Matrix A = [69−4k] such that A2=[0000]\begin{bmatrix*}[r] 6 & 9 \\ -4 & k \end{bmatrix*} \text{ such that } A^2 = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*}[6−49k] such that A2=[0000]. Then k is :
6
-6
36
±6
Given, A2=[0000]A^2 = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*}A2=[0000].
∴[69−4k][69−4k]=[0000]⇒[6×6+9×(−4)6×9+9×k−4×6+k×(−4)(−4)×9+k×k]=[0000]⇒[36+(−36)54+9k−24−4k−36+k2]=[0000]⇒[054+9k−24−4k−36+k2]=[0000]⇒54+9k=0⇒9k=−54⇒k=−549=−6.\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.∴[6−49k][6−49k]=[0000]⇒[6×6+9×(−4)−4×6+k×(−4)6×9+9×k(−4)×9+k×k]=[0000]⇒[36+(−36)−24−4k54+9k−36+k2]=[0000]⇒[0−24−4k54+9k−36+k2]=[0000]⇒54+9k=0⇒9k=−54⇒k=−954=−6.
Hence, Option 2 is the correct option.
