If A = $\begin{bmatrix} 3 & x \ 0 & 1 \end{bmatrix}$ and B = $\begin{bmatrix} 9 & -16 \ 0 & -y \end{bmatrix}$, find x and y when A² = B.

Given,

A = $\begin{bmatrix} 3 & x \ 0 & 1 \end{bmatrix}$ and B = $\begin{bmatrix} 9 & -16 \ 0 & -y \end{bmatrix}$

Solving A²:

$\Rightarrow A^2 = \begin{bmatrix} 3 & x \ 0 & 1 \end{bmatrix} \times \begin{bmatrix} 3 & x \ 0 & 1 \end{bmatrix} \\[1em] = \begin{bmatrix} (3)(3) + (x)(0) & (3)(x) + (x)(1) \ (0)(3) + (1)(0) & (0)(x) + (1)(1) \end{bmatrix} \\[1em] = \begin{bmatrix} 9 + 0 & 3x + x \ 0 + 0 & 0 + 1 \end{bmatrix} \\[1em] = \begin{bmatrix} 9 & 4x \ 0 & 1 \end{bmatrix}.$

Solving for x and y:

A² = B

$\Rightarrow \begin{bmatrix} 9 & 4x \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 9 & -16 \ 0 & -y \end{bmatrix}$

∴ 4x = -16

⇒ x = $\dfrac{-16}{4}$

⇒ x = -4.

∴ -y = 1

⇒ y = -1.

Hence, x = -4 and y = -1.