## Precalculus (6th Edition) Blitzer

The inverse is, \begin{align} & {{\left( AB \right)}^{-1}}=\left[ \begin{matrix} -23 & 16 \\ 13 & -9 \\ \end{matrix} \right] \\ & {{A}^{-1}}{{B}^{-1}}=\left[ \begin{matrix} -3 & 11 \\ 8 & -29 \\ \end{matrix} \right] \\ & {{B}^{-1}}{{A}^{-1}}=\left[ \begin{matrix} -23 & 16 \\ 13 & -9 \\ \end{matrix} \right] \\ \end{align}.
Consider the given matrix $A=\left[ \begin{matrix} 2 & 1 \\ 3 & 1 \\ \end{matrix} \right],B=\left[ \begin{matrix} 4 & 7 \\ 1 & 2 \\ \end{matrix} \right]$. \begin{align} & \left[ AB \right]=\left[ \begin{matrix} 2 & 1 \\ 3 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 4 & 7 \\ 1 & 2 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 8+1 & 14+2 \\ 12+1 & 21+2 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 9 & 16 \\ 13 & 23 \\ \end{matrix} \right] \end{align} Now, the inverse of matrix $\left[ AB \right]$ is equal to $A{{B}^{-1}}=\frac{1}{ad-bc}\left[ \begin{matrix} d & -c \\ -d & a \\ \end{matrix} \right]$. Now, we will compare the matrix to the original matrix, \begin{align} & a=9 \\ & b=16 \\ & c=13 \\ & d=23 \end{align} Now, the inverse is, ${{A}^{-1}}=\frac{1}{ad-bc}\left[ \begin{matrix} d & -c \\ -b & a \\ \end{matrix} \right]$ Substitute the values to get, \begin{align} & {{A}^{-1}}=\frac{1}{ad-bc}\left[ \begin{matrix} d & -c \\ -b & a \\ \end{matrix} \right] \\ & =\frac{1}{-1}\left[ \begin{matrix} 23 & -16 \\ -13 & 9 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -23 & 16 \\ 13 & -9 \\ \end{matrix} \right] \end{align} Therefore, the inverse of the matrix $\left[ AB \right]$ is $\left[ \begin{matrix} -23 & 16 \\ 13 & -9 \\ \end{matrix} \right]$. Now, for the inverse of $A$ and inverse of $B$. Then, consider the matrix, $A=\left[ \begin{matrix} 2 & 1 \\ 3 & 1 \\ \end{matrix} \right]$ And $B=\left[ \begin{matrix} 4 & 7 \\ 1 & 2 \\ \end{matrix} \right]$ Now, the inverse is, ${{A}^{-1}}=\frac{1}{ad-bc}\left[ \begin{matrix} d & -c \\ -d & a \\ \end{matrix} \right]$ So compare the values to get, \begin{align} & {{A}^{-1}}={{\left[ \begin{matrix} 2 & 1 \\ 3 & 1 \\ \end{matrix} \right]}^{-1}} \\ & =\frac{1}{-1}\left[ \begin{matrix} 1 & -1 \\ -3 & 2 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -1 & 1 \\ 3 & -2 \\ \end{matrix} \right] \end{align} it can be further simplified as: \begin{align} & {{B}^{-1}}=\frac{1}{1}\left[ \begin{matrix} 2 & -7 \\ -1 & 4 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 2 & -7 \\ -1 & 4 \\ \end{matrix} \right] \end{align} Then, \begin{align} & {{A}^{-1}}{{B}^{-1}}=\left[ \begin{matrix} -1 & 1 \\ 3 & -2 \\ \end{matrix} \right]\left[ \begin{matrix} 2 & -7 \\ -1 & 4 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -23 & 16 \\ 13 & -9 \\ \end{matrix} \right] \end{align} Therefore, the inverse of $A$ and inverse of $B$ is $\left[ \begin{matrix} -23 & 16 \\ 13 & -9 \\ \end{matrix} \right]$. So, ${{\left( AB \right)}^{-1}}={{B}^{-1}}{{A}^{-1}}$