## Precalculus (6th Edition) Blitzer

The inverse is, \begin{align} & {{\left( AB \right)}^{-1}}=\left[ \begin{matrix} -11 & 26 \\ 19 & -45 \\ \end{matrix} \right] \\ & {{A}^{-1}}{{B}^{-1}}=\left[ \begin{matrix} -79 & 101 \\ -18 & 23 \\ \end{matrix} \right] \\ & {{B}^{-1}}{{A}^{-1}}=\left[ \begin{matrix} -11 & 26 \\ 19 & -45 \\ \end{matrix} \right] \\ \end{align}
Consider the given matrix $A=\left[ \begin{matrix} 2 & -9 \\ 1 & -4 \\ \end{matrix} \right],B=\left[ \begin{matrix} 9 & 5 \\ 7 & 4 \\ \end{matrix} \right]$. \begin{align} & \left[ AB \right]=\left[ \begin{matrix} 2 & -9 \\ 1 & -4 \\ \end{matrix} \right]\left[ \begin{matrix} 9 & 5 \\ 7 & 4 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 18-63 & 10-36 \\ 9-16 & 5-16 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -45 & -26 \\ -19 & -11 \\ \end{matrix} \right] \end{align} Now, the inverse of matrix $\left[ AB \right]$ is equal to $A{{B}^{-1}}=\frac{1}{ad-bc}\left[ \begin{matrix} d & -b \\ -c & a \\ \end{matrix} \right]$. Now, we will compare the matrix to the original matrix. So, \begin{align} & a=-45 \\ & b=-26 \\ & c=-19 \\ & d=-11 \end{align} Now, the inverse is: ${{A}^{-1}}=\frac{1}{ad-bc}\left[ \begin{matrix} d & -b \\ -c & a \\ \end{matrix} \right]$ Substitute the values to get, \begin{align} & {{A}^{-1}}=\frac{1}{ad-bc}\left[ \begin{matrix} d & -b \\ -c & a \\ \end{matrix} \right] \\ & =\frac{1}{-1}\left[ \begin{matrix} -11 & 26 \\ 19 & -45 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -11 & 26 \\ 19 & -45 \\ \end{matrix} \right] \end{align} Therefore, the inverse of the matrix $\left[ AB \right]$ is $\left[ \begin{matrix} -11 & 26 \\ 19 & -45 \\ \end{matrix} \right]$. Now, for the inverse $A$ and inverse $B$ Then, consider the matrix $A=\left[ \begin{matrix} 2 & -9 \\ 1 & -4 \\ \end{matrix} \right]$ And $B=\left[ \begin{matrix} 9 & 5 \\ 7 & 4 \\ \end{matrix} \right]$ Now, the inverse is: ${{A}^{-1}}=\frac{1}{ad-bc}\left[ \begin{matrix} d & -b \\ -c & a \\ \end{matrix} \right]$. So compare the values \begin{align} & {{A}^{-1}}={{\left[ \begin{matrix} 2 & -9 \\ 1 & -4 \\ \end{matrix} \right]}^{-1}} \\ & =\frac{1}{-1}\left[ \begin{matrix} -4 & -\left( -9 \right) \\ -1 & 2 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -4 & 9 \\ -1 & 2 \\ \end{matrix} \right] \end{align} And, \begin{align} & {{B}^{-1}}=\frac{1}{1}\left[ \begin{matrix} 4 & -5 \\ -7 & 9 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 4 & -5 \\ -7 & 9 \\ \end{matrix} \right] \end{align} Then, \begin{align} & {{A}^{-1}}{{B}^{-1}}=\left[ \begin{matrix} -4 & 9 \\ -1 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} 4 & -5 \\ -7 & 9 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} \left( -4 \right)\times 4+9\times \left( -7 \right) & \left( -4 \right)\left( -5 \right)+81 \\ \left( -1 \right)\times 4+2\times \left( -7 \right) & \left( -1 \right)\left( -5 \right)+18 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -79 & 101 \\ -18 & 23 \\ \end{matrix} \right] \end{align} Now, the inverse of $B$ and inverse of $A$ is $\left[ \begin{matrix} -11 & 26 \\ 19 & -45 \\ \end{matrix} \right]$