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!

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.
