# Question 23

\begin{align} & \text{Find the inverse of }\left[ \begin{matrix} 5 & 3 \\ 6 & 4 \\\end{matrix} \right] \\ & (A)\left[ \begin{matrix} 2 & \tfrac{3}{2} \\ -3 & \tfrac{5}{2} \\\end{matrix} \right] \\ & (B)\left[ \begin{matrix} 2 & -\tfrac{3}{2} \\ -3 & -\tfrac{5}{2} \\\end{matrix} \right] \\ & (C)\left[ \begin{matrix} 2 & -\tfrac{3}{2} \\ -3 & \tfrac{5}{2} \\\end{matrix} \right] \\ & (D)\,\left[ \begin{matrix} 2 & \tfrac{3}{2} \\ -3 & -\tfrac{5}{2} \\\end{matrix} \right] \\\end{align}
\begin{align} & \text{Given that }A=\left[ \begin{matrix} a & b \\ c & d \\\end{matrix} \right] \\ & {{A}^{-1}}=\frac{1}{ad-bc}\left[ \begin{matrix} d & -b \\ -c & a \\\end{matrix} \right] \\ & A=\left[ \begin{matrix} 5 & 3 \\ 6 & 4 \\\end{matrix} \right] \\ & {{A}^{-1}}=\frac{1}{5\times 4-3\times 6}\left[ \begin{matrix} 4 & -3 \\ -6 & 5 \\\end{matrix} \right]=\frac{1}{20-18}\left[ \begin{matrix} 4 & -3 \\ -6 & 5 \\\end{matrix} \right] \\ & {{A}^{-1}}=\frac{1}{2}\left[ \begin{matrix} 4 & -3 \\ -6 & 5 \\\end{matrix} \right]=\left[ \begin{matrix} \tfrac{4}{2} & -\tfrac{3}{2} \\ -\tfrac{6}{2} & \tfrac{5}{2} \\\end{matrix} \right]=\left[ \begin{matrix} 2 & -\tfrac{3}{2} \\ -3 & \tfrac{5}{2} \\\end{matrix} \right] \\\end{align}