Answer
The provided statement is False.
Work Step by Step
The provided expression is False as the correct statement is: ${{\left[ AB \right]}^{-1}}={{B}^{-1}}{{A}^{-1}}$ not ${{\left[ AB \right]}^{-1}}={{A}^{-1}}{{B}^{-1}}$
Assuming the matrix,
$A=\left[ \begin{matrix}
1 & 1 \\
2 & 3 \\
\end{matrix} \right]$
Now, the multiplicative inverse is given by,
$\begin{align}
& A=\left[ \begin{matrix}
a & b \\
c & d \\
\end{matrix} \right] \\
& =\frac{1}{ab-bc}\left[ \begin{matrix}
d & -b \\
-c & a \\
\end{matrix} \right]
\end{align}$
Then, compare the value, to get,
$\left[ A \right]=\frac{1}{\left( 1 \right)\left( 3 \right)-\left( 1 \right)\left( 2 \right)}\left[ \begin{matrix}
3 & -1 \\
-2 & 1 \\
\end{matrix} \right]$
$\begin{align}
& \left[ A \right]=\frac{1}{1}\left[ \begin{matrix}
3 & -1 \\
-2 & 1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
3 & -1 \\
-2 & 1 \\
\end{matrix} \right]
\end{align}$
So, the inverse of matrix $\left[ A \right]=\left[ \begin{matrix}
3 & -1 \\
-2 & 1 \\
\end{matrix} \right]$; invertible matrix.
Consider, $B=\left[ \begin{matrix}
2 & 3 \\
4 & 5 \\
\end{matrix} \right]$
Compare the value, to get,
$\begin{align}
& a=2 \\
& b=3 \\
& c=4 \\
& d=5
\end{align}$
Now, the inverse of $B$ is given by,
${{\left[ B \right]}^{-1}}=\frac{1}{ad-bc}\left[ \begin{matrix}
d & -b \\
-c & a \\
\end{matrix} \right]$
Substitute the values to get,
$\begin{align}
& {{\left[ B \right]}^{-1}}=\frac{1}{ad-bc}\left[ \begin{matrix}
d & -b \\
-c & a \\
\end{matrix} \right] \\
& =\frac{1}{-2}\left[ \begin{matrix}
5 & -3 \\
-4 & 2 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-\frac{5}{2} & \frac{3}{2} \\
2 & -1 \\
\end{matrix} \right]
\end{align}$
Therefore, the inverse of the matrix $\left[ B \right]$ is $\left[ \begin{matrix}
-\frac{5}{2} & \frac{3}{2} \\
2 & -1 \\
\end{matrix} \right]$; invertible matrix. So, the multiplicative of $AB$ is:
$\begin{align}
& \left[ AB \right]=\left[ \begin{matrix}
1 & 1 \\
2 & 3 \\
\end{matrix} \right]\left[ \begin{matrix}
2 & 3 \\
4 & 5 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
6 & 8 \\
16 & 21 \\
\end{matrix} \right]
\end{align}$
Now, the inverse $AB$ is,
${{\left[ AB \right]}^{-1}}=\frac{1}{ad-bc}\left[ \begin{matrix}
d & -b \\
-c & a \\
\end{matrix} \right]$
So, compare the value to get,
$\begin{align}
& a=6 \\
& b=8 \\
& c=16 \\
& d=21 \\
\end{align}$
Then,
$\begin{align}
& {{\left[ AB \right]}^{-1}}={{\left[ \begin{matrix}
6 & 8 \\
16 & 21 \\
\end{matrix} \right]}^{-1}} \\
& =\frac{1}{\left( 6 \right)\left( 21 \right)-\left( 16 \right)\left( 8 \right)}\left[ \begin{matrix}
21 & -8 \\
-16 & 6 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-10.5 & 4 \\
8 & -3 \\
\end{matrix} \right]
\end{align}$
So, $AB$ is invertible and ${{\left[ AB \right]}^{-1}}=\left[ \begin{matrix}
10.5 & 4 \\
8 & -3 \\
\end{matrix} \right]$
Now,
$\begin{align}
& \left[ {{A}^{-1}}{{B}^{-1}} \right]=\left[ \begin{matrix}
3 & -1 \\
-2 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
-\frac{5}{2} & \frac{3}{2} \\
2 & -1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-9.5 & 5.5 \\
7 & -7 \\
\end{matrix} \right]
\end{align}$
Therefore, the provided expression is False. The True expression is ${{\left[ AB \right]}^{-1}}=\left[ {{B}^{-1}}{{A}^{-1}} \right]$
$\begin{align}
& \left[ {{B}^{-1}}{{A}^{-1}} \right]=\left[ \begin{matrix}
-\frac{5}{2} & \frac{3}{2} \\
2 & -1 \\
\end{matrix} \right]\left[ \begin{matrix}
3 & -1 \\
-2 & 1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-10.5 & 4 \\
8 & -3 \\
\end{matrix} \right]
\end{align}$
Therefore, ${{\left[ AB \right]}^{-1}}=\left[ {{B}^{-1}}{{A}^{-1}} \right]$
