If A = $\begin{bmatrix$}[r] 2 & 3 \ 1 & 2 \end{bmatrix}, find x, y so that A² = xA + yI.

Given,

$\text{A } = \begin{bmatrix$}[r] 2 & 3 \ 1 & 2 \end{bmatrix}, \text{ I } = \begin{bmatrix}[r] 1 & 0 \ 0 & 1 \end{bmatrix} \\[1em] \text{A}^2 = x\text{A} + y\text{I } \\[1em] \Rightarrow \begin{bmatrix}[r] 2 & 3 \ 1 & 2 \end{bmatrix} \begin{bmatrix}[r] 2 & 3 \ 1 & 2 \end{bmatrix} = x \begin{bmatrix}[r] 2 & 3 \ 1 & 2 \end{bmatrix} + y \begin{bmatrix}[r] 1 & 0 \ 0 & 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 2 \times 2 + 3 \times 1 & 2 \times 3 + 3 \times 2 \ 1 \times 2 + 2 \times 1 & 1 \times 3 + 2 \times 2 \end{bmatrix} = \begin{bmatrix}[r] 2x & 3x \ x & 2x \end{bmatrix} + \begin{bmatrix}[r] y & 0 \ 0 & y \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 4 + 3 & 6 + 6 \ 2 + 2 & 3 + 4 \end{bmatrix} = \begin{bmatrix}[r] 2x + y & 3x + 0 \ x + 0 & 2x + y \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 7 & 12 \ 4 & 7 \end{bmatrix} = \begin{bmatrix}[r] 2x + y & 3x + 0 \ x + 0 & 2x + y \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 7 & 12 \ 4 & 7 \end{bmatrix} = \begin{bmatrix}[r] 2x + y & 3x \ x & 2x + y \end{bmatrix} \\[1em]

By definition of equality of matrices we get,

3x = 12 or x = 4     (…Eq 1)

2x + y = 7              (…Eq 2)

Putting value of x from Eq1 in Eq 2,

⇒ 2(4) + y = 7
⇒ 8 + y = 7
⇒ y = 7 - 8
⇒ y = -1.

Hence, the value of x = 4 and y = -1.