Let A = [210−2], B =[41−3−2] and C =[−32−14],\begin{bmatrix}[r] 2 & 1 \ 0 & -2 \end{bmatrix}, \text{ B } = \begin{bmatrix}[r] 4 & 1 \ -3 & -2 \end{bmatrix} \text{ and C } = \begin{bmatrix}[r] -3 & 2 \ -1 & 4 \end{bmatrix},[201−2], B =[4−31−2] and C =[−3−124], find A2 + AC - 5B.
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A2=[210−2][210−2]=[2×2+1×02×1+1×(−2)0×2+(−2)×00×1+(−2)×(−2)]=[4+02−20+00+4]=[4004].AC=[210−2][−32−14]=[2×(−3)+1×(−1)2×2+1×40×(−3)+(−2)×(−1)0×2+(−2)×4]=[−6+(−1)4+40+20+(−8)]=[−782−8].5B=5[41−3−2]=[205−15−10]∴A2+AC−5B=[4004]+[−782−8]−[205−15−10]=[4+(−7)−200+8−50+2−(−15)4+(−8)−(−10)]=[−233176].A^2 = \begin{bmatrix}[r] 2 & 1 \ 0 & -2 \end{bmatrix} \begin{bmatrix}[r] 2 & 1 \ 0 & -2 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 2 \times 2 + 1 \times 0 & 2 \times 1 + 1 \times (-2) \ 0 \times 2 + (-2) \times 0 & 0 \times 1 + (-2) \times (-2) \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 4 + 0 & 2 - 2 \ 0 + 0 & 0 + 4 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 4 & 0 \ 0 & 4 \end{bmatrix}. \\[1.5em] AC = \begin{bmatrix}[r] 2 & 1 \ 0 & -2 \end{bmatrix} \begin{bmatrix}[r] -3 & 2 \ -1 & 4 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 2 \times (-3) + 1 \times (-1) & 2 \times 2 + 1 \times 4 \ 0 \times (-3) + (-2) \times (-1) & 0 \times 2 + (-2) \times 4 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] -6 + (-1) & 4 + 4 \ 0 + 2 & 0 + (-8) \end{bmatrix} \\[1em] = \begin{bmatrix}[r] -7 & 8 \ 2 & -8 \end{bmatrix}. \\[1.5em] 5B = 5 \begin{bmatrix}[r] 4 & 1 \ -3 & -2 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 20 & 5 \ -15 & -10 \end{bmatrix} \\[1.5em] \therefore A^2 + AC - 5B = \begin{bmatrix}[r] 4 & 0 \ 0 & 4 \end{bmatrix} + \begin{bmatrix}[r] -7 & 8 \ 2 & -8 \end{bmatrix} - \begin{bmatrix}[r] 20 & 5 \ -15 & -10 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 4 + (-7) - 20 & 0 + 8 - 5 \ 0 + 2 - (-15) & 4 + (-8) - (-10) \end{bmatrix} \\[1em] = \begin{bmatrix}[r] -23 & 3 \ 17 & 6 \end{bmatrix} .A2=[201−2][201−2]=[2×2+1×00×2+(−2)×02×1+1×(−2)0×1+(−2)×(−2)]=[4+00+02−20+4]=[4004].AC=[201−2][−3−124]=[2×(−3)+1×(−1)0×(−3)+(−2)×(−1)2×2+1×40×2+(−2)×4]=[−6+(−1)0+24+40+(−8)]=[−728−8].5B=5[4−31−2]=[20−155−10]∴A2+AC−5B=[4004]+[−728−8]−[20−155−10]=[4+(−7)−200+2−(−15)0+8−54+(−8)−(−10)]=[−231736].
Hence, the matrix A2 + AC - 5B = [−233176].\begin{bmatrix}[r] -23 & 3 \ 17 & 6 \end{bmatrix} .[−231736].
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