Answer
The provided statement is False.
Work Step by Step
The given statement false. The inverse of a $2\times 2$ matrix may not exist.
Consider the matrix,
$A=\left[ \begin{matrix}
a & b \\
c & d \\
\end{matrix} \right]$
Now, compare the matrix to the given matrix.
$\begin{align}
& a=1 \\
& b=4 \\
& c=2 \\
& d=8
\end{align}$
Now, the inverse is given by,
${{\left[ A \right]}^{-1}}=\frac{1}{ad-bc}\left[ \begin{matrix}
d & -b \\
-c & a \\
\end{matrix} \right]$
Substituting the values, we get,
$\begin{align}
& {{\left[ A \right]}^{-1}}=\frac{1}{ad-bc}\left[ \begin{matrix}
d & -b \\
-c & a \\
\end{matrix} \right] \\
& =\frac{1}{\left( 1 \right)\left( 8 \right)-\left( 4 \right)\left( 2 \right)}\left[ \begin{matrix}
8 & -4 \\
-2 & 1 \\
\end{matrix} \right] \\
& =\frac{1}{0}\left[ \begin{matrix}
8 & -4 \\
-2 & 1 \\
\end{matrix} \right]
\end{align}$
So, $ad-bc=0$, which shows the matrix is not invertible. Hence, the statement is False.
