## Precalculus (6th Edition) Blitzer

Consider the given matrix $A=\left[ \begin{matrix} 1 & -3 \\ -1 & 3 \\ \end{matrix} \right]$. Now, we check if the matrix is invertible or not. Consider the matrix, $A=\left[ \begin{matrix} 1 & -3 \\ -1 & 3 \\ \end{matrix} \right]$ The inverse of matrix $\left[ A \right]$ is equal to, ${{\left[ A \right]}^{-1}}=\frac{1}{ad-bc}\left[ \begin{matrix} d & -b \\ -c & a \\ \end{matrix} \right]$ Compare the matrix to the original matrix. So, \begin{align} & a=1 \\ & b=-3 \\ & c=-1 \\ & d=3 \end{align} Substitute the values to get, \begin{align} & {{\left[ A \right]}^{-1}}=\frac{1}{ad-bc}\left[ \begin{matrix} d & -b \\ -c & a \\ \end{matrix} \right] \\ & =\frac{1}{3-3}\left[ \begin{matrix} 3 & 3 \\ 1 & 1 \\ \end{matrix} \right] \\ & =\frac{1}{0}\left[ \begin{matrix} 3 & 3 \\ 1 & 1 \\ \end{matrix} \right] \end{align} So, $ad-bc=0$, which shows that the matrix is not invertible. Hence, the statement is False. If $ad-bc\ne 0$ Then, the matrix is invertible and the statement will be true.