Answer
For $a=-2$ or $a=3$,
the matrix is not invertible.
Work Step by Step
If $A=\left[\begin{array}{ll}
a & b\\
c & d
\end{array}\right]$, then$ A$ is invertible if and only if $ad-bc\neq 0.$
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So, if
$1(4)-(a+1)(a-2)=0,$
the matrix will not be invertible.
Solve for a.
$1(4)-(a+1)(a-2)=0,$
$4-(a^{2}-a-2)=0$
$4-a^{2}+a+2=0\qquad/\times(-1)$
$a^{2}-a-6=0$
... factor the LHS by
... finding two factors of $-6$ whose sum is $-1...$
... $-3$ and $+2$ ...
$(a+2)(a-3)=0$
For $a=-2$ or $a=3$,
the matrix is not invertible.
