If A = $\begin{bmatrix} 2 & -3 \ a & b \end{bmatrix}$, find a and b so that A² = I.

Solving A²,

$\Rightarrow A^2 = \begin{bmatrix} 2 & -3 \ a & b \end{bmatrix} \times \begin{bmatrix} 2 & -3 \ a & b \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2 \times 2 + (-3) \times a & 2 \times (-3) + (-3) \times b \ a \times 2 + b \times a & a \times (-3) + b \times b \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 - 3a & -6 - 3b \ 2a + ab & -3a + b^2 \end{bmatrix}.$

Given,

A² = I,

$\begin{bmatrix} 4 - 3a & -6 - 3b \ 2a + ab & -3a + b^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}.$

∴ 4 - 3a = 1,

⇒ -3a = -3

⇒ a = 1.

∴ -6 - 3b = 0,

⇒ -3b = 6

⇒ b = -2.

Hence, a = 1 and b = -2.