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(D) Given $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$.Since $A$ satisfies $x^2 + k = 0$,we have $A^2 + kI = O$,where $I$ is the identity matrix

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both $(A)$ and $(C)$

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(D) Given $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$.Since $A$ satisfies $x^2 + k = 0$,we have $A^2 + kI = O$,where $I$ is the identity matrix and $O$ is the zero matrix.$A^2 = \begin{bmatrix} a & b \ c & d \end{bmatrix} \begin{bmatrix} a & b \ c & d \end{bmatrix} = \begin{bmatrix} a^2 + bc & ab + bd \ ac + cd & bc + d^2 \end{bmatrix}$.Substituting into $A^2 + kI = O$:$\begin{bmatrix} a^2 + bc + k & b(a + d) \ c(a + d) & bc + d^2 + k \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix}$.Comparing elements,we get $b(a + d) = 0$ and $c(a + d) = 0$. Since $bc 
eq 0$,we must have $a + d = 0$,which implies $a = -d$.From the diagonal elements,$a^2 + bc + k = 0$ and $d^2 + bc + k = 0$.Since $a = -d$,$a^2 = d^2$. Thus,$k = -(a^2 + bc)$.The determinant $|A| = ad - bc$. Since $a = -d$,$|A| = (-d)d - bc = -(d^2 + bc)$.Therefore,$k = -(-(d^2 + bc)) = |A|$.Thus,both $a + d = 0$ and $k = |A|$ are correct.

If $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ (where $bc \neq 0$) satisfies the equation $x^2 + k = 0$,then:

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