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(D) Given the matrix equation $AX = B$: $\begin{bmatrix} 1 & -1 & 2 \ 2 & 0 & 1 \ 3 & 2 & 1 \end{bmatrix} \begin{bmatrix} a \ b \ c \end{bmatrix}

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$4$

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(D) Given the matrix equation $AX = B$: $\begin{bmatrix} 1 & -1 & 2 \ 2 & 0 & 1 \ 3 & 2 & 1 \end{bmatrix} \begin{bmatrix} a \ b \ c \end{bmatrix} = \begin{bmatrix} 3 \ 1 \ 4 \end{bmatrix}$ This gives the system of linear equations: $a - b + 2c = 3$ $(i)$ $2a + c = 1$ $(ii)$ $3a + 2b + c = 4$ $(iii)$ From $(ii)$,we have $c = 1 - 2a$. Substitute $c$ into $(i)$: $a - b + 2(1 - 2a) = 3 \implies a - b + 2 - 4a = 3 \implies -3a - b = 1 \implies b = -3a - 1$. Substitute $a, b, c$ into $(iii)$: $3a + 2(-3a - 1) + (1 - 2a) = 4$ $3a - 6a - 2 + 1 - 2a = 4$ $-5a - 1 = 4$ $-5a = 5 \implies a = -1$. Now find $b$ and $c$: $b = -3(-1) - 1 = 3 - 1 = 2$. $c = 1 - 2(-1) = 1 + 2 = 3$. Finally,calculate $2a - 3b + 4c$: $2(-1) - 3(2) + 4(3) = -2 - 6 + 12 = 4$.

Let $X = \begin{bmatrix} a \\ b \\ c \end{bmatrix}$,$A = \begin{bmatrix} 1 & -1 & 2 \\ 2 & 0 & 1 \\ 3 & 2 & 1 \end{bmatrix}$,and $B = \begin{bmatrix} 3 \\ 1 \\ 4 \end{bmatrix}$. If $AX = B$,then the value of $2a - 3b + 4c$ is:

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