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!
 A: 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.
A: This type of problem is easier to analyse directly from the matrix avoiding going back to the system of linear equations as follows, this avoids some confusion.
The system can be reduced by row as
$$\begin{bmatrix} 1 & 2 & -3 & | & 4\\ 3 & -1 & 5 & | & 2\\ 4 & 1 & (a^{2}-14) & | & a+2\end{bmatrix} \sim \cdots \sim \begin{bmatrix} 1 & 2 & -3 & | & 4\\ 0 & -7 & 14 & | & -10\\ 0 & 0 & a^{2}-16 & | & a-4 \end{bmatrix}$$
Now, since $a^{2}-16=(a-4)(a+4)$ we have

*

*If $a^{2}-16=0$ and $a-4=0$ we have infinitely many solutions, i.e., $a=4$.

*If $a^{2}-16=0$ and $a-4\not=0$ we have no solution, i.e., $a=-4$.

*If $a^{2}-16\not=0 $ and $a-4\not=0$ we have one solution, i.e. $a\not=\pm4$.


More details about above
There is a way to look at this a little more quickly then reduce by row in an informal way by following the scheme below.

*

*Infinitely many solution $\to$ $\sim \cdots \sim \begin{bmatrix} * & * & * & | & *\\ 0 & * &* &|& *\\ 0 & 0 & 0 & |   &0 \end{bmatrix}$.


*Have no solution $\to$ $\sim \cdots \sim \begin{bmatrix} * & * & * & | & *\\ 0 & * &* &|& *\\ 0 & 0 & 0 & |   &* \end{bmatrix}$.


*have one solution $\to$ $\sim \cdots \sim \begin{bmatrix} * & * & * & | & *\\ 0 & * &* &|& *\\ 0 & 0 & * & |   &* \end{bmatrix}$.
The important part in the above scheme is the last row. We can achieve consistency or inconsistency by making a $0$ or a number other than zero $*$ (whether or not zeros appear in rows one and two does not matter) and another important detail is that we must get the parameter in this case "$a$" in the last row to apply the scheme. However, do not consider this as a general rule, there are cases where it is important to do additional analysis. But in this problem the scheme work well.
A: 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.
A: Rather than proceeding with Gauss-Jordan elimination....
For what values of $a$ are the vectors $(1,2,-3),(3,-1,5),(4,1,a^2-14)$ linearly independent or dependent?
If they are independent we have a unique solution.
If they are dependent, either the system of equations is consistent and we have infinitely many solutions, or the system is inconsistent and we no solution.
Adding the first two vectors, $(1,2,-3)+(3,-1,5) = (4,1,2).$ We have dependency if $a^2-14 = 2$ or $a = \pm 4$
Now to check for consistency... From the sum of the first two equations.
$4x + y + 2z = 6$
and from the third
$4x + y+ 4z = \pm 4 + 2$
If we choose positive 4 our equations are consistent, and if we choose $-4$ they are not.
