<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] You cannot just look at the final product if you did not carefully note steps in which you were assuming facts about the value of $a$. So let us take a careful look at the Gaussian elimination process.
Starting from
$$\left(\begin{array}{rrr|r}
1 & 2 & -3 & 4\\
3 & -1 & 5 & 2\\
4 & 1 & a^2-14 & a+2
\end{array}\right)$$
we first subtract three times the first row from the second, and four times the first row from the third row. We get:
$$\left(\begin{array}{rrr|r}
1 & 2 & -3 & 4\\
0 & -7 & 14 & -10\\
0 & -7 & a^2-2 & a-14
\end{array}\right)$$
Then subtracting the second row from the third row, we obtain:
$$\left(\begin{array}{rrr|r}
1 & 2 & -3 & 4\\
0 & -7 & 26 & -10\\
0 & 0 & a^2-16 & a-4
\end{array}\right).$$
At this point:  if $a=-4$, then the last row becomes
$$\left(\begin{array}{rrr|r}
0 & 0 & 0 & -8
\end{array}\right).$$
So the system has no solutions.
If $a=4$, on the other hand, the last row is
$$\left(\begin{array}{ccc|c}
0 & 0 & 0 & 0
\end{array}\right)$$
and your matrix has rank $2$, giving you infinitely many solutions.
And if $a\neq 4$ and $a\neq -4$, then you have a matrix of rank $3$, so you will get exactly one solution. This matches what the solutions say.
I suspect what happened is that you proceeded to divide the second row by $-7$ (no problem there):
$$\left(\begin{array}{rrr|r}
1 & 2 & -3 & 4\\
0 & 1 & -\frac{26}{7} & \frac{10}{7}\\
0 & 0 & a^2-16 & a-4
\end{array}\right)$$
and then divided the last row by $a^2-16$. But this last step requires the assumption that $a^2-16\neq 0$. Thus, you are implicitly saying "and by the way, $a\neq 4$ and also $a\neq -4$." Nothing you get after that can be used in the case where $a=4$ or where $a=-4$. You need to consider those cases separately.
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