Answer
The value is: $a=-2,a=3$.
Work Step by Step
Consider the given matrix: $\left[ \begin{matrix}
1 & a+1 \\
a-2 & 4 \\
\end{matrix} \right]$
The determinant of the matrix is given by:
$\left| \begin{matrix}
1 & a+1 \\
a-2 & 4 \\
\end{matrix} \right|$
The matrix will not be invertible when the value of its determinant is zero.
Now, we will solve the determinant:
$\begin{align}
& \left| \begin{matrix}
1 & a+1 \\
a-2 & 4 \\
\end{matrix} \right|=4-\left( a-2 \right)\left( a+1 \right) \\
& =4-\left( {{a}^{2}}+a-2a-2 \right) \\
& =4-{{a}^{2}}+a+2 \\
& =-{{a}^{2}}+a+6
\end{align}$
Solve the quadratic equation by splitting its middle term:
$\begin{align}
& {{a}^{2}}-a-6=0 \\
& {{a}^{2}}-3a+2a-6=0 \\
& a\left( a-3 \right)+2\left( a-3 \right)=0 \\
& \left( a-3 \right)\left( a+2 \right)=0
\end{align}$
Either,
$\begin{align}
& \left( a-3 \right)=0 \\
& a=3
\end{align}$
Or,
$\begin{align}
& \left( a+2 \right)=0 \\
& a=-2
\end{align}$
