## Precalculus (6th Edition) Blitzer

Consider the two matrices $A$ and $B$ be the multiplicative inverses; that is, the product of the matrix AB and $BA$ is equal to the identity matrix. As example: Let matrix $A=\left[ \begin{matrix} 4 & -3 \\ -5 & 4 \\ \end{matrix} \right]$ and $B=\left[ \begin{matrix} 4 & 3 \\ 5 & 4 \\ \end{matrix} \right]$ If $B$ is the multiplicative inverse of $A$, then both $AB$ and $BA$ must result in the identity matrix ${{I}_{2}}=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]$ Checking whether $AB={{I}_{2}}$ \begin{align} & AB={{I}_{2}} \\ & AB=\left[ \begin{matrix} 4 & -3 \\ -5 & 4 \\ \end{matrix} \right]\left[ \begin{matrix} 4 & 3 \\ 5 & 4 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right] \\ & ={{I}_{2}} \end{align} Now, check whether $BA={{I}_{2}}$ \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} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right] \\ & ={{I}_{2}} \end{align} Thus, matrix $B$ is the multiplicative inverse of matrix $A$. Hence, $AB={{I}_{n}}$ and $BA={{I}_{n}}$.