If A = begin{bmatrix} 1 & -2 3 & 0 end{bmatr | vedclass

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(B) Given $A = \begin{bmatrix} 1 & -2 \ 3 & 0 \end{bmatrix}$,$B = \begin{bmatrix} -1 & 4 \ 2 & 3 \end{bmatrix}$,and $C = \begin{bmatrix} 0 & -1 \ 1

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$\begin{bmatrix} 8 & -20 \ 7 & -9 \end{bmatrix}$

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(B) Given $A = \begin{bmatrix} 1 & -2 \ 3 & 0 \end{bmatrix}$,$B = \begin{bmatrix} -1 & 4 \ 2 & 3 \end{bmatrix}$,and $C = \begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix}$.We need to calculate $5A - 3B - 2C$.First,calculate the scalar multiplications:$5A = 5 \begin{bmatrix} 1 & -2 \ 3 & 0 \end{bmatrix} = \begin{bmatrix} 5 & -10 \ 15 & 0 \end{bmatrix}$$3B = 3 \begin{bmatrix} -1 & 4 \ 2 & 3 \end{bmatrix} = \begin{bmatrix} -3 & 12 \ 6 & 9 \end{bmatrix}$$2C = 2 \begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -2 \ 2 & 0 \end{bmatrix}$Now,perform the matrix subtraction:$5A - 3B - 2C = \begin{bmatrix} 5 & -10 \ 15 & 0 \end{bmatrix} - \begin{bmatrix} -3 & 12 \ 6 & 9 \end{bmatrix} - \begin{bmatrix} 0 & -2 \ 2 & 0 \end{bmatrix}$$= \begin{bmatrix} 5 - (-3) - 0 & -10 - 12 - (-2) \ 15 - 6 - 2 & 0 - 9 - 0 \end{bmatrix}$$= \begin{bmatrix} 5 + 3 & -22 + 2 \ 7 & -9 \end{bmatrix}$$= \begin{bmatrix} 8 & -20 \ 7 & -9 \end{bmatrix}$.

If $A = \begin{bmatrix} 1 & -2 \\ 3 & 0 \end{bmatrix}$,$B = \begin{bmatrix} -1 & 4 \\ 2 & 3 \end{bmatrix}$,$C = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$,then $5A - 3B - 2C = $

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