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(B) Given $B = A^{-1}$,therefore $BA = I$.$\begin{bmatrix} -2 & 0 & b \ 7 & -1 & -2 \ c & 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \ 1 & a & 3

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$14$

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(B) Given $B = A^{-1}$,therefore $BA = I$.$\begin{bmatrix} -2 & 0 & b \ 7 & -1 & -2 \ c & 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \ 1 & a & 3 \ 3 & 2 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$Multiplying the matrices on the left side:Row $1$,Column $1$: $(-2)(1) + (0)(1) + (b)(3) = -2 + 3b = 1 \Rightarrow 3b = 3 \Rightarrow b = 1$.Row $2$,Column $2$: $(7)(1) + (-1)(a) + (-2)(2) = 7 - a - 4 = 3 - a = 1 \Rightarrow a = 2$.Row $3$,Column $3$: $(c)(1) + (1)(3) + (1)(2) = c + 3 + 2 = c + 5 = 1 \Rightarrow c = -4$.Now,calculate the value of $4a + 2b - c$:$4(2) + 2(1) - (-4) = 8 + 2 + 4 = 14$.

If $A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & a & 3 \\ 3 & 2 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} -2 & 0 & b \\ 7 & -1 & -2 \\ c & 1 & 1 \end{bmatrix}$ and if matrix $B$ is the inverse of matrix $A$,then the value of $4a + 2b - c$ is:

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