## Precalculus (6th Edition) Blitzer

The product of $AB={{I}_{2}}$, product of $BA={{I}_{2}}$ and $B$ is the multiplicative inverse of $A$ i.e. $B={{A}^{-1}}$.
The given expression is $A=\left[ \begin{matrix} 4 & -3 \\ -5 & 4 \\ \end{matrix} \right],B=\left[ \begin{matrix} 4 & 3 \\ 5 & 4 \\ \end{matrix} \right]$ Now, compute the matrix as $\left[ AB \right]$ \begin{align} & AB=\left[ \begin{matrix} 4 & -3 \\ -5 & 4 \\ \end{matrix} \right]\left[ \begin{matrix} 4 & 3 \\ 5 & 4 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 4\times 4+\left( -3 \right)\times 5 & 4\times 3+\left( -3 \right)\times 4 \\ \left( -5 \right)\times 4+4\times 5 & \left( -5 \right)\times 3+4\times 4 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right] \\ & ={{I}_{2}} \end{align} Now we will compute the matrix as $\left[ BA \right]$ \begin{align} & BA=\left[ \begin{matrix} 4 & 3 \\ 5 & 4 \\ \end{matrix} \right]\left[ \begin{matrix} 4 & -3 \\ -5 & 4 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 4\times 4+3\times \left( -5 \right) & 4\times \left( -3 \right)+3\times 4 \\ 5\times 4+4\times \left( -5 \right) & 5\times \left( -3 \right)+4\times 4 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right] \\ & ={{I}_{2}} \end{align} Since both matrix $AB$ and $BA$ are equal to the identity matrix. Now, we will compute the matrix $B={{A}^{-1}}$. $B=\left[ \begin{matrix} 4 & 3 \\ 5 & 4 \\ \end{matrix} \right]$ And $A=\left[ \begin{matrix} 4 & -3 \\ -5 & 4 \\ \end{matrix} \right]$ 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, $a=4,b=-3,c=-5,d=4$ So, \begin{align} & {{A}^{-1}}=\frac{1}{\left| 16-15 \right|}\left[ \begin{matrix} 4 & 3 \\ 5 & 4 \\ \end{matrix} \right] \\ & =\frac{1}{1}\left[ \begin{matrix} 4 & 3 \\ 5 & 4 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 4 & 3 \\ 5 & 4 \\ \end{matrix} \right] \end{align} Yes, $B$ is the multiplicative inverse of $A$.