Precalculus (6th Edition) Blitzer

The product of $AB={{I}_{2}}$, product of $BA={{I}_{2}}$ and $B$ is the multiplicative inverse of $A$ $B={{A}^{-1}}$.
The given expression is $A=\left[ \begin{matrix} 4 & 5 \\ 2 & 3 \\ \end{matrix} \right],B=\left[ \begin{matrix} \frac{3}{2} & -\frac{5}{2} \\ -1 & 2 \\ \end{matrix} \right]$ Now, we will compute the matrix as $\left[ AB \right]$ \begin{align} & AB=\left[ \begin{matrix} 4 & 5 \\ 2 & 3 \\ \end{matrix} \right]\left[ \begin{matrix} \frac{3}{2} & -\frac{5}{2} \\ -1 & 2 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 4\times \frac{3}{2}+5\times \left( -1 \right) & 4\times \left( -\frac{5}{2} \right)+5\times 2 \\ 2\times \frac{3}{2}+3\times \left( -1 \right) & 2\times \left( -\frac{5}{2} \right)+3\times 2 \\ \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} \frac{3}{2} & -\frac{5}{2} \\ -1 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} 4 & 5 \\ 2 & 3 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} \frac{3}{2}\times 4+\left( -\frac{5}{2} \right)\times 2 & \frac{3}{2}\times 5+\left( -\frac{5}{2} \right)\times 3 \\ \left( -1 \right)\times 4+2\times 2 & \left( -1 \right)\times 5+2\times 3 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right] \\ & ={{I}_{2}} \end{align} Both matrix $AB$ and matrix $BA$ are equal to the identity matrix Now, we will compute the matrix $B={{A}^{-1}}$. $B=\left[ \begin{matrix} \frac{3}{2} & -\frac{5}{2} \\ -1 & 2 \\ \end{matrix} \right]$ And, $A=\left[ \begin{matrix} 4 & 5 \\ 2 & 3 \\ \end{matrix} \right]$ 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 values to get, \begin{align} & a=4 \\ & b=5 \\ & c=2 \\ & d=3 \\ \end{align} So, \begin{align} & {{A}^{-1}}=\frac{1}{\left| 12-10 \right|}\left[ \begin{matrix} 3 & -5 \\ -2 & 4 \\ \end{matrix} \right] \\ & =\frac{1}{2}\left[ \begin{matrix} 3 & -5 \\ -2 & 4 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} \frac{3}{2} & \frac{-5}{2} \\ -1 & 2 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} \frac{3}{2} & \frac{-5}{2} \\ -1 & 2 \\ \end{matrix} \right] \end{align} Therefore, $B={{A}^{-1}}$.