## Precalculus (6th Edition) Blitzer

The product of $AB=\left[ \begin{matrix} -3 & -4 \\ 0 & 1 \\ \end{matrix} \right]$, product of $BA=\left[ \begin{matrix} -3 & 0 \\ -4 & 1 \\ \end{matrix} \right]$ and $B$ is not the multiplicative inverse of $A$ i.e. $B\ne {{A}^{-1}}$
The given expression is: $A=\left[ \begin{matrix} -2 & -1 \\ -1 & 1 \\ \end{matrix} \right],B=\left[ \begin{matrix} 1 & 1 \\ 1 & 2 \\ \end{matrix} \right]$ Now, we will compute the matrix as $\left[ AB \right]$ \begin{align} & AB=\left[ \begin{matrix} -2 & -1 \\ -1 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 1 \\ 1 & 2 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} \left( -2 \right)\times 1+\left( -1 \right)\times 1 & \left( -2 \right)\times 1+\left( -1 \right)\times 2 \\ \left( -1 \right)\times 1+1\times 1 & \left( -1 \right)\times 1+1\times 2 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -3 & -4 \\ 0 & 1 \\ \end{matrix} \right] \end{align} Now we will compute the matrix as $\left[ BA \right]$ \begin{align} & BA=\left[ \begin{matrix} 1 & 1 \\ 1 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} -2 & -1 \\ -1 & 1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1\times \left( -2 \right)+1\times \left( -1 \right) & 1\times \left( -1 \right)+1\times 1 \\ 1\times \left( -2 \right)+2\times \left( -1 \right) & 1\times \left( -1 \right)+2\times 1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} -3 & 0 \\ -4 & 1 \\ \end{matrix} \right] \end{align} Neither matrix $AB$ nor matrix $BA$ are equal to the identity matrix. Now, we will compute the matrix $B={{A}^{-1}}$ $B=\left[ \begin{matrix} 1 & 1 \\ 1 & 2 \\ \end{matrix} \right]$ And, $A=\left[ \begin{matrix} -2 & -1 \\ -1 & 1 \\ \end{matrix} \right]$ Determine the inverse of matrix $A$. Now, using the 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=-1 \\ & c=-1 \\ & d=1 \\ \end{align} So, \begin{align} & {{A}^{-1}}=\frac{1}{\left| -2-1 \right|}\left[ \begin{matrix} 1 & 1 \\ 1 & -2 \\ \end{matrix} \right] \\ & =\frac{1}{-3}\left[ \begin{matrix} 1 & 1 \\ 1 & -2 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} \frac{1}{-3} & \frac{1}{-3} \\ \frac{1}{-3} & \frac{2}{3} \\ \end{matrix} \right] \end{align} Therefore, no $B$ is not the multiplicative inverse of $A$.