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(D) Given the equation $2A + 3B - 5C = 0$.Rearranging for $C$,we get $5C = 2A + 3B$.Substitute the matrices $A$ and $B$:$5C = 2\left[\begin{array}{ccc

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$\left[\begin{array}{cccc}2 & 1 & 6/5 & 7/5 \ 1 & -7/5 & 2 & 3/5\end{array}ight]$

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(D) Given the equation $2A + 3B - 5C = 0$.Rearranging for $C$,we get $5C = 2A + 3B$.Substitute the matrices $A$ and $B$:$5C = 2\left[\begin{array}{cccc}2 & 1 & 3 & -1 \ 1 & -2 & 2 & -3\end{array}ight] + 3\left[\begin{array}{cccc}2 & 1 & 0 & 3 \ 1 & -1 & 2 & 3\end{array}ight]$$5C = \left[\begin{array}{cccc}4 & 2 & 6 & -2 \ 2 & -4 & 4 & -6\end{array}ight] + \left[\begin{array}{cccc}6 & 3 & 0 & 9 \ 3 & -3 & 6 & 9\end{array}ight]$$5C = \left[\begin{array}{cccc}4+6 & 2+3 & 6+0 & -2+9 \ 2+3 & -4-3 & 4+6 & -6+9\end{array}ight]$$5C = \left[\begin{array}{cccc}10 & 5 & 6 & 7 \ 5 & -7 & 10 & 3\end{array}ight]$Dividing by $5$,we get $C = \left[\begin{array}{cccc}10/5 & 5/5 & 6/5 & 7/5 \ 5/5 & -7/5 & 10/5 & 3/5\end{array}ight] = \left[\begin{array}{cccc}2 & 1 & 6/5 & 7/5 \ 1 & -7/5 & 2 & 3/5\end{array}ight]$.

If $A=\left[\begin{array}{cccc}2 & 1 & 3 & -1 \\ 1 & -2 & 2 & -3\end{array}\right]$,$B=\left[\begin{array}{cccc}2 & 1 & 0 & 3 \\ 1 & -1 & 2 & 3\end{array}\right]$ and $2A+3B-5C=0$,then $C=$

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