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.
\align
assumes you have multiple equations, so the second&
is supposed to separate one equation from another, and the third gives the location of the second equal sign). Also, I think you were missing az
in the last equation. Please check that this is correct as edited. $\endgroup$