The inverse of the matrix $\left[ \begin{matrix} 2 & 1 \\ 1 & 1 \\\end{matrix} \right]$is
$\left[ \begin{matrix} 1 & 1 \\ 1 & 2 \\\end{matrix} \right]$
$\left[ \begin{matrix} 1 & 1 \\ -1 & 2 \\\end{matrix} \right]$
$\left[ \begin{matrix} 1 & -1 \\ -1 & 2 \\\end{matrix} \right]$
$\left[ \begin{matrix} 1 & -1 \\ 1 & 2 \\\end{matrix} \right]$
$\begin{align} & \text{Given a 2}\times \text{2 matrix such that} \\ & A=\left[ \begin{matrix} a & b \\ c & d \\\end{matrix} \right] \\ & {{A}^{-1}}=\frac{1}{ad-cb}\left[ \begin{matrix} d & -b \\ -c & a \\\end{matrix} \right] \\ & \text{Where }{{A}^{-1}}=inverse\text{ of }A \\ & A=\left[ \begin{matrix} 2 & 1 \\ 1 & 1 \\\end{matrix} \right] \\ & {{A}^{-1}}=\frac{1}{2-1}\left[ \begin{matrix} 1 & -1 \\ -1 & 2 \\\end{matrix} \right]=\left[ \begin{matrix} 1 & -1 \\ -1 & 2 \\\end{matrix} \right] \\\end{align}$