If matrix A = $\begin{bmatrix$}[r] 1 & 2 \ 2 & 1 \end{bmatrix} and I is a unit matrix of order 2; find the value of m so that :

A² - 2A - mI = 0

Substituting value of A in A² - 2A - mI = 0, we get :

$\Rightarrow \begin{bmatrix$}[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} - m\begin{bmatrix}[r] 1 & 0 \ 0 & 1 \end{bmatrix} = \begin{bmatrix}[r] 0 & 0 \ 0 & 0 \end{bmatrix} \\[1em] \Rightarrow \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] m & 0 \ 0 & m \end{bmatrix} = \begin{bmatrix}[r] 0 & 0 \ 0 & 0 \end{bmatrix} \\[1em] \Rightarrow \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] m & 0 \ 0 & m \end{bmatrix} = \begin{bmatrix}[r] 0 & 0 \ 0 & 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 5 & 4 \ 4 & 5 \end{bmatrix} - \begin{bmatrix}[r] 2 & 4 \ 4 & 2 \end{bmatrix} - \begin{bmatrix}[r] m & 0 \ 0 & m \end{bmatrix} = \begin{bmatrix}[r] 0 & 0 \ 0 & 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 5 - 2 - m & 4 - 4 - 0 \ 4 - 4 - 0 & 5 - 2 - m \end{bmatrix} = \begin{bmatrix}[r] 0 & 0 \ 0 & 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 3 - m & 0 \ 0 & 3 - m \end{bmatrix} = \begin{bmatrix}[r] 0 & 0 \ 0 & 0 \end{bmatrix} \\[1em] \Rightarrow 3 - m = 0 \\[1em] \Rightarrow m = 3.

Hence, m = 3.