there exists a positive invariant function that is unique up to multiplication by constants. In this case, the reversal of \( \bs X \) is unique, and is the chain subordinate to the Poisson process with rate \( r \) whose jump chain is the reversal of \( \bs Y \).
Uniform Chains
In the construction above for a Markov chain \( \bs X = \{X_t: t \in [0, \infty)\} \) that is subordinate to the Poisson process with rate \( r \) and jump transition kernel \( Q \), we assumed of course that \( Q(x, x) = 0 \) for every \( x \in S \). So there are no absorbing states and the sequence \( (\tau_1, \tau_2, \ldots) \) of arrival times of the Poisson process are the jump times of the chain \( \bs X \). However in our introduction to continuous-time chains, we saw that the general construction of a chain starting with the function \( \lambda \) and the transition matrix \( Q \) works without this assumption on \( Q \), although the exponential parameters and transition probabilities change. The same idea works here.
Suppose that \( \bs N = \{N_t: t \in [0, \infty)\} \) is a counting Poisson process with rate \( r \in (0, \infty) \) and that \( \bs Y = \{Y_n: n \in \N\} \) is a discrete-time Markov chain with transition matrix \( Q \) on \( S \times S \) satisfying \( Q(x, x) \lt 1 \) for \( x \in S \). Assume also that \( \bs N \) and \( \bs Y \) are independent. Define \( X_t = Y_{N_t} \) for \( t \in [0, \infty) \). Then \( \bs X = \{X_t: t \in [0, \infty)\} \) is a continuous-Markov chain with exponential parameter function \( \lambda(x) = r [1 - Q(x, x)] \) for \( x \in S \) and jump transition matrix \( \tilde Q \) given by \[ \tilde Q(x, y) = \frac{Q(x, y)}{1 - Q(x, x)}, \quad (x, y) \in S^2, \, x \ne y \]
Proof
This follows from the result in the introduction.
The Markov chain constructed above is no longer a chain subordinate to the Poisson process by our definition above , since the exponential parameter function is not constant, and the transition times of \( \bs X \) are no longer the arrival times of the Poisson process. Nonetheless, many of the basic results above still apply.
Let \( \bs X = \{X_t: t \in [0, \infty)\} \) be the Markov chain constructed in the previous theorem. Then
- For \( t \in [0, \infty) \), the transition matrix \( P_t \) is given by \[ P_t = \sum_{n=0}^\infty e^{- r t} \frac{(r t)^n}{n!} Q^n \]
- For \( \alpha \in (0, \infty) \), the \( \alpha \) potential matrix is given by \[ U_\alpha = \frac{1}{\alpha + r} \sum_{n=0}^\infty \left(\frac{r}{\alpha + r}\right)^n Q^n \]
- The generator matrix is \( G = r (Q - I) \)
- \( g: S \to (0, \infty) \) is invariant for \( \bs X \) if and only if \( g \) is invariant for \( \bs Y \).
Proof
The proofs are just as before.
It's a remarkable fact that every continuous-time Markov chain with bounded exponential parameters can be constructed as in the last theorem, a process known as uniformization . The name comes from the fact that in the construction, the exponential parameters become constant, but at the expense of allowing the embedded discrete-time chain to jump from a state back to that state. To review the definition, suppose that \( \bs X = \{X_t: t \in [0, \infty)\} \) is a continuous-time Markov chain on \( S \) with transition semigroup \( \bs P = \{P_t: t \in [0, \infty)\} \), exponential parameter function \( \lambda \) and jump transition matrix \( Q \). Then \( \bs P \) is uniform if \( P_t(x, x) \to 1 \) as \( t \downarrow 0 \) uniformly in \( x \), or equivalently if \( \lambda \) is bounded.
Suppose that \( \lambda: S \to (0, \infty) \) is bounded and that \( Q \) is a transition matrix on \( S \) with \( Q(x, x) = 0 \) for every \( x \in S \). Let \( r \in (0, \infty) \) be an upper bound on \( \lambda \) and \( \bs N = \{N_t: t \in [0, \infty)\} \) a Poisson counting process with rate \( r \). Define the transition matrix \( \hat Q \) on \( S \) by \begin{align*} \hat Q(x, x) & = 1 - \frac{\lambda(x)}{r} \quad x \in S \\ \hat Q(x, y) & = \frac{\lambda(x)}{r} Q(x, y) \quad (x, y) \in S^2, \, x \ne y \end{align*} and let \( \bs Y = \{Y_n: n \in \N\} \) be a discrete-time Markov chain with transition matrix \( \hat Q \), independent of \( \bs N \). Define \( X_t = Y_{N_t} \) for \( t \in [0, \infty) \). Then \( \bs X = \{X_t: t \in [0, \infty)\} \) is a continuous-time Markov chain with exponential parameter function \( \lambda \) and jump transition matrix \( Q \).
Proof
Note that \( \hat Q(x, y) \ge 0 \) for every \( (x, y) \in S^2 \) and \( \sum_{y \in S} \hat Q(x, y) = 1 \) for every \( x \in S \). Thus \( \hat Q \) is a transition matrix on \( S \). Note also that \( \hat Q(x, x) \lt 1 \) for every \( x \in S \). By construction, \( \lambda(x) = r[1 - \hat Q(x, x)] \) for \( x \in S \) and \[ Q(x, y) = \frac{\hat Q(x, y)}{1 - \hat Q(x, x)}, \quad (x, y) \in S^2, \, x \ne y \] So the result now follows from the theorem above .
Note in particular that if the state space \( S \) is finite then of course \( \lambda \) is bounded so the previous theorem applies. The theorem is useful for simulating a continuous-time Markov chain, since the Poisson process and discrete-time chains are simple to simulate. In addition, we have nice representations for the transition matrices, potential matrices, and the generator matrix.
Suppose that \( \bs X = \{X_t: t \in [0, \infty\} \) is a continuous-time Markov chain on \( S \) with bounded exponential parameter function \( \lambda: S \to (0, \infty) \) and jump transition matrix \( Q \). Define \( r \) and \( \hat Q \) as in the last theorem. Then
- For \( t \in [0, \infty) \), the transition matrix \( P_t \) is given by \[ P_t = \sum_{n=0}^\infty e^{- r t} \frac{(r t)^n}{n!} \hat Q^n \]
- For \( \alpha \in (0, \infty) \), the \( \alpha \) potential matrix is given by \[ U_\alpha = \frac{1}{\alpha + r} \sum_{n=0}^\infty \left(\frac{r}{\alpha + r}\right)^n \hat Q^n \]
- The generator matrix is \( G = r (\hat Q - I) \)
- \( g: S \to (0, \infty) \) is invariant for \( \bs X \) if and only if \( g \) is invariant for \( \hat Q \).
Proof
These results follow from the theorem above .
Examples
The Two-State Chain
The following exercise applies the uniformization method to the two-state chain.
Consider the continuous-time Markov chain \( \bs X = \{X_t: t \in [0, \infty)\} \) on \( S = \{0, 1\} \) with exponential parameter function \( \lambda = (a, b) \), where \( a, \, b \in (0, \infty) \).  Show that
- \( \hat Q = \frac{1}{a + b} \left[\begin{matrix} b & a \\ b & a \end{matrix} \right] \)
- \( G = \left[\begin{matrix} -a & a \\ b & - b \end{matrix}\right] \)
- \( P_t = \hat Q - \frac{1}{a + b} e^{-(a + b) t} G \) for \( t \in [0, \infty) \)
- \( U_\alpha = \frac{1}{\alpha} \hat Q - \frac{1}{(\alpha + a + b)(a + b)} G \) for \( \alpha \in (0, \infty) \)
Proof
The form of \( \hat Q \) follows easily from the definition above . Note that the rows of \( \hat Q \) are the invariant PDF. It then follows that \( \hat Q^n = \hat Q \) for \( n \in \N_+ \). The results for the transition matrix \( P_t \) and the potential \( U_\alpha \) then follow easily from the theorem above .
Although we have obtained all of these results for the two-state chain before, the derivation based on uniformization is the easiest.


11.E: Molecular Biology of Gene Expression (Exercises)
11.1: The Structure of DNA
Multiple Choice
Which of the following does cytosine pair with?
A. guanine
B. thymine
C. adenine
D. a pyrimidine
- Answer
-
A
Prokaryotes contain a ________chromosome, and eukaryotes contain ________ chromosomes.
A. single-stranded circular; single-stranded linear
B. single-stranded linear; single-stranded circular
C. double-stranded circular; double-stranded linear
D. double-stranded linear; double-stranded circular
- Answer
-
C
Free Response
Describe the organization of the eukaryotic chromosome.
- Answer
-
The DNA is wound around proteins called histones. The histones then stack together in a compact form that creates a fiber that is 30-nm thick. The fiber is further coiled for greater compactness. During metaphase of mitosis, the chromosome is at its most compact to facilitate chromosome movement. During interphase, there are denser areas of chromatin, called heterochromatin, that contain DNA that is not expressed, and less dense euchromatin that contains DNA that is expressed.
Describe the structure and complementary base pairing of DNA.
- Answer
-
A single strand of DNA is a polymer of nucleic acids joined covalently between the phosphate group of one and the deoxyribose sugar of the next to form a “backbone” from which the nitrogenous bases stick out. In its natural state, DNA has two strands wound around each other in a double helix. The bases on each strand are bonded to each other with hydrogen bonds. Only specific bases bond with each other; adenine bonds with thymine, and cytosine bonds with guanine.
11.2: DNA Replication
Multiple Choice
DNA replicates by which of the following models?
A. conservative
B. semiconservative
C. dispersive
D. none of the above
- Answer
-
B
The initial mechanism for repairing nucleotide errors in DNA is ________.
A. mismatch repair
B. DNA polymerase proofreading
C. nucleotide excision repair
D. thymine dimers
- Answer
-
B
Free Response
How do the linear chromosomes in eukaryotes ensure that its ends are replicated completely?
- Answer
-
Telomerase has an inbuilt RNA template that extends the 3' end, so a primer is synthesized and extended. Thus, the ends are protected.
11.3: Transcription
Multiple Choice
A promoter is ________.
A. a specific sequence of DNA nucleotides
B. a specific sequence of RNA nucleotides
C. a protein that binds to DNA
D. an enzyme that synthesizes RNA
- Answer
-
A
Portions of eukaryotic mRNA sequence that are removed during RNA processing are ________.
A. exons
B. caps
C. poly-A tails
D. introns
- Answer
-
D
11.4: Translation
Multiple Choice
The RNA components of ribosomes are synthesized in the ________.
A. cytoplasm
B. nucleus
C. nucleolus
D. endoplasmic reticulum
- Answer
-
C
How long would the peptide be that is translated from this MRNA sequence: 5'-AUGGGCUACCGA-3'?
A. 0
B. 2
C. 3
D. 4
- Answer
-
D
Free Response
Transcribe and translate the following DNA sequence (nontemplate strand): 5'-ATGGCCGGTTATTAAGCA-3'
- Answer
-
The mRNA would be: 5'-AUGGCCGGUUAUUAAGCA-3'. The protein would be: MAGY. Even though there are six codons, the fifth codon corresponds to a stop, so the sixth codon would not be translated.
11.5: How Genes Are Regulated
Multiple Choice
Control of gene expression in eukaryotic cells occurs at which level(s)?
A. only the transcriptional level
B. epigenetic and transcriptional levels
C. epigenetic, transcriptional, and translational levels
D. epigenetic, transcriptional, post-transcriptional, translational, and post-translational levels
- Answer
-
D
Post-translational control refers to:
A. regulation of gene expression after transcription
B. regulation of gene expression after translation
C. control of epigenetic activation
D. period between transcription and translation
- Answer
-
B
Free Response
Describe how controlling gene expression will alter the overall protein levels in the cell.
- Answer
-
The cell controls which protein is expressed, and to what level that protein is expressed, in the cell. Prokaryotic cells alter the transcription rate to turn genes on or off. This method will increase or decrease protein levels in response to what is needed by the cell. Eukaryotic cells change the accessibility (epigenetic), transcription, or translation of a gene. This will alter the amount of RNA, and the lifespan of the RNA, to alter the amount of protein that exists. Eukaryotic cells also change the protein’s translation to increase or decrease its overall levels. Eukaryotic organisms are much more complex and can manipulate protein levels by changing many stages in the process.
Contributors and Attributions
Remixed and/or curated from the following works:
Fowler, S., Roush, R., & Wise, J. (2013). 9.E Molecular Biology (Exercises) . In Concepts of Biology . OpenStax (CC BY 4.0; Access for free at https://openstax.org/books/concepts-biology/pages/1-introduction ).


3.1: Invertibility
\(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\twovec}[2]{\begin{pmatrix} #1 \\ #2 \end{pmatrix} } \) \(\newcommand{\threevec}[3]{\begin{pmatrix} #1 \\ #2 \\ #3 \end{pmatrix} } \) \(\newcommand{\fourvec}[4]{\begin{pmatrix} #1 \\ #2 \\ #3 \\ #4 \end{pmatrix} } \) \(\newcommand{\fivevec}[5]{\begin{pmatrix} #1 \\ #2 \\ #3 \\ #4 \\ #5 \end{pmatrix} } \)In previous sections, we have found solutions to linear systems using the Gaussian elimination algorithm. We will now investigate another way of finding solutions to a specific type of equation \(A\mathbf x=\mathbf b\) when the matrix \(A\) has the same number of rows and columns. To get started, let's look at some familiar examples.
Preview Activity 3.1.1.
- Explain how you would solve the equation \(3x = 5\) without using the concept of division.
- Find the \(2\times2\) matrix \(A\) that rotates vectors counterclockwise by \(90^\circ\text{.}\)
- Find the \(2\times2\) matrix \(B\) that rotates vectors clockwise by \(90^\circ\text{.}\)
- What do you expect the product \(BA\) to be? Explain the reasoning behind your expectation and then compute \(BA\) to verify it.
-
Solve the equation \(A\mathbf x = \twovec{3}{-2}\) using Gaussian elimination.
- Explain why your solution may also be found by computing \(\mathbf x = B\twovec{3}{-2}\text{.}\)
Invertible matrices
The preview activity began with a familiar type of equation, \(3x = 5\text{,}\) and asked for a strategy to solve it. One possible response is to divide both sides by 3; instead, let's rephrase this as multiplying by \(3^{-1} = \frac 13\text{,}\) the multiplicative inverse of 3.
Now that we are interested in solving equations of the form \(A\mathbf x = \mathbf b\text{,}\) we might try to find a similar approach. Is there a matrix \(A^{-1}\) that plays the role of the multiplicative inverse? Of course, we can't expect every matrix to have a multiplicative inverse; after all, the real number \(0\) doesn't have an inverse. We will see, however, that many matrices do.
An \(n\times n\) matrix \(A\) is called invertible if there is a matrix \(B\) such that \(BA = I_n\text{,}\) where \(I_n\) is the \(n\times n\) identity matrix. The matrix \(B\) is called the inverse of \(A\) and denoted \(A^{-1}\text{.}\)
In the preview activity, we considered the matrices
since \(A\) rotates vectors in \(\mathbb R^2\) by \(90^\circ\) and \(B\) rotates vectors by \(-90^\circ\text{.}\) It's easy to check that
This shows that \(B = A^{-1}\text{.}\)
The preview also indicates the use of matrix inverses. Since we have \(A^{-1}A = I\text{,}\) we can solve the equation \(A\mathbf x = \mathbf b\) by multiplying both sides on the left by \(A^{-1}\text{:}\)
Notice that this is similar to finding the solution to \(3x=5\) as \(x=\frac13 5\text{.}\)
Activity 3.1.2.
Let's consider the matrices
-
Define these matrices in Sage and verify that \(BA = I\) so that \(B=A^{-1}\text{.}\)
- Find the solution to the equation \(A\mathbf x = \threevec{4}{-1}{4}\) using \(A^{-1}\text{.}\)
- Using your Sage cell above, multiply \(A\) and \(B\) in the opposite order; that is, what do you find when you evaluate \(AB\text{?}\)
- Suppose that \(A\) is an \(n\times n\) invertible matrix with inverse \(A^{-1}\text{.}\) This means that every equation of the form \(A\mathbf x=\mathbf b\) has a solution, namely, \(\mathbf x = A^{-1}\mathbf b\text{.}\) What can you conclude about the span of the columns of \(A\text{?}\)
- What can you conclude about the pivot positions of the matrix \(A\text{?}\)
- If \(A\) is an invertible \(4\times4\) matrix, what is its reduced row echelon form?
This activity demonstrates a few important things. First, we said that \(A\) is invertible if there is a matrix \(B\) such that \(BA = I\text{.}\) In general, multiplying matrices requires care because the product depends on the order in which the matrices are multiplied. However, in this case, we can check that \(BA = I\) implies that \(AB = I\) as well. This means that \(B\) is also invertible and that \(A=B^{-1}\text{.}\) This is the subject of Exercise <IP_ADDRESS>.
Also, if the matrix \(A\) is invertible, then every equation \(A\mathbf x = \mathbf b\) has a solution \(\mathbf x = A^{-1}\mathbf b\text{.}\) This means that the span of the columns of \(A\) is \(\mathbb R^n\) so that \(A\) has a pivot in every row. Since the matrix \(A\) has \(n\) rows and \(n\) columns, there must be a pivot in every row and every column. Therefore, the reduced row echelon form of \(A\) is
This provides us with a useful characterization of invertible matrices.
Constructing a matrix inverse
We have seen that an invertible matrix \(A\) has the property that its reduced row echelon form is the identity; that is, \(A\sim I\text{.}\) Here, we will use this fact to construct the inverse of a matrix \(A\text{.}\)
Activity 3.1.3.
In this activity, we will begin with the matrix
and construct its inverse \(A^{-1}\text{.}\) For the time being, let's denote the inverse by \(B\) so that \(B=A^{-1}\text{.}\)
-
We know that \(AB = I\text{.}\) If we write \(B = \left[\begin{array}{rr}\mathbf b_1& \mathbf b_2\end{array}\right]\text{,}\) then we have
\begin{equation*} AB = \left[\begin{array}{rr} A\mathbf b_1 & A\mathbf b_2 \end{array}\right] = \left[\begin{array}{rr} \mathbf e_1 & \mathbf e_2 \end{array}\right] = I\text{.} \end{equation*}
This means that we need to solve the equations
\begin{equation*} \begin{aligned} A\mathbf b_1 & {}={} \mathbf e_1 \\ A\mathbf b_2 & {}={} \mathbf e_2 \\ \end{aligned}\text{.} \end{equation*}Using the Sage cell below, solve these equations for the columns of \(B\text{.}\)
-
What is the matrix \(B\text{?}\) Check that \(AB = I\) and \(BA = I\text{.}\)
-
To find the columns of \(B\text{,}\) we solved two equations, \(A\mathbf b_1=\mathbf e_1\) and \(A\mathbf b_2=\mathbf e_2\text{.}\) We could do this by augmenting \(A\) two separate times, forming matrices
\begin{equation*} \begin{aligned} \left[\begin{array}{r|r} A & \mathbf e_1 \end{array}\right] & \\ \left[\begin{array}{r|r} A & \mathbf e_2 \end{array}\right] & \\ \end{aligned} \end{equation*}
and finding their reduced row echelon forms. But instead of solving these two equations separately, we could also solve them together by forming the augmented matrix \(\left[\begin{array}{r|rr} A & \mathbf e_1 & \mathbf e_2 \end{array}\right]\) and finding the row reduced echelon form. In other words, we augment \(A\) by the matrix \(I\) to form \(\left[\begin{array}{r|r} A & I \end{array} \right] \text{.}\)
Form this augmented matrix and find its reduced row echelon form to find \(A^{-1}\text{.}\)
Assuming \(A\) is invertible, we have shown that
\begin{equation*} \left[\begin{array}{r|r} A & I \end{array}\right] \sim \left[\begin{array}{r|r} I & A^{-1} \end{array}\right]\text{.} \end{equation*} -
If you have defined a matrix \(A\) in Sage, you can find it's inverse as
A.inverse()
. Use Sage to find the inverse of the matrix\begin{equation*} A = \left[\begin{array}{rrr} 1 & -2 & -1 \\ -1 & 5 & 6 \\ 5 & -4 & 6 \\ \end{array}\right]\text{.} \end{equation*} -
What happens when we try to find the inverse of the matrix
\begin{equation*} \left[\begin{array}{rr} -4 & 2 \\ -2 & 1 \\ \end{array}\right]\text{?} \end{equation*}
- Suppose that \(n\times n\) matrices \(C\) and \(D\) are both invertible. What do you find when you simplify the product \((D^{-1}C^{-1})(CD)\text{?}\) Explain why the product \(CD\) is invertible and \((CD)^{-1} = D^{-1}C^{-1}\text{.}\)
Finding the inverse of an \(n\times n\) matrix \(A\) requires us to solve \(n\) equations. If we write the inverse as
then we need to solve
We can, of course, solve each equation separately, but it is more efficient to bundle the equations together by forming the augmented matrix \(\left[\begin{array}{r|r} A & I \end{array}\right]\) and finding its row reduced echelon form. We then find
We saw earlier that, if \(A\) has an inverse, then \(A\sim I\text{.}\) We have now seen that, if \(A\sim I\text{,}\) then \(A\) has an inverse.
Finally, we see that the product of two invertible matrices \(A\) and \(B\) is also invertible. This is because
Therefore, we have \((AB)^{-1} = B^{-1}A^{-1}\text{.}\) Because the matrix product depends on the order in which we multiply matrices, use care when applying this relationship. The inverse of a product is the product of the inverses with the order of multiplication reversed.
Properties of invertible matrices.
- An \(n\times n\) matrix \(A\) is invertible if and only if \(A\sim I\text{.}\)
- If \(A\) is invertible, then the solution to the equation \(A\mathbf x = \mathbf b\) is given by \(\mathbf x = A^{-1}\mathbf b\text{.}\)
-
We can find \(A^{-1}\) by finding the reduced row echelon form of \(\left[\begin{array}{r|r} A & I \end{array}\right]\text{;}\) namely,
\begin{equation*} \left[\begin{array}{r|r} A & I \end{array}\right] \sim \left[\begin{array}{r|r} I & A^{-1} \end{array}\right]\text{.} \end{equation*}
- If \(A\) and \(B\) are two invertible \(n\times n\) matrices, then their product \(AB\) is also invertible and \((AB)^{-1} = B^{-1}A^{-1}\text{.}\)
There is a simple formula for finding the inverse of a \(2\times2\) matrix:
which can be easily checked. The condition that \(A\) be invertible is, in this case, reduced to the condition that \(ad-bc\neq 0\text{.}\) We will understand this condition better once we have explored determinants in Section 3.4 . There is a similar formula for the inverse of a \(3\times 3\) matrix, but there is not a good reason to write it here.
Triangular matrices and Gaussian elimination
Generally speaking, solving an equation \(A\mathbf x=\mathbf b\) by first finding \(A^{-1}\) and then evaluating \(\mathbf x = A^{-1}\mathbf b\) is not the best strategy since row reducing the augmented matrix \(\left[\begin{array}{r|r} A & \mathbf b \end{array}\right]\) involves considerably less work. This becomes clear once we remember that finding the inverse \(A^{-1}\) requires us to solve \(n\) equations of this form.
For the class of triangular matrices, however, finding inverses is relatively efficient and useful, as we will see in Section 5.1 .
We say that a matrix \(A\) is lower triangular if all its entries above the diagaonal are zero. Similarly, \(A\) is upper triangular if all the entries below the diagonal are zero.
For example, the matrix \(L\) below is a lower triangular matrix while \(U\) is an upper triangular one.
We can develop a simple test to determine whether an \(n\times n\) lower triangular matrix is invertible. Let's use Gaussian elimination to find the reduced row echelon form of the lower triangular matrix
Because the entries on the diagonal are nonzero, we find a pivot position in every row, which tells us that the matrix is invertible. If, however, there is a zero entry on the diagonal, the matrix cannot be invertible. Considering the matrix below, we see that having a zero on the diagonal leads to a row without a pivot position.
An \(n\times n\) triangular matrix is invertible if and only if the entries on the diagonal are all nonzero.
Up to this point, our primary tool for studying linear systems, sets of vectors, and matrices has been Gaussian elimination. As the next activity demonstrates, we can express the row operations performed in Gaussian elimination in terms of matrix multiplication. In Section 5.1 , we will use this observation to create an efficient way to solve equations of the form \(A\mathbf x=\mathbf b\text{.}\)
Activity 3.1.4.
As an example, we will consider the matrix
When performing Gaussian elimination on \(A\text{,}\) we first apply a row replacement operation in which we multiply the first row by \(-2\) and add to the second row. After this step, we have a new matrix \(A_1\text{.}\)
-
Show that multiplying \(A\) by the lower triangular matrix
\begin{equation*} L_1 = \left[\begin{array}{rrr} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array}\right] \end{equation*}
has the same effect as this row operation; that is, show that \(L_1A = A_1\text{.}\)
-
Explain why \(L_1\) is invertible and find its inverse \(L_1^{-1}\text{.}\)
- You should see that there is a simple relationship between \(L_1\) and \(L_1^{-1}\text{.}\) Describe this relationship and explain why it holds.
-
To continue the Gaussian elimination algorithm, we need to apply two more row replacements to bring \(A\) into a triangular form \(U\) where
\begin{equation*} A = \left[\begin{array}{rrr} 1 & 2 & 1 \\ 2 & 0 & -2 \\ -1 & 2 & -1 \\ \end{array}\right] \sim \left[\begin{array}{rrr} 1 & 2 & 1 \\ 0 & -4 & -4 \\ 0 & 0 & -4 \\ \end{array}\right] = U\text{.} \end{equation*}
Find the matrices \(L_2\) and \(L_3\) that perform these row replacement operations so that \(L_3L_2L_1 A = U\text{.}\)
-
Explain why the matrix product \(L_3L_2L_1\) is invertible and use this fact to write \(A = LU\text{.}\) What is the matrix \(L\) that you find? Why do you think we denote it by \(L\text{?}\)
-
Row replacement operations may always be performed by multiplying by a lower triangular matrix. It turns out the other two row operations, scaling and interchange, may also be performed using matrix multiplication. For instance, consider the two matrices
\begin{equation*} S = \left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 1 \\ \end{array}\right], \hspace{24pt} P = \left[\begin{array}{rrr} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ \end{array}\right]\text{.} \end{equation*}
Show that multiplying \(A\) by \(S\) performs a scaling operation and that multiplying by \(P\) performs a row interchange.
- Explain why the matrices \(S\) and \(P\) are invertible and state their inverses.
We will demonstrate the ideas in this activity again using the matrix
After performing three row replacement operations, we find the row equivalent upper triangular matrix \(U\text{:}\)
The first row replacement operation multiplies the first row by \(3\) and adds the result to the second row. We can perfom this operation by multiplying \(A\) by the lower triangular matrix \(L_1\) where
The next two row replacement operations are performed by the matrices
so that \(L_3L_2L_1A = U\text{.}\)
Notice that the inverse of \(L_1\) has the simple form:
This makes sense; if we want to undo the operation of multiplying the first row by \(3\) and adding to the second row, we should multiply the first row by \(-3\) and add it to the second row. This is the effect of \(L_1^{-1}\text{.}\)
The other row operations we use in implementing Gaussian elimination can also be performed by multiplying by an invertible matrix. In particular, if we scale a row by a nonzero number \(s\text{,}\) we can undo this operation by scaling by \(\frac 1s\text{.}\) This leads to the invertible diagonal matrices, such as
Similarly, a row interchange leads to a matrix \(P\text{,}\) which is its own inverse. An example is
Summary
In this section, we found conditions guaranteeing that a matrix has an inverse. When these conditions hold, we also found an algorithm for finding the inverse.
- The \(n\times n\) matrix \(A\) is invertible if and only if it is row equivalent to \(I_n\text{,}\) the \(n\times n\) identity matrix.
- If a matrix \(A\) is invertible, then the solution to the equation \(A\mathbf x = \mathbf b\) is \(\mathbf x = A^{-1}\mathbf b\text{.}\)
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If a matrix \(A\) is invertible, we can use Gaussian elimination to find its inverse:
\begin{equation*} \left[\begin{array}{r|r} A & I \end{array}\right] \sim \left[\begin{array}{r|r} I & A^{-1} \end{array}\right]\text{.} \end{equation*}
- The row operations used in performing Gaussian elimination can be performed by multiplying by invertible matrices. More specifically, a row replacement operation may be performed by multiplying by an invertible lower triangular matrix.
Exercises 3.1.5Exercises
Consider the matrix
- Explain why \(A\) has an inverse.
- Find the inverse of \(A\) by augmenting by the identity \(I\) to form \(\left[\begin{array}{r|r}A & I \end{array}\right]\text{.}\)
- Use your inverse to solve the equation \(A\mathbf x = \fourvec{3}{2}{-3}{-1}\text{.}\)
In this exercise, we will consider \(2\times 2\) matrices as defining linear transformations.
- Write the matrix \(A\) that performs a \(45^\circ\) rotation. What geometric operation undoes this rotation? Find the matrix that perform this operation and verify that it is \(A^{-1}\text{.}\)
- Write the matrix \(A\) that performs a \(180^\circ\) rotation. Verify that \(A^2 = I\) so that \(A^{-1} = A\text{,}\) and explain geometrically why this is the case.
- Find three more matrices \(A\) that satisfy \(A^2 = I\text{.}\)
Suppose that \(A\) is an \(n\times n\) matrix.
- Suppose that \(A^2 = AA\) is invertible with inverse \(B\text{.}\) This means that \(BA^2 = BAA = I\text{.}\) Explain why \(A\) must be invertible with inverse \(BA\text{.}\)
- Suppose that \(A^{100}\) is invertible with inverse \(B\text{.}\) Explain why \(A\) is invertible. What is \(A^{-1}\) in terms of \(A\) and \(B\text{?}\)
Our definition of an invertible matrix requires that \(A\) be a square \(n\times n\) matrix. Let's examine what happens when \(A\) is not square. For instance, suppose that
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Verify that \(BA = I_2\text{.}\) In this case, we say that \(B\) is a
left
inverse of \(A\text{.}\)
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If \(A\) has a left inverse \(B\text{,}\) we can still use it to find solutions to linear equations. If we know there is a solution to the equation \(A\mathbf x = \mathbf b\text{,}\) we can multiply both sides of the equation by \(B\) to find \(\mathbf x = B\mathbf b\text{.}\)
Suppose you know there is a solution to the equation \(A\mathbf x = \threevec{-1}{-3}{6}\text{.}\) Use the left inverse \(B\) to find \(\mathbf x\) and verify that it is a solution.
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Now consider the matrix
\begin{equation*} C = \left[\begin{array}{rrr} 1 & -1 & 0 \\ -2 & 1 & 0 \\ \end{array}\right] \end{equation*}
and verify that \(C\) is also a left inverse of \(A\text{.}\) This shows that the matrix \(A\) may have more than one left inverse.
- When \(A\) is a square matrix, we said that \(BA=I\) implies that \(AB=I\text{.}\) In this problem, we have a non-square matrix \(A\) with \(BA = I\text{.}\) What happens when we compute \(AB\text{?}\)
If a matrix \(A\) is invertible, there is a sequence of row operations that transform \(A\) into the identity matrix \(I\text{.}\) We have seen that every row operation can be performed by matrix multiplication. If the \(j^{th}\) step in the Gaussian elimination process is performed by multiplying by \(E_j\text{,}\) then we have
which means that
For each of the following matrices, find a sequence of row operations that transforms the matrix to the identity \(I\text{.}\) Write the matrices \(E_j\) that perform the steps and use them to find \(A^{-1}\text{.}\)
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\begin{equation*} A = \left[\begin{array}{rrr} 0 & 2 & 0 \\ -3 & 0 & 0 \\ 0 & 0 & 1 \\ \end{array}\right]\text{.} \end{equation*}
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\begin{equation*} A = \left[\begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 2 & 1 & 0 & 0 \\ 0 & -3 & 1 & 0 \\ 0 & 0 & 2 & 1 \\ \end{array}\right]\text{.} \end{equation*}
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\begin{equation*} A = \left[\begin{array}{rrr} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 2 \\ \end{array}\right]\text{.} \end{equation*}
Determine whether the following statements are true or false and explain your reasoning.
- If \(A\) is invertible, then the columns of \(A\) are linearly independent.
- If \(A\) is a square matrix whose diagonal entries are all nonzero, then \(A\) is invertible.
- If \(A\) is an invertible \(n\times n\) matrix, then the columns of \(A\) span \(\mathbb R^n\text{.}\)
- If \(A\) is invertible, then there is a nontrivial solution to the homogeneous equation \(A\mathbf x = \zerovec\text{.}\)
- If \(A\) is an \(n\times n\) matrix and the equation \(A\mathbf x = \mathbf b\) has a solution for every vector \(\mathbf b\text{,}\) then \(A\) is invertible.
Provide a justification for your response to the following questions.
- Suppose that \(A\) is a square matrix with two identical columns. Can \(A\) be invertible?
- Suppose that \(A\) is a square matrix with two identical rows. Can \(A\) be invertible?
- Suppose that \(A\) is an invertible matrix and that \(AB = AC\text{.}\) Can you conclude that \(B = C\text{?}\)
- Suppose that \(A\) is an invertible \(n\times n\) matrix. What can you say about the span of the columns of \(A^{-1}\text{?}\)
- Suppose that \(A\) is an invertible matrix and that \(B\) is row equivalent to \(A\text{.}\) Can you guarantee that \(B\) is invertible?
Suppose that we start with the \(3\times3\) matrix \(A\) and perform the following sequence of row operations:
- Multiply row 1 by -2 and add to row 2.
- Multiply row 1 by 4 and add to row 3.
- Scale row 2 by \(1/2\text{.}\)
- Multiply row 2 by -1 and add to row 3.
Suppose we arrive at the upper triangular matrix
- Write the matrices \(E_1\text{,}\) \(E_2\text{,}\) \(E_3\text{,}\) and \(E_4\) that perform the four row operations.
- Find the matrix \(E = E_4E_3E_2E_1\text{.}\)
- We then have \(E_4E_3E_2E_1 A = EA = U\text{.}\) Now that we have the matrix \(E\text{,}\) find the original matrix \(A = E^{-1}U\text{.}\)
We defined an \(n\times n\) matrix to be invertible if there is a matrix \(B\) such that \(BA=I_n\text{.}\) In this exercise, we will explain why \(B\) is also invertible and that \(AB = I\text{.}\) This means that, if \(B=A^{-1}\text{,}\) then \(A = B^{-1}\text{.}\)
- Given the fact that \(BA = I_n\text{,}\) explain why the matrix \(B\) must also be a square \(n\times n\) matrix.
- Suppose that \(\mathbf b\) is a vector in \(\mathbb R^n\text{.}\) Since we have \(BA = I\text{,}\) it follows that \(B(A\mathbf b) = \mathbf b\text{.}\) Use this to explain why the columns of \(B\) span \(\mathbb R^n\text{.}\) What does this say about the pivot positions of \(B\text{?}\)
- Explain why the equation \(B\mathbf x = \zerovec\) has only the trivial solution.
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Beginning with the equation, \(BA=I\text{,}\) multiply both sides by \(B\) to obtain \(BAB = B\text{.}\) We will rearrange this equation:
\begin{equation*} \begin{aligned} BAB & {}={} B \\ BAB - B& {}={} 0 \\ B(AB-I) & {}={} 0\text{.} \\ \end{aligned} \end{equation*}
Since the homogeneous equation \(B\mathbf x =\zerovec\) has only the trivial solution, explain why \(AB-I = 0\) and therefore, \(AB = I\text{.}\)


4.7: Soil Types
SOIL TEXTURE AND COMPOSITION
The inorganic portion of soil is made of many different size particles, and these different size particles are present in different proportions. The combination of these two factors determines some of the properties of the soil.
Figure 1. A loam field.
- A permeable soil allows water to flow through it easily because the spaces between the inorganic particles are large and well connected. Sandy or silty soils are considered ‘light’ soils because they are permeable, water-draining types of soils.
- Soils that have lots of very small spaces are water-holding soils. For example, when clay is present in a soil, the soil is heavier, holds together more tightly, and holds water.
- When a soil contains a mixture of grain sizes, the soil is called a loam (figure 1).
When soil scientists want to precisely determine soil type, they measure the percentage of sand, silt, and clay. They plot this information on a triangular diagram, with each size particle at one corner (figure 2). The soil type can then be determined from the location on the diagram. At the top, a soil would be clay; at the left corner, it would be sand, and at the right corner it would be silt. Soils in the lower middle with less than 50% clay are loams.
Figure 2. Soil types by particle size.
Using the chart as a guide, what is the composition of a sandy clay loam? If you would like to determine soil type by feel, here’s a chart from the USDA to help you.
SOIL HORIZONS AND PROFILES
A residual soil forms over many years, as mechanical and chemical weathering slowly change solid rock into soil. The development of a residual soil may go something like this.
- The bedrock fractures because of weathering from ice wedging or another physical process.
- Water, oxygen, and carbon dioxide seep into the cracks to cause chemical weathering.
- Plants, such as lichens or grasses, become established and produce biological weathering.
- Weathered material collects until there is soil.
- The soil develops soil horizons , as each layer becomes progressively altered. The greatest degree of weathering is in the top layer. Each successive, lower layer is altered just a little bit less. This is because the first place where water and air come in contact with the soil is at the top.
Figure 3. Soil is an important resource. Each soil horizon is distinctly visible in this photograph.
A cut in the side of a hillside shows each of the different layers of soil. All together, these are called a soil profile (figure 3).
The simplest soils have three horizons.
Topsoil
Called the A horizon, the topsoil is usually the darkest layer of the soil because it has the highest proportion of organic material. The topsoil is the region of most intense biological activity: insects, worms, and other animals burrow through it and plants stretch their roots down into it. Plant roots help to hold this layer of soil in place. In the topsoil, minerals may dissolve in the fresh water that moves through it to be carried to lower layers of the soil. Very small particles, such as clay, may also get carried to lower layers as water seeps down into the ground.
Subsoil
The B horizon or subsoil is where soluble minerals and clays accumulate. This layer is lighter brown and holds more water than the topsoil because of the presence of iron and clay minerals. There is less organic material. Look at figure 4.
Figure 4. A soil profile is the complete set of soil layers. Each layer is called a horizon.
C Horizon
The C horizon is a layer of partially altered bedrock. There is some evidence of weathering in this layer, but pieces of the original rock are seen and can be identified.
Not all climate regions develop soils, and not all regions develop the same horizons. Some areas develop as many as five or six distinct layers, while others develop only very thin soils or perhaps no soils at all.
TYPES OF SOILS
Although soil scientists recognize thousands of types of soil—each with its own specific characteristics and name—let’s consider just three soil types. This will help you to understand some of the basic ideas about how climate produces a certain type of soil, but there are many exceptions to what we will learn right now (figure 5).
Figure 5. Just some of the thousands of soil types.
Pedalfer
Deciduous trees, the trees that lose their leaves each winter, need at least 65 cm of rain per year. These forests produce soils called pedalfers , which are common in many areas of the temperate, eastern part of the United States (figure 6). The word pedalfer comes from some of the elements that are commonly found in the soil. The Al in ped al fer is the chemical symbol of the element aluminum, and the Fe in pedal fe r is the chemical symbol for iron. Pedalfers are usually a very fertile, dark brown or black soil. Not surprising, they are rich in aluminum clays and iron oxides. Because a great deal of rainfall is common in this climate, most of the soluble minerals dissolve and are carried away, leaving the less soluble clays and iron oxides behind.
Figure 6. A pedalfer is the dark, fertile type of soil that will form in a forested region.
Pedocal
Pedocal soils form in drier, temperate areas where grasslands and brush are the usual types of vegetation (figure 7). The climates that form pedocals have less than 65 cm rainfall per year, so compared to pedalfers, there is less chemical weathering and less water to dissolve away soluble minerals so more soluble minerals are present and fewer clay minerals are produced. It is a drier region with less vegetation, so the soils have lower amounts of organic material and are less fertile.
Figure 7. A pedocal is the alkaline type of soil that forms in grassland regions.
A pedocal is named for the calcite enriched layer that forms. Water begins to move down through the soil layers, but before it gets very far, it begins to evaporate. Soluble minerals, like calcium carbonate, concentrate in a layer that marks the lowest place that water was able to reach. This layer is called caliche.
Laterite
In tropical rainforests where it rains literally every day, laterite soils form (figure 8). In these hot, wet, tropical regions, intense chemical weathering strips the soils of their nutrients. There is practically no humus. All soluble minerals are removed from the soil and all plant nutrients are carried away. All that is left behind are the least soluble materials, like aluminum and iron oxides. These soils are often red in color from the iron oxides. Laterite soils bake as hard as a brick if they are exposed to the sun.
Figure 8. A laterite is the type of thick, nutrient poor soil that forms in the rainforest.
Many climates types have not been mentioned here. Each produces a distinctive soil type that forms in the particular circumstances found there. Where there is less weathering, soils are thinner but soluble minerals may be present. Where there is intense weathering, soils may be thick but nutrient poor. Soil development takes a very long time, it may take hundreds or even thousands of years for a good fertile topsoil to form. Soil scientists estimate that in the very best soil-forming conditions, soil forms at a rate of about 1mm/year. In poor conditions, soil formation may take thousands of years!
SOIL CONSERVATION
Soil is only a renewable resource if it is carefully managed. Drought, insect plagues, or outbreaks of disease are natural cycles of events that can negatively impact ecosystems and the soil, but there are also many ways in which humans neglect or abuse this important resource.
Figure 9. Organic material can be added to soil to help increase its fertility.
One harmful practice is removing the vegetation that helps to hold soil in place. Sometimes just walking or riding your bike over the same place will kill the grass that normally grows there. Land is also deliberately cleared or deforested for wood. The loose soils then may be carried away by wind or running water. In many areas of the world, the rate of soil erosion is many times greater than the rate at which it is forming. Soils can also be contaminated if too much salt accumulates in the soil or where pollutants sink into the ground. There are many practices that can protect and preserve soil resources. Adding organic material to the soil in the form of plant or animal waste, such as compost or manure, increases the fertility of the soil and improves its ability to hold onto water and nutrients (figure 9). Inorganic fertilizer can also temporarily increase the fertility of a soil and may be less expensive or time consuming, but it does not provide the same long-term improvements as organic materials.
Figure 10. Steep slopes can be terraced to make level planting areas and decrease surface water runoff and erosion.
Agricultural practices such as rotating crops, alternating the types of crops planted in each row, and planting nutrient rich cover crops all help to keep soil more fertile as it is used season after season. Planting trees as windbreaks, plowing along contours of the field, or building terraces into steeper slopes will all help to hold soil in place (figure 10). No-till or low-tillage farming helps to keep soil in place by disturbing the ground as little as possible when planting.
LESSON SUMMARY
- Soil texture and composition, plus the amount of organic material in a soil, determine a soil’s qualities and fertility.
- Given enough time, rock is weathered to produce a layered soil, called a soil profile.
- Each type of climate can ultimately produce a unique type of soil.
REFLECTION QUESTIONS
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Contributors and Attributions
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Original content from Kimberly Schulte (Columbia Basin College) and supplemented by Lumen Learning . The content on this page is copyrighted under a Creative Commons Attribution 4.0 International license.


2.3.5: Advanced Directives
Advanced Directives are legal documents that allow a person to choose the type of medical care they wish to receive should they become unable to state their choices and make decisions in the future (Leahy, Fuzy & Grafe, 2013)
An advanced directive is a legal document that tells a healthcare provider the type of treatments a patient wishes and does not wish to receive. They also allow a patient to select an individual(s) to make decisions for them should they be unable to do so. There are many types of advanced directives, which can be further explored by visiting MedlinePlus at http :// www . nlm . nih . gov / medlineplus / advancedirectives . html
Living Wills
Living wills are a type of an advanced directive. They explain the type of medical care a patient wants and does not want to receive should they be unable in the future to make decisions for themselves.
A living will is not the same as a last will and testament. A last will and testament is a legal document that delegates a person to be an executor of their estate and explains how a person wishes to have their property and possessions disposed of after their death. A living will is only about medical care and treatment.
Laws about living wills vary from state to state. Examples of what a person may include in their living will might be decisions about:
- Being artificially fed through a vein or stomach
- Having a tube inserted down their throat for mechanical ventilation
- CPR
- Blood transfusions
- Dialysis
Durable P ower of A ttorney for H ealth C are ( a.k.a. H ealth C are P roxy )
A health care proxy is someone who the patient designates to make decisions for them in the future should they be unable to do so. A health care proxy is a type of advanced directive. A health care proxy is not the same as an executor of a will . A patient can have one person as the executor of their will and a completely different person designated as their health care proxy if they choose. A health care proxy can make decisions for a patient’s health care if the patient is unable to do so. A health care proxy is responsible for carrying out a patient’s wishes and for following a living will. If the patient has not made a living will, the health care proxy is responsible for making decisions on behalf of the patient according to what they believe is in the best interests of the patient.
Laws about health care proxies and the decisions they can make vary from state to state (Health care agents, 2014). Most states do not allow a doctor or other health care provider to be a patient’s health care proxy. Home Health Aides/Personal Care Aides should not agree to be their patient’s health care proxy. This would be a boundary violation.
Do -N ot -R esuscitate ( DNR ) O rder
A Do–not – resuscitate (DNR) is another type of advanced directive. A patient who has a DNR has made a decision to not receive CPR (cardiopulmonary resuscitation). This means that should the patient ’ s heart stop , or if the patient stops breathing , that the health care providers are not allowed to perform CPR . A DNR order is an order that is written by a physician after the patient has discussed their wishes with their doctor (Do-not-resuscitate order, 2014). If a patient has a DNR order and a health care provider performs CPR knowing there is a DNR order, legal consequences can occur against the health care provider. It is important to know whether your patient has a DNR order in place.
Having a DNR order does not affect any other treatments the patient receives. This means that a patient can continue to receive treatments such as medications, surgeries, physical therapy, and all other medical interventions. A DNR order only applies to whether or not your patient would like CPR performed should their heart stop or if they stop breathing.
A health care proxy or family member cannot have a DNR order removed if it was already in place, based on the patient’s wishes and discussion with their doctor. This means that even a patient’s health care proxy cannot revoke (have removed) a DNR order if the patient had one in place before they became unable to make their own choices.
Decide whether the statement is True or False
Patients have a right to considerate and respectful care. _______
Patients have a right to review their medical records and to control who can access their records and medical information. _______
You can share medical information about your patient with any family member who wants to know how the patient is doing. ______
Patients have a right to refuse medical treatment even if you think they should have the treatment. ______
The patient has a responsibility to inform their healthcare team about who their health care proxy is and if there are any changes in their medical condition._______
A patient who has a DNR order can still continue to receive home health services, medications, and treatments. _______
A DNR order is a physician order based on the patient’s wishes, which states should the patient stop breathing, or their heart stop beating, CPR will not be performed.________
If a patient has a DNR order, the patient’s family can get it removed. _________
A health care proxy is someone the patient selects who can make decisions for the patient should the patient become unable to do so. _______
A health care proxy can have a DNR order removed because they know what is best for the patient. ________
- Answer
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1. True
2. True
3. False
4. True
5. True
6. True
7. True
8. False
9. True
10. False
Feedback :
1. According to the Patient Bill of Rights, patients have a right to considerate and respectful care.
2. According to the Patient Bill of Rights, patients have a right to request and receive a copy of their medical record. This is also a federal law under HIPAA guidelines.
3. You cannot share any information about the patient without the patient’s specific consent.
4. According to the Patient Bill of Rights, patients have a right to refuse treatment.
5. According to the American Hospital Association, 1992, Patient Responsibilities include informing their healthcare team about who their health care proxy is and if there are any changes in their medical condition.
6. A patient who has a DNR order can still continue to receive home health services, medications, and treatments. A DNR order will only prohibit CPR and artificial respiration to occur should the patient choose to obtain this physician order.
7. A DNR order is a physician order based on the patient’s wishes, which states should the patient stop breathing, or their heart stop beating, CPR will not be performed.
8. It is a patient decision about whether or not to have a DNR order. If the patient is conscious and oriented, the family cannot have a DNR order removed.
9. A health care proxy is someone the patient selects who can make decisions for the patient should the patient become unable to do so.
10. If the patient is conscious and oriented, it is a patient decision and not up to the family about whether or not to have a DNR order.


24.2: Standing Waves and Resonance
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Imagine you have a sinusoidal traveling wave of the form ( 12.1.5 ), only traveling to the left, incident from the right on a “fixed end” at \(x\) = 0. The incident wave will go as \(\xi_{0} \sin [2 \pi(x+c t) / \lambda]\); the reflected wave should be flipped left to right and upside down, so change \(x\) to \(−x\) and put an overall minus sign on the displacement, to get \(-\xi_{0} \sin [2 \pi(-x+c t) / \lambda]\). The sum of the two waves in the region \(x > 0\) is then
\[ \xi(x, t)=\xi_{0} \sin \left[\frac{2 \pi}{\lambda}(x+c t)\right]-\xi_{0} \sin \left[\frac{2 \pi}{\lambda}(-x+c t)\right]=2 \xi_{0} \sin \left(\frac{2 \pi x}{\lambda}\right) \cos (2 \pi f t) \label{eq:12.16} \]
using a trigonometrical identity for \(\sin(a + b)\), and \(f = c/\lambda\).
The result on the right-hand side of Equation (\ref{eq:12.16}) is called a standing wave . It does not travel anywhere, it just oscillates “in place”: every point x behaves like a separate oscillator with an amplitude \(2 \xi_{0} \sin (2 \pi x / \lambda)\). This amplitude is zero at special points, where \(2x/\lambda\) is equal to an integer. These points are called nodes .
We could think of “confining” a wave of this sort to a string fixed at both ends, if we make the string have an end at \(x\) = 0 and the other one at one of these points where the amplitude is zero; this means we want the length \(L\) of the string to satisfy
\[ 2L = n \lambda \label{eq:12.17} \]
where \(n\) = 1, 2,.... Alternatively, we can think of \(L\) as being fixed and Equation (\ref{eq:12.17}) as giving us the possible values of \(\lambda\) that will give us standing waves: \(\lambda = 2L/n\). Since \(f = c/\lambda\), we see that all these possible standing waves, for fixed \(L\) and \(c\), have different frequencies that we can write as
\[ f_{n}=\frac{n c}{2 L}, \quad n=1,2,3, \ldots \label{eq:12.18} .\]
Note that these are all multiples of the frequency \(f_1 = c/2L\). We call this the fundamental frequency of oscillation of a string fixed at both ends. The period corresponding to this fundamental frequency is the roundtrip time of a wave pulse around the string, \(2L/c\).
The first three standing waves are plotted in Figure \(\PageIndex{1}\). Their wave functions are given by the right-hand side of Equation (\ref{eq:12.16}), for 0 ≤ \(x\) ≤ \(L\), with \(\lambda = 2L/n\) (\(n\) = 1, 2, 3), and \(f = f_n = nc/2L\). The amplitude is arbitrary; in the figure I have set it equal to 1 for convenience. Calling the corresponding function \(u_n(x, t)\) is more or less common practice in other contexts:
\[ u_{n}(x, t)=\sin \left(\frac{n \pi x}{L}\right) \cos \left(2 \pi f_{n} t\right) \label{eq:12.19} .\]
These functions are called the normal modes of vibration of the string. In Figure \(\PageIndex{1}\) I have shown, for each of them, the displacement at the initial time, \(t\) = 0, as a solid line, and then half a period later as a dashed line. In addition to this, notice that the wave function vanishes identically (the string is flat) at the quarter-period intervals, \(t = 1/4f_n\) and \(t = 3/4f_n\). At those times, the wave has no elastic potential energy (since the string is unstretched): as with a simple oscillator passing through the equilibrium position, all its energy is kinetic. For \(n > 1\), there are also nodes (places where the oscillation amplitude is always zero) at points other than the ends. Including the endpoints, the \(n\)-th normal mode has \(n + 1\) nodes. The places where the oscillation amplitude is largest are called antinodes .
Animations of these standing waves can be found in many places; one I particularly like is here: http://newt.phys.unsw.edu.au/jw/strings.html#standing . It also shows graphically how the standing wave can be considered as a superposition of two oppositely-directed traveling waves, as in Equation (\ref{eq:12.16}).
If we initially bent the string into one of the shapes shown in Figure \(\PageIndex{1}\), and then released it, it would oscillate at the corresponding frequency \(f_n\), keeping the same shape, only scaling it up and down by a factor \(\cos(2\pi f_{n}t)\) as time elapsed. So, another way to think of standing waves is as the natural modes of vibration of an extended system —the string, in this case, although standing waves can be produced in any medium that can carry a traveling wave.
What I mean by a “natural mode of vibration” is the following: a single oscillator, say, a pendulum, has a single “natural” frequency; if you displace it or hit it, it just oscillates at that frequency with a constant amplitude. An extended system, like the string, can be viewed as a collection of coupled oscillators, which may in general oscillate in many different and complicated ways; yet, there is a specific set of frequencies—for the string with two ends fixed, the sequence \(f_n\) of Equation (\ref{eq:12.18})—and associated shapes that will result in all the parts of the string performing simple harmonic motion, in synchrony, all at the same frequency.
Of course, to produce just one of these specific modes of oscillation requires some care (“driving” the string at the right frequency is probably the easiest way; see next paragraph); however, if you simply hit or pluck the string in any random way, a remarkable thing happens: the resulting motion will be, mathematically, described as a sum of sinusoidal standing waves, each with one of the frequencies \(f_n\), and each with a different amplitude \(A_n\). In a musical instrument, this will eventually generate a superposition of sound waves with frequencies \(f_1 = c/2L\), \(f_2 = 2f_1\), \(f_3 = 3f_1\) ... (called, in this context, the fundamental , \(f_1\), and its overtones , \(f_n = nf_1\)). Each one of these frequencies corresponds to a different pitch, or musical note, and the result will sound a little like a chord, although not nearly as pronounced—we will mostly hear only the fundamental, which corresponds to the root note of the chord, but all the notes in a major triad are in fact present in the vibration of a single guitar or piano string 4 .
But wait, there’s more! Suppose that you try to get the string to oscillate by “driving” it: that is to say, grabbing a hold of one end and shaking it at some frequency, only with a very small amplitude, so the displacement at that end remains always close to zero. In that case, you will typically get only very small amplitude oscillations, until the driving frequency hits one of the special frequencies \(f_n\), at which point you will get a large oscillation with the shape of the corresponding standing wave. This is a phenomenon known as resonance , and the \(f_n\) are the resonant frequencies of this system.
Note that the effect I just described is essentially the same as you experience when you are “pumping,” or simply pushing, a swing. Unless you do it at the right frequency, you do not get very far; but if you do it at the right frequency (which is the swing’s natural frequency, the one at which it will swing on its own), you can get huge amplitude oscillations. So, the frequencies (\ref{eq:12.18}) may be said to be the string’s natural oscillation frequencies in the same two ways: they are the ones at which it will oscillate if you just pluck it, and they are the ones at which you have to drive it if you want to get large oscillations.
Pretty much everything I have just shown you above for standing waves on a string applies to sound waves inside a tube or pipe open at both ends. In that case, however, it is not the displacement, but the pressure (or density) wave that must have zeros at the ends (since the ends are open, the pressure there must be just the average atmospheric pressure; note that the pressure or density waves in a sound wave do not give the absolute pressure or density, but the deviation, positive or negative, from the average). The math, however, is identical, and one finds the same set of normal modes and resonance frequencies as above. These are then the frequencies that would be produced when blowing in a flute or an organ pipe open at both ends. So, both from pipes and strings we get the same “harmonic series” of frequencies (\ref{eq:12.18}) that has been the foundation of Western music since at least the time of Pythagoras.
4 It works like this: say \(f_1\) corresponds to a C, then \(f_2\) is the C above that, \(f_3\) the G above that, \(f_4\) the C above that, and \(f_5\) the E above that.


3.4.2: Characteristics of Nonmetals
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Skills to Develop
- List some characteristics of nonmetals
- Contrast metals and nonmetals
Remember that non-metals are on the right and top of the periodic table. Based on the periodic trends in the last 4 sections, this means that they are usually smaller, more likely to gain electrons, and less likely to lose electrons, than the metals.
Elemental Properties
In the elemental form, non-metals can be gas, liquid or solid. They aren't shiny (lustrous) and they don't conduct heat or electricity well. Usually their melting points are lower than for metals, although there are exceptions. The solids usually break easily, and can't bend like metals. It's a general pattern that the closer an atom is to the noble gas electron configuration, the fewer bonds it makes. Non-metals are close to the noble gas configuration, so they usually make a few bonds to a few neighbors. The noble gases make no bonds, and are monatomic (single atoms); halogens make 1 bond to 1 other atom, etc. This means that they don't usually form extended structures (except diamond and graphite). Instead, they form separate molecules. These molecules aren't held together tightly, so solids can easily melt or break. The electrons are held tightly by just 1 - 2 atoms, so they can't conduct electricity.
Reaction Patterns
Non-metals can react with each other to form compounds in which electrons are shared. These compounds have some of the same characteristics as the elementals forms: usually they melt or boil at relatively low temperature and don't conduct heat or electricity. When non-metals react with metals, they usually gain electrons to form anions. The cations are then attracted to the anions, so the result are ionic or sort of ionic compounds. The more a non-metal wants to gain electrons, the more reactive it is. Thus, the halogens are all reactive, but iodine is pretty safe, while bromine, chlorine and especially fluorine are really nasty and dangerous! Oxygen only seems safe and friendly to us because we are adapted to it. When oxygen first appeared in the atmosphere due to photosynthesis, most of the early life forms probably died from it; we descended from the survivors.


1.4: Types of Statistical Studies (2 of 4)
Learning Objectives
- Determine if a study is an experiment or an observational study.
- From a description of a statistical study, determine the goal of the study.
In general, there are two types of statistical studies: observational studies and experiments.
An observational study observes individuals and measures variables of interest. The main purpose of an observational study is to describe a group of individuals or to investigate an association between two variables. We can answer questions about a population with an observational study. We can also investigate a relationship between two variables. But in an observational study, researchers do not attempt to manipulate one variable to cause an effect in another variable. For this reason, an observational study does not provide convincing evidence of a cause-and-effect relationship.
An experiment intentionally manipulates one variable in an attempt to cause an effect on another variable. The primary goal of an experiment is to provide evidence for a cause-and-effect relationship between two variables. But the experiment has to be well-designed to provide convincing evidence of a cause-and-effect relationship. We study experiment design in the section “Collecting Data – Conducting an Experiment.”
For now, our goal is to distinguish between these two types of studies. We focus on the connection between the research question, the type of study, and the kinds of conclusions we can make.
Example
Music and Learning
Many students listen to music while studying. Does listening to music improve learning? Students in a statistics class decide to investigate this question. They write more specific research questions related to the topic of music and learning. Then they design the following three studies:
Study 1
Specific research questions: Do the majority of college students listen to music while they study? Do the majority of college students believe that listening to music improves their learning?
To investigate these questions, the statistics students conduct a survey in their other classes. They ask these two questions:
- Do you listen to music while you study?
- Do you think listening to music improves your concentration and memory?
This is an observational study designed to answer two questions about a population of college students. Each question contains a claim about the population of college students. We can use data from this study to see if these claims are true. But data from this study cannot provide evidence of a cause-and-effect relationship between listening to music while studying and improvements in learning.
Study 2
Specific research question: When we compare students who study with music to students who study in a quiet environment, which group gives higher ratings for understanding what they studied?
To investigate this question, the instructor divides the class into two groups: (1) those who listen to music when they study and (2) those who do not listen to music when they study. The students keep a journal for a week. Each time they study, they record the following information:
- Length of study session (in minutes)
- A rating of how well they understood what they studied, on a scale of 1–10: 1 = no understanding, 10 = excellent understanding.
This investigation is also an observational study. It compares two populations: (1) college students who listen to music when studying and (2) college students who do not listen to music when studying. We can also view this as an observational study of one population (college students) that investigates the relationship between two variables: listening to music while studying and perceived understanding of material studied . From this study, we might learn something interesting about the connection between college students’ study habits and their perception of their learning. But since this is an observational study, data from this study cannot provide evidence of a cause-and-effect relationship between listening to music while studying and improvements in learning.
Study 3
Specific research question: Does listening to music improve students’ ability to quickly identify information?
To investigate this question, the instructor uses word-search puzzles. She divides the class into two groups. Students on one side of the room do a word puzzle for 3 minutes while listening to music on an iPod. Students on the other side of the room do a word puzzle for 3 minutes without music. The instructor calculates the average number of words found by each group.
This study is an experiment. The instructor manipulates music to investigate the impact on puzzle completion. Data from this study can provide evidence of a cause-and-effect relationship between listening to music while studying and improvements in learning. But the improvement in learning is more narrowly defined as the ability to quickly identify information, such as words in a puzzle.
Contributors and Attributions
- Concepts in Statistics. Provided by : Open Learning Initiative. Located at : oli.cmu.edu. License : CC BY: Attribution