If A = begin{bmatrix} 8 & 0 4 & -2 3 & 6 en | vedclass

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(D) Given the equation $2A + 3X = 5B$. Subtract $2A$ from both sides: $3X = 5B - 2A$. Thus,$X = \frac{1}{3}(5B - 2A)$. Substitute the matrices $A$ and

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$\begin{bmatrix} -2 & -\frac{10}{3} \ 4 & \frac{14}{3} \ -\frac{31}{3} & -\frac{7}{3} \end{bmatrix}$

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(D) Given the equation $2A + 3X = 5B$. Subtract $2A$ from both sides: $3X = 5B - 2A$. Thus,$X = \frac{1}{3}(5B - 2A)$. Substitute the matrices $A$ and $B$: $X = \frac{1}{3} \left( 5 \begin{bmatrix} 2 & -2 \ 4 & 2 \ -5 & 1 \end{bmatrix} - 2 \begin{bmatrix} 8 & 0 \ 4 & -2 \ 3 & 6 \end{bmatrix} ight)$ $X = \frac{1}{3} \left( \begin{bmatrix} 10 & -10 \ 20 & 10 \ -25 & 5 \end{bmatrix} - \begin{bmatrix} 16 & 0 \ 8 & -4 \ 6 & 12 \end{bmatrix} ight)$ $X = \frac{1}{3} \begin{bmatrix} 10-16 & -10-0 \ 20-8 & 10-(-4) \ -25-6 & 5-12 \end{bmatrix}$ $X = \frac{1}{3} \begin{bmatrix} -6 & -10 \ 12 & 14 \ -31 & -7 \end{bmatrix}$ $X = \begin{bmatrix} -2 & -\frac{10}{3} \ 4 & \frac{14}{3} \ -\frac{31}{3} & -\frac{7}{3} \end{bmatrix}$

If $A = \begin{bmatrix} 8 & 0 \\ 4 & -2 \\ 3 & 6 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & -2 \\ 4 & 2 \\ -5 & 1 \end{bmatrix}$,then find the matrix $X$ such that $2A + 3X = 5B$.

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