If begin{bmatrix} a_1+a_2 & 4 3 & a_3+a_4 en | vedclass

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(D) Given the matrix equation: $\begin{bmatrix} a_1+a_2 & 4 \ 3 & a_3+a_4 \end{bmatrix} - \begin{bmatrix} 3a_2 & 3a_1 \ 3a_4 & 3a_3 \end{bmatrix} =

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$16$

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(D) Given the matrix equation: $\begin{bmatrix} a_1+a_2 & 4 \ 3 & a_3+a_4 \end{bmatrix} - \begin{bmatrix} 3a_2 & 3a_1 \ 3a_4 & 3a_3 \end{bmatrix} = \begin{bmatrix} -6 & -a_1 \ -2a_4 & 1 \end{bmatrix}$.Performing the subtraction,we get: $\begin{bmatrix} a_1-2a_2 & 4-3a_1 \ 3-3a_4 & a_3-2a_4 \end{bmatrix} = \begin{bmatrix} -6 & -a_1 \ -2a_4 & 1 \end{bmatrix}$.Comparing the corresponding elements:$1) \; a_1 - 2a_2 = -6$$2) \; 4 - 3a_1 = -a_1 \implies 4 = 2a_1 \implies a_1 = 2$$3) \; 3 - 3a_4 = -2a_4 \implies a_4 = 3$$4) \; a_3 - 2a_4 = 1 \implies a_3 - 2(3) = 1 \implies a_3 = 7$From equation $(1)$,substitute $a_1 = 2$: $2 - 2a_2 = -6 \implies -2a_2 = -8 \implies a_2 = 4$.Now,calculate the sum $\sum_{i=1}^4 a_i = a_1 + a_2 + a_3 + a_4 = 2 + 4 + 7 + 3 = 16$.

If $\begin{bmatrix} a_1+a_2 & 4 \\ 3 & a_3+a_4 \end{bmatrix} - \begin{bmatrix} 3a_2 & 3a_1 \\ 3a_4 & 3a_3 \end{bmatrix} = \begin{bmatrix} -6 & -a_1 \\ -2a_4 & 1 \end{bmatrix}$,then $\sum_{i=1}^4 a_i = $ . . . . . .

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