## Elementary Linear Algebra 7th Edition

$B=A^{-1}$
$A=\begin{pmatrix}2&1\\5&3\end{pmatrix}\quad B=\begin{pmatrix}3&-1\\-5&2\end{pmatrix}$ To verify that $B$ is the inverse of $A$ we simply need to perform matrix multiplication: $AB=\begin{pmatrix}2&1\\5&3\end{pmatrix}\begin{pmatrix}3&-1\\-5&2\end{pmatrix}=\begin{pmatrix}2\times3+1\times(-5)&2\times(-1)+1\times2\\5\times3+3\times(-5)&5\times(-1)+3\times2\end{pmatrix}$ $=\begin{pmatrix}1&0\\0&1\end{pmatrix}$ $BA=\begin{pmatrix}3&-1\\-5&2\end{pmatrix}\begin{pmatrix}2&1\\5&3\end{pmatrix}=\begin{pmatrix}3\times2+(-1)\times5&3\times1+(-1)\times3\\(-5)\times2+2\times5&(-5)\times1+2\times3\end{pmatrix}$ $=\begin{pmatrix}1&0\\0&1\end{pmatrix}$ and so $B=A^{-1}$