If $$I$$ be the $$2×2$$ identity matrix and $$A = \begin{pmatrix} 2 & -1 \\-1 & 2\end{pmatrix}$$, then $$A^2$$ is equal to​

Correct option is A

Given:

$$\ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \quad A = \begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix}$$​​

Solution:

​$$\text{Compute } A^2: \\A^2 = A \cdot A =\begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix}\cdot\begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix}=\begin{pmatrix}4 + 1 & -2 - 2 \\-2 - 2 & 1 + 4\end{pmatrix}=\begin{pmatrix}5 & -4 \\-4 & 5\end{pmatrix} \\$$

​​$$\text{Try } 4A - 3I: \\4A = \begin{pmatrix} 8 & -4 \\ -4 & 8 \end{pmatrix}, \quad3I = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} \\4A - 3I = \begin{pmatrix} 5 & -4 \\ -4 & 5 \end{pmatrix} = A^2 \\\boxed{\text{Answer: (A) } A^2 = 4A - 3I}$$

​​