If $A = \begin{bmatrix$}[r] 3 & 3 \ p & q \end{bmatrix} \text{ and } A^2 = O, find p and q.

Given, A² = O.

$\Rightarrow \begin{bmatrix$}[r] 3 & 3 \ p & q \end{bmatrix} \begin{bmatrix}[r] 3 & 3 \ p & q \end{bmatrix} = \begin{bmatrix}[r] 0 & 0 \ 0 & 0 \end{bmatrix} \\[0.5em] \Rightarrow \begin{bmatrix}[r] 3 \times 3 + 3 \times p & 3 \times 3 + 3 \times q \ p \times 3 + q \times p & p \times 3 + q \times q \end{bmatrix} = \begin{bmatrix}[r] 0 & 0 \ 0 & 0 \end{bmatrix} \\[0.5em] \Rightarrow \begin{bmatrix}[r] 9 + 3p & 9 + 3q \ 3p + qp & 3p + q^2 \end{bmatrix} = \begin{bmatrix}[r] 0 & 0 \ 0 & 0 \end{bmatrix} \\[0.5em]

By definition of equality of matrices we get,

⇒ 9 + 3p = 0 or 3p = -9 or p = -3

⇒ 9 + 3q = 0 or 3q = -9 or q = -3

⇒ 3p + qp = 0     (Eq 1)

⇒ 3p + q² = 0     (Eq 2)

Checking whether p = -3 and q = -3 satisfy Eq 1,

⇒ 3p + qp = 0

L.H.S. = 3(-3) + (-3)(-3) = -9 + 9 = 0 = R.H.S.

Checking whether p = -3 and q = -3 satisfy Eq 2,

⇒ 3p + q² = 0

L.H.S. = 3(-3) + (-3)² = -9 + 9 = 0 = R.H.S.

Since, p = -3 and q = -3 satisfies Eq 1 and Eq 2,

∴ p = -3 and q = -3.

Hence, the values are p = -3 and q = -3.