<s>[INST] Explanation for a solution: Howard Anton, Elementary Linear Algebra I am currently working with the 1st edition of Howard Anton's "Elementary Linear Algebra". I tried the following problem:
Excercise Set 1.2 (p. 17), Problem 12:
For which values of $a$ will the following system have no solutions? Exactly one solution? Infinitely many solutions?
$$\begin{array}{rccccl} 
x &+& 2y &-& 3z &=& 4 \\ 
3x &-& y &+& 5z &=& 2 \\
4x &+& y &+& (a^2 - 14)z &=& a + 2\end{array}$$
By using Gauss-Jordan-Elimination (and Gaussian for a double check), I found the following solution set:
$$\begin{align*} x &= \frac{8}{7} + \frac{-a+4}{a^2-16} \\ 
y &= \frac{10}{7} + \frac{2a-8}{a^2-16} \\
z &= \frac{a-4}{a^2-16}\end{align*}$$
The solution in the textbook says, that the system has no solution if $a=-4$ an one solution if $a\neq\pm4$. This part I understand, since $z=\frac{a-4}{a^2-16}$ as well as others terms in the formulas for $y$ and $z$ are not defined for $a=-4$ but the formulas for $x,y,z$ will yield unambiguous values for $a\neq\pm4$. However, the solution also says, that the system has infinitely many solutions for $a=4$ and this is the point, which I don't understand. Isn't for instance $z=\frac{a-4}{a^2-16}$ still undefined for $a=-4$ or can I simply put in $z = \frac{0}{0} = 0$. And if I can, isn't $z=0$ still a unique value. I can find no room for different values of $x,y,z$ to satisfy the system of equations.
Can somebody explain this to me? Thanks in advance!
 [/INST] Alternative approach:
Compute the determinant of the matrix.
If the determinant is non-zero, then there will automatically be exactly $1$ solution.
If the determinant is $0$, then there will either be $0$ solutions or an infinite number of solutions, depending on whether there is an inconsistency (explained at the end of this answer) among the values to the right of the equal signs.

The determinant is
$(14 - a^2 - 5) + [(-2)(3a^2 - 62)] + [(-3)(7)] = 112 - 7a^2 = 7(16 - a^2).$
So, if $(16 - a^2) \neq 0$, then you know immediately that there is exactly one solution.
The problem then reduces to a consideration of $a^2 = 16.$

Here, I will explain the idea of inconsistency among linear equations as follows:
Consider the following two pairs of linear equations:

*

*$x + 2y = 6, ~2x + 4y = 12.$

*$x + 2y = 6, ~2x + 4y = 13.$
Both pairs of equations will have a determinant of $(1 \times 4) - (2 \times 2) = 0.$
However, the first pair of equations above is consistent, and therefore permits an infinite number of solutions.  The second pair of equations above is inconsistent, so there will be $0$ solutions.

The first thing to notice is that the superficial judgement that the evaluation of $a=4$ and $a=-4$ will yield identical results is wrong.  It is true that the LHS makes no distinction between $a=4,$ and $a=-4$, since the LHS only features an $a^2$ item.
However the RHS features an $(a+2)$ value which requires that $a=4$ and $a=-4$ be evaluated separately, to determine which values of $a$ (if any) results in consistent values that yield an infinite number of solutions, and which values of $a$ (if any) result in inconsistent values that yield $0$ solutions.

With $a$ equal to either $+4$ or $-4$, the equations become
$1x + 2y - 3z = 4$ 
$3x - 1y + 5z = 2$ 
$4x + 1y + 2z = 6 ~\text{or}~ -2.$
Adding the 2nd and 3rd equations above together yields:
$7x + 7z = 8 ~\text{or}~ 0.$
Multiplying the 2nd equation above by $(2)$ and adding it to the 1st equation yields:
$7x + 7z = 8.$
At this point, you know immediately that $a=-4$ causes an inconsistency, and that therefore, if $a=-4$, there are no solutions.
However, the value of $a=4$ must be explored further, to determine whether it yields consistent results.
Treating $z$ as a fixed (unknown) value, results in :
$1x + 2y = 4 + 3z$ 
$3x - 1y = 2 - 5z$ 
$4x + 1y=  6 - 2z$.
Adding the 2nd and 3rd of these revised equations together yields
$\displaystyle 7x = 8 - 7z \implies x = \frac{8}{7} - z.$
Using the 3rd equation above, this implies that
$\displaystyle y = (6 - 2z) - 4\left(\frac{8}{7} - z\right) = \frac{10}{7} + 2z.$
It only remains to verify that the following values also satisfy the 1st and 2nd equations above, which they do:

*

*$\displaystyle x = \frac{8}{7} - z, ~y = \frac{10}{7} + 2z.$
Therefore, $a=4$ yields consistent results.
Therefore, $a=4$ yields an infinite number of solutions.
</s>