Show that $\begin{bmatrix$}[r] 1 & 2 \ 2 & 1 \end{bmatrix} is a solution of the matrix equation X² - 2X - 3I = 0 where I is the unit matrix of order 2.

$I = \begin{bmatrix$}[r] 1 & 0 \ 0 & 1 \end{bmatrix}, X = \begin{bmatrix}[r] 1 & 2 \ 2 & 1 \end{bmatrix} \\[1em]

Given,

X² - 2X - 3I = 0

Putting value of X and I in above equation we get,

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

Since, L.H.S. = $\begin{bmatrix$}[r] 0 & 0 \ 0 & 0 \end{bmatrix} = R.H.S. Hence, proved that X² - 2X -3I = 0.

∴$\begin{bmatrix$}[r] 1 & 2 \ 2 & 1 \end{bmatrix} is a solution of the matrix equation X² - 2X - 3I = 0