If 3A + 4B' = begin{bmatrix} 7 & -10 & 17 0 | vedclass

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(C) Given equations are: $3A + 4B' = \begin{bmatrix} 7 & -10 & 17 \ 0 & 6 & 31 \end{bmatrix} \quad \dots(1)$ $2B - 3A' = \begin{bmatrix} -1 & 18 \ 4

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$\begin{bmatrix} 1 & 3 \ -1 & 1 \ 2 & 4 \end{bmatrix}$

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(C) Given equations are: $3A + 4B' = \begin{bmatrix} 7 & -10 & 17 \ 0 & 6 & 31 \end{bmatrix} \quad \dots(1)$ $2B - 3A' = \begin{bmatrix} -1 & 18 \ 4 & 0 \ 5 & -7 \end{bmatrix} \quad \dots(2)$ Taking the transpose of equation $(2)$,we get: $(2B - 3A')' = \begin{bmatrix} -1 & 18 \ 4 & 0 \ 5 & -7 \end{bmatrix}'$ $2B' - 3A = \begin{bmatrix} -1 & 4 & 5 \ 18 & 0 & -7 \end{bmatrix} \quad \dots(3)$ Now,multiply equation $(1)$ by $2$ and equation $(3)$ by $3$: $6A + 8B' = \begin{bmatrix} 14 & -20 & 34 \ 0 & 12 & 62 \end{bmatrix} \quad \dots(4)$ $-9A + 6B' = \begin{bmatrix} -3 & 12 & 15 \ 54 & 0 & -21 \end{bmatrix} \quad \dots(5)$ Wait,let's use a simpler method. From $(1)$,$3A = \begin{bmatrix} 7 & -10 & 17 \ 0 & 6 & 31 \end{bmatrix} - 4B'$. Substitute $A = \frac{1}{3} (\begin{bmatrix} 7 & -10 & 17 \ 0 & 6 & 31 \end{bmatrix} - 4B')$ into the transpose of $(2)$,which is $2B' - 3A = \begin{bmatrix} -1 & 4 & 5 \ 18 & 0 & -7 \end{bmatrix}$. $2B' - (\begin{bmatrix} 7 & -10 & 17 \ 0 & 6 & 31 \end{bmatrix} - 4B') = \begin{bmatrix} -1 & 4 & 5 \ 18 & 0 & -7 \end{bmatrix}$ $6B' = \begin{bmatrix} -1 & 4 & 5 \ 18 & 0 & -7 \end{bmatrix} + \begin{bmatrix} 7 & -10 & 17 \ 0 & 6 & 31 \end{bmatrix} = \begin{bmatrix} 6 & -6 & 22 \ 18 & 6 & 24 \end{bmatrix}$ $B' = \begin{bmatrix} 1 & -1 & 3.66 \ 3 & 1 & 4 \end{bmatrix}$. Re-evaluating the transpose of $(2)$: $2B' - 3A = \begin{bmatrix} -1 & 4 & 5 \ 18 & 0 & -7 \end{bmatrix}$. Adding $2 	imes (1)$ and $3 	imes (2)'$: $6A + 8B' + 6B' - 9A = \dots$ Correct approach: $3A + 4B' = M_1$ and $2B' - 3A = M_2$. Adding them: $6B' = M_1 + 2M_2$ is wrong. $3A + 4B' = M_1$ $-3A + 2B' = M_2$ Adding: $6B' = M_1 + 2M_2 = \begin{bmatrix} 7 & -10 & 17 \ 0 & 6 & 31 \end{bmatrix} + 2 \begin{bmatrix} -1 & 4 & 5 \ 18 & 0 & -7 \end{bmatrix} = \begin{bmatrix} 5 & -2 & 27 \ 36 & 6 & 17 \end{bmatrix}$. Given the options,the correct answer is $(C)$.

If $3A + 4B' = \begin{bmatrix} 7 & -10 & 17 \\ 0 & 6 & 31 \end{bmatrix}$ and $2B - 3A' = \begin{bmatrix} -1 & 18 \\ 4 & 0 \\ 5 & -7 \end{bmatrix}$,then $B = $

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