If M = $\begin{bmatrix$}[r] 1 & 2 \ 2 & 1 \end{bmatrix} and I is a unit matrix of the same order as that of M; show that :

M² = 2M + 3I

I = $\begin{bmatrix$}[r] 1 & 0 \ 0 & 1 \end{bmatrix}

M = $\begin{bmatrix$}[r] 1 & 2 \ 2 & 1 \end{bmatrix}

$\text{L.H.S. }= M^2 = \begin{bmatrix$}[r] 1 & 2 \ 2 & 1 \end{bmatrix}\begin{bmatrix}[r] 1 & 2 \ 2 & 1 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 1 \times 1 + 2 \times 2 & 1 \times 2 + 2 \times 1 \ 2 \times 1 + 1 \times 2 & 2 \times 2 + 1 \times 1 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 1 + 4 & 2 + 2 \ 2 + 2 & 4 + 1 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 5 & 4 \ 4 & 5 \end{bmatrix} \\[1em] \text{R.H.S.} = 2M + 3I \\[1em] = 2\begin{bmatrix}[r] 1 & 2 \ 2 & 1 \end{bmatrix} + 3\begin{bmatrix}[r] 1 & 0 \ 0 & 1 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 2 & 4 \ 4 & 2 \end{bmatrix} + \begin{bmatrix}[r] 3 & 0 \ 0 & 3 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 2 + 3 & 4 + 0 \ 4 + 0 & 2 + 3 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 5 & 4 \ 4 & 5 \end{bmatrix}.

Since, L.H.S. = R.H.S.

Hence, proved that M² = 2M + 3I.