## Precalculus (6th Edition) Blitzer

The product of $AB=\left[ \begin{matrix} 8 & -16 \\ -2 & 7 \\ \end{matrix} \right]$, product of $BA=\left[ \begin{matrix} 12 & 12 \\ 1 & 3 \\ \end{matrix} \right]$ and $B$ is not multiplicative inverse of $A$ $B\ne {{A}^{-1}}$
The given expression is $A=\left[ \begin{matrix} -4 & 0 \\ 1 & 3 \\ \end{matrix} \right],B=\left[ \begin{matrix} -2 & 4 \\ 0 & 1 \\ \end{matrix} \right]$ Now, we will compute the matrix as $\left[ AB \right]$ \begin{align} & AB=\left[ \begin{matrix} -4 & 0 \\ 1 & 3 \\ \end{matrix} \right]\left[ \begin{matrix} -2 & 4 \\ 0 & 1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} \left( -4 \right)\times 1+0\times 0 & \left( -4 \right)\times 4+0\times 1 \\ 1\times \left( -2 \right)+3\times 0 & 1\times 4+3\times 1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 8 & -16 \\ -2 & 7 \\ \end{matrix} \right] \end{align} And, we will compute the matrix as $\left[ BA \right]$ \begin{align} & BA=\left[ \begin{matrix} -2 & 4 \\ 0 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} -4 & 0 \\ 1 & 3 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} \left( -2 \right)\times \left( -4 \right)+4\times 1 & \left( -2 \right)\times 0+4\times 3 \\ 0\times \left( -4 \right)+1\times 1 & 0\times 0+1\times 3 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 12 & 12 \\ 1 & 3 \\ \end{matrix} \right] \end{align} Since both the matrix $AB=\left[ \begin{matrix} 8 & -16 \\ -2 & 7 \\ \end{matrix} \right]$ and $BA=\left[ \begin{matrix} 12 & 12 \\ 1 & 3 \\ \end{matrix} \right]$ Now, we will compute the matrix $B={{A}^{-1}}$. $B=\left[ \begin{matrix} -2 & 4 \\ 0 & 1 \\ \end{matrix} \right]$ And $A=\left[ \begin{matrix} -4 & 0 \\ 1 & 3 \\ \end{matrix} \right]$ Determine the inverse of matrix A. Now, using inverse formula ${{A}^{-1}}=\frac{1}{\left| ad-bc \right|}\left[ \begin{matrix} d & -b \\ -c & a \\ \end{matrix} \right]$ Substitute the value to get, $a=-4,b=0,c=1,d=3$ So, \begin{align} & {{A}^{-1}}=\frac{1}{\left| -12-0 \right|}\left[ \begin{matrix} 3 & -0 \\ -1 & -4 \\ \end{matrix} \right] \\ & =\frac{1}{-12}\left[ \begin{matrix} 3 & -0 \\ -1 & -4 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} \frac{3}{-12} & \frac{0}{12} \\ \frac{1}{12} & \frac{4}{12} \\ \end{matrix} \right] \end{align} Therefore, $B\ne {{A}^{-1}}$