Matrix 4x4 calculation Let $$A = \begin{pmatrix}
1 &  a+3 &  8 &  2 \\
0 &  0 & 2 & 0 \\
0 & a-2 &  5 & 0 \\
0 & 10 & a &  a+3
\end{pmatrix}\begin{pmatrix}
x \\
y \\
z \\
w \\
\end{pmatrix}  = \begin{pmatrix}
0 \\
7 \\
-2 \\
0 \\
\end{pmatrix}$$
I'm a bit stuck on this question... I need to find values of a for which this matrix has unique solutions.
So far I found that:
$z=\dfrac{7}{2}$
$y= \dfrac{-39}{2(a-2)}$
$w= \dfrac{(390-(7a^2)+14a)}{ (2(a-2)(a+3))}$
Can someone help to solve it completely?
 A: $$\begin{align}\det A &= 1\cdot\det\left(\begin{array}{ccc}0 & 2 & 0\\a-2 & 3 & 0\\10 & a & a+3\end{array}\right)\quad (\text{expanded down first column})\\ &= 1\cdot(a+3)\cdot\det\left(\begin{array}{cc}0 & 2\\a-2 & 5\end{array}\right)\quad(\text{expanded down third column})\\ &= 1\cdot(a+3)\cdot\bigl(0-2(a-2)\bigr)\\ &= -2(a+3)(a-2).\end{align}$$
You'll have unique solutions if and only if $\det A\neq 0.$

In the comments below, you've come to see that as long as $a\neq 2$ and $a\neq -3$, then there will be a unique solution. Another way to see that is simply to look at the equations you've already generated: $$z=\frac72\tag{1}$$ $$y=\frac{-39}{2(a-2)}\tag{2}$$ $$w=\frac1{a+3}\left(\frac{195}{a-2}-\frac{7a}2\right)\tag{3}$$ Now $(2)$ doesn't even make sense when $a=2$, so in that case, we have no solutions at all. Likewise, $(3)$ makes no sense when $a=-3$, so we've no solutions in that case, either.
Otherwise, $(1)$ through $(3)$ make sense, and plugging the values thus determined into $$x=-(a+3)y-8z-2z\tag{4}$$ gives us the value of $x$ needed for the solution. Since $x,y,z,w$ are determined once we've plugged in our appropriate $a$, then the solution thus determined is unique.
