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(D) We are given the matrix equation $AX = B$, where $A = \begin{bmatrix} 1 & -1 & 0 \ 0 & 1 & -1 \ 1 & 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix

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$\begin{bmatrix} 4 \ 2 \ 1 \end{bmatrix}$

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(D) We are given the matrix equation $AX = B$, where $A = \begin{bmatrix} 1 & -1 & 0 \ 0 & 1 & -1 \ 1 & 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 2 \ 1 \ 7 \end{bmatrix}$.We can verify the options by performing matrix multiplication $AX$ for each option.Let $X = \begin{bmatrix} x \ y \ z \end{bmatrix}$.Testing option $B$: $X = \begin{bmatrix} 1 \ 2 \ 4 \end{bmatrix}$.$AX = \begin{bmatrix} 1 & -1 & 0 \ 0 & 1 & -1 \ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} 1 \ 2 \ 4 \end{bmatrix} = \begin{bmatrix} (1)(1) + (-1)(2) + (0)(4) \ (0)(1) + (1)(2) + (-1)(4) \ (1)(1) + (1)(2) + (1)(4) \end{bmatrix} = \begin{bmatrix} 1 - 2 + 0 \ 0 + 2 - 4 \ 1 + 2 + 4 \end{bmatrix} = \begin{bmatrix} -1 \ -2 \ 7 \end{bmatrix} 
eq B$.Testing option $D$: $X = \begin{bmatrix} 4 \ 2 \ 1 \end{bmatrix}$.$AX = \begin{bmatrix} 1 & -1 & 0 \ 0 & 1 & -1 \ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} 4 \ 2 \ 1 \end{bmatrix} = \begin{bmatrix} (1)(4) + (-1)(2) + (0)(1) \ (0)(4) + (1)(2) + (-1)(1) \ (1)(4) + (1)(2) + (1)(1) \end{bmatrix} = \begin{bmatrix} 4 - 2 + 0 \ 0 + 2 - 1 \ 4 + 2 + 1 \end{bmatrix} = \begin{bmatrix} 2 \ 1 \ 7 \end{bmatrix} = B$.Thus, the correct matrix is $X = \begin{bmatrix} 4 \ 2 \ 1 \end{bmatrix}$.

Let $A = \begin{bmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 1 & 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 2 \\ 1 \\ 7 \end{bmatrix}$. For the equation $AX = B$, find the matrix $X$.

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