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Mathematics

If A = [2357], B =[04−17] and C =[10−14],\begin{bmatrix}[r] 2 & 3 \ 5 & 7 \end{bmatrix}, \text{ B } = \begin{bmatrix}[r] 0 & 4 \ -1 & 7 \end{bmatrix} \text{ and C } = \begin{bmatrix}[r] 1 & 0 \ -1 & 4 \end{bmatrix}, find AC + B2 - 10C.

Matrices

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Answer

AC=[2357][10−14]=[2×1+3×(−1)2×0+3×45×1+7×(−1)5×0+7×4]=[2−30+125−70+28]=[−112−228].AC = \begin{bmatrix}[r] 2 & 3 \ 5 & 7 \end{bmatrix} \begin{bmatrix}[r] 1 & 0 \ -1 & 4 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 2 \times 1 + 3 \times (-1) & 2 \times 0 + 3 \times 4 \ 5 \times 1 + 7 \times (-1) & 5 \times 0 + 7 \times 4 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 2 - 3 & 0 + 12 \ 5 - 7 & 0 + 28 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] -1 & 12 \ -2 & 28 \end{bmatrix}.

B2=[04−17][04−17]=[0×0+4×(−1)0×4+4×7(−1)×0+7×(−1)(−1)×4+7×7]=[0−40+280−7−4+49]=[−428−745].B^2 = \begin{bmatrix}[r] 0 & 4 \ -1 & 7 \end{bmatrix} \begin{bmatrix}[r] 0 & 4 \ -1 & 7 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 0 \times 0 + 4 \times (-1) & 0 \times 4 + 4 \times 7 \ (-1) \times 0 + 7 \times (-1) & (-1) \times 4 + 7 \times 7 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 0 - 4 & 0 + 28 \ 0 - 7 & -4 + 49 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] -4 & 28 \ -7 & 45 \end{bmatrix}.

10C=10[10−14]=[100−1040].∴AC+B2−10C=[−112−228]+[−428−745]−[100−1040]=[−1+(−4)−1012+28−0−2+(−7)−(−10)28+45−40]=[−1540133]10C = 10 \begin{bmatrix}[r] 1 & 0 \ -1 & 4 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 10 & 0 \ -10 & 40 \end{bmatrix}. \\[1em] \therefore AC + B^2 - 10C = \begin{bmatrix}[r] -1 & 12 \ -2 & 28 \end{bmatrix} + \begin{bmatrix}[r] -4 & 28 \ -7 & 45 \end{bmatrix} - \begin{bmatrix}[r] 10 & 0 \ -10 & 40 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] -1 + (-4) - 10 & 12 + 28 - 0 \ -2 + (-7) - (-10) & 28 + 45 - 40 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] -15 & 40 \ 1 & 33 \end{bmatrix}

Hence, the matrix AC + B2 - 10C = [−1540133]\begin{bmatrix}[r] -15 & 40 \ 1 & 33 \end{bmatrix}.

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