## Precalculus (6th Edition) Blitzer

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, ${{\left[ AB \right]}^{-1}}=\left[ {{B}^{-1}}{{A}^{-1}} \right]$