Matrix A = $\begin{bmatrix$}[r] 6 & 9 \ -4 & k \end{bmatrix} \text{ such that } A^2 = \begin{bmatrix}[r] 0 & 0 \ 0 & 0 \end{bmatrix}. Then k is :

1. 6

2. -6

3. 36

4. ±6

Given, $A^2 = \begin{bmatrix$}[r] 0 & 0 \ 0 & 0 \end{bmatrix}.

$\therefore \begin{bmatrix$}[r] 6 & 9 \ -4 & k \end{bmatrix}\begin{bmatrix}[r] 6 & 9 \ -4 & k \end{bmatrix} = \begin{bmatrix}[r] 0 & 0 \ 0 & 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 6 \times 6 + 9 \times (-4) & 6 \times 9 + 9 \times k \ -4 \times 6 + k \times (-4) & (-4) \times 9 + k \times k \end{bmatrix} = \begin{bmatrix}[r] 0 & 0 \ 0 & 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 36 + (-36) & 54 + 9k \ -24 - 4k & -36 + k^2 \end{bmatrix} = \begin{bmatrix}[r] 0 & 0 \ 0 & 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 0 & 54 + 9k \ -24 - 4k & -36 + k^2 \end{bmatrix} = \begin{bmatrix}[r] 0 & 0 \ 0 & 0 \end{bmatrix} \\[1em] \Rightarrow 54 + 9k = 0 \\[1em] \Rightarrow 9k = -54 \\[1em] \Rightarrow k = -\dfrac{54}{9} = -6.

Hence, Option 2 is the correct option.