Answer
The provided statement is False.
Work Step by Step
Consider the given matrix $A=\left[ \begin{matrix}
1 & -3 \\
-1 & 3 \\
\end{matrix} \right]$.
Now, we check if the matrix is invertible or not.
Consider the matrix,
$A=\left[ \begin{matrix}
1 & -3 \\
-1 & 3 \\
\end{matrix} \right]$
The inverse of matrix $\left[ A \right]$ is equal to,
${{\left[ A \right]}^{-1}}=\frac{1}{ad-bc}\left[ \begin{matrix}
d & -b \\
-c & a \\
\end{matrix} \right]$
Compare the matrix to the original matrix. So,
$\begin{align}
& a=1 \\
& b=-3 \\
& c=-1 \\
& d=3
\end{align}$
Substitute the values to get,
$\begin{align}
& {{\left[ A \right]}^{-1}}=\frac{1}{ad-bc}\left[ \begin{matrix}
d & -b \\
-c & a \\
\end{matrix} \right] \\
& =\frac{1}{3-3}\left[ \begin{matrix}
3 & 3 \\
1 & 1 \\
\end{matrix} \right] \\
& =\frac{1}{0}\left[ \begin{matrix}
3 & 3 \\
1 & 1 \\
\end{matrix} \right]
\end{align}$
So, $ad-bc=0$, which shows that the matrix is not invertible. Hence, the statement is False.
If $ad-bc\ne 0$
Then, the matrix is invertible and the statement will be true.
