## Precalculus (6th Edition) Blitzer

The product of $AB=\left[ \begin{matrix} -6 & -12 \\ 3 & 6 \\ \end{matrix} \right]$, product of $BA=\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \\ \end{matrix} \right]$ and $B\ne {{A}^{-1}}$ because $ad-bc=0$ and the matrix is not invertible.
The given expression is $A=\left[ \begin{matrix} -2 & 4 \\ 1 & -2 \\ \end{matrix} \right],B=\left[ \begin{matrix} 1 & 2 \\ -1 & -2 \\ \end{matrix} \right]$ Now, we will compute the matrix as $\left[ AB \right]$ \begin{align} & AB=\left[ \begin{matrix} -2 & 4 \\ 1 & -2 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 \\ -1 & -2 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} \left( -2 \right)\times 1+4\times \left( -1 \right) & \left( -2 \right)\times 2+4\times \left( -2 \right) \\ 1\times 1+\left( -2 \right)\times \left( -1 \right) & 1\times 2+\left( -2 \right)\times \left( -2 \right) \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -6 & -12 \\ 3 & 6 \\ \end{matrix} \right] \end{align} And, we will compute the matrix as $\left[ BA \right]$ \begin{align} & BA=\left[ \begin{matrix} 1 & 2 \\ -1 & -2 \\ \end{matrix} \right]\left[ \begin{matrix} -2 & 4 \\ 1 & -2 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1\times \left( -2 \right)+2\times 1 & 1\times 4+2\times \left( -2 \right) \\ \left( -1 \right)\times \left( -2 \right)+\left( -2 \right)\times 1 & \left( -1 \right)\times 4+\left( -2 \right)\times \left( -2 \right) \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \\ \end{matrix} \right] \end{align} Thus $AB=\left[ \begin{matrix} -6 & -12 \\ 3 & 6 \\ \end{matrix} \right]$ and matrix $BA=\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \\ \end{matrix} \right]$ Now, compute the matrix $B={{A}^{-1}}$. $B=\left[ \begin{matrix} 1 & 2 \\ -1 & -2 \\ \end{matrix} \right]$ And $A=\left[ \begin{matrix} -2 & 4 \\ 1 & -2 \\ \end{matrix} \right]$ 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 \begin{align} & a=-2 \\ & b=4 \\ & c=1 \\ & d=-2 \\ \end{align} So, \begin{align} & {{A}^{-1}}=\frac{1}{\left| 4-4 \right|}\left[ \begin{matrix} -2 & -4 \\ -1 & -2 \\ \end{matrix} \right] \\ & =\frac{1}{0}\left[ \begin{matrix} -2 & -4 \\ -1 & -2 \\ \end{matrix} \right] \end{align} Here, $ad-bc=0$ So, the matrix is not invertible. Thus, $B$ is not the multiplicative inverse of $A$.