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(D) Given $A = \begin{bmatrix} 2 & -2 & -4 \ -1 & 3 & 4 \ 1 & -2 & x \end{bmatrix}$ and $A^2 = A$.Calculating $A^2$:$A^2 = \begin{bmatrix} 2 & -2 &

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(D) Given $A = \begin{bmatrix} 2 & -2 & -4 \ -1 & 3 & 4 \ 1 & -2 & x \end{bmatrix}$ and $A^2 = A$.Calculating $A^2$:$A^2 = \begin{bmatrix} 2 & -2 & -4 \ -1 & 3 & 4 \ 1 & -2 & x \end{bmatrix} \begin{bmatrix} 2 & -2 & -4 \ -1 & 3 & 4 \ 1 & -2 & x \end{bmatrix} = \begin{bmatrix} 4+2-4 & -4-6+8 & -8-8-4x \ -2-3+4 & 2+9-8 & 4+12+4x \ 2+2+x & -2-6-2x & -4-12+x^2 \end{bmatrix} = \begin{bmatrix} 2 & -2 & -16-4x \ -1 & 3 & 16+4x \ 4+x & -8-2x & -16+x^2 \end{bmatrix}$.Since $A^2 = A$,comparing the elements,we have $4+x = 1$,which gives $x = -3$.Substituting $x = -3$ into $A$:$A = \begin{bmatrix} 2 & -2 & -4 \ -1 & 3 & 4 \ 1 & -2 & -3 \end{bmatrix}$.The determinant $|A| = 2(-9+8) + 2(3-4) - 4(2-3) = 2(-1) + 2(-1) - 4(-1) = -2 - 2 + 4 = 0$.Since $|A| = 0$,the rank $r Checking the minor of order $2$: $\begin{vmatrix} 2 & -2 \ -1 & 3 \end{vmatrix} = 6 - 2 = 4 
eq 0$.Thus,the rank $r = 2$.Therefore,$r + x = 2 + (-3) = -1$.

Let $A = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & x \end{bmatrix}$ and $A^2 = A$. If $r$ is the rank of $A$,then $r + x =$

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