If $P=\left( \begin{matrix} 2 & 1 \\ -3 & 0 \\\end{matrix} \right)$ and I is a 2 × 2 unit matrix. Evaluate ${{p}^{2}}-2p+4I$
$\left( \begin{matrix} 9 & 4 \\ -12 & 1 \\\end{matrix} \right)$
$\left( \begin{matrix} -3 & 0 \\ 0 & -3 \\\end{matrix} \right)$
$\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\\end{matrix} \right)$
$\left( \begin{matrix} 1 & 4 \\ 4 & 1 \\\end{matrix} \right)$
$\begin{align} & P=\left( \begin{matrix} 2 & 1 \\ -3 & 0 \\\end{matrix} \right) \\ & {{P}^{2}}=\left( \begin{matrix} 2 & 1 \\ -3 & 0 \\\end{matrix} \right)\left( \begin{matrix} 2 & 1 \\ -3 & 0 \\\end{matrix} \right)=\left( \begin{matrix} 2\times 2+1\times (-3) & 2\times 1+1\times 0 \\ -3\times 2+0\times (-3) & -3\times 1+0\times 0 \\\end{matrix} \right) \\ & {{P}^{2}}=\left( \begin{matrix} 4-3 & 2+0 \\ -6+0 & -3+0 \\\end{matrix} \right)=\left( \begin{matrix} 1 & 2 \\ -6 & -3 \\\end{matrix} \right) \\ & {{P}^{2}}-2P+4I=\left( \begin{matrix} 1 & 2 \\ -6 & -3 \\\end{matrix} \right)+2\left( \begin{matrix} 2 & 1 \\ -3 & 0 \\\end{matrix} \right)+4\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\\end{matrix} \right) \\ & {{P}^{2}}-2P+4I=\left( \begin{matrix} 1 & 2 \\ -6 & -3 \\\end{matrix} \right)+\left( \begin{matrix} 4 & 2 \\ -6 & 0 \\\end{matrix} \right)+\left( \begin{matrix} 4 & 0 \\ 0 & 4 \\\end{matrix} \right) \\ & {{P}^{2}}-2P+4I=\left( \begin{matrix} 1-4+4 & 2-2+0 \\ -6+6+0 & -3+0+4 \\\end{matrix} \right) \\ & {{P}^{2}}-2P+4I=\left( \begin{matrix} 1 & 0 \\ 0 & 1 \\\end{matrix} \right) \\\end{align}$