If X = begin{bmatrix} x y z end{bmatrix} is | vedclass

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(D) We know that $X = A^{-1}B = \frac{	ext{adj } A}{|A|} B$.First, we find the determinant $|A|$. We know that $|	ext{adj } A| = |A|^{n-1}$, where $

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$2$

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(D) We know that $X = A^{-1}B = \frac{	ext{adj } A}{|A|} B$.First, we find the determinant $|A|$. We know that $|	ext{adj } A| = |A|^{n-1}$, where $n=3$.$|	ext{adj } A| = 4(0 - (-10)) - 2(-15 - 5) + 2(10 - 0) = 4(10) - 2(-20) + 2(10) = 40 + 40 + 20 = 100$.So, $|A|^2 = 100$, which implies $|A| = \pm 10$.Now, $X = \pm \frac{1}{10} \begin{bmatrix} 4 & 2 & 2 \ -5 & 0 & 5 \ 1 & -2 & 3 \end{bmatrix} \begin{bmatrix} 4 \ 0 \ 2 \end{bmatrix} = \pm \frac{1}{10} \begin{bmatrix} 16 + 0 + 4 \ -20 + 0 + 10 \ 4 + 0 + 6 \end{bmatrix} = \pm \frac{1}{10} \begin{bmatrix} 20 \ -10 \ 10 \end{bmatrix} = \pm \begin{bmatrix} 2 \ -1 \ 1 \end{bmatrix}$.Thus, $x = \pm 2, y = \mp 1, z = \pm 1$.Then $|x + y + z| = |\pm(2 - 1 + 1)| = |\pm 2| = 2$.

If $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ is a solution of the system of equations $AX = B$, where $\text{adj } A = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$, then $|x + y + z|$ is equal to:

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