Let A = [1021] and B =[23−10],\begin{bmatrix} 1 & 0 \ 2 & 1 \end{bmatrix} \text{ and B } = \begin{bmatrix}[r] 2 & 3 \ -1 & 0 \end{bmatrix},[1201] and B =[2−130], find A2 + AB + B2.
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A2=[1021][1021]=[1×1+0×21×0+0×12×1+1×22×0+1×1]=[1+00+02+20+1]=[1041]AB=[1021][23−10]=[1×2+0×(−1)1×3+0×02×2+1×(−1)2×3+1×0]=[2+03+04−16+0]=[2336]B2=[23−10][23−10]=[2×2+3×(−1)2×3+3×0(−1)×2+0×(−1)(−1)×3+0×0]=[4−36+0−2+0−3+0]=[16−2−3].∴A2+AB+B2=[1041]+[2336]+[16−2−3]=[1+2+10+3+64+3+(−2)1+6+(−3)]=[4954].A^2 = \begin{bmatrix} 1 & 0 \ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \ 2 & 1 \end{bmatrix} \\[1em] = \begin{bmatrix} 1 \times 1 + 0 \times 2 & 1 \times 0 + 0 \times 1 \ 2 \times 1 + 1 \times 2 & 2 \times 0 + 1 \times 1 \end{bmatrix} \\[1em] = \begin{bmatrix} 1 + 0 & 0 + 0 \ 2 + 2 & 0 + 1 \end{bmatrix} \\[1em] = \begin{bmatrix} 1 & 0 \ 4 & 1 \end{bmatrix} \\[1em] AB = \begin{bmatrix} 1 & 0 \ 2 & 1 \end{bmatrix} \begin{bmatrix}[r] 2 & 3 \ -1 & 0 \end{bmatrix} \\[1em] = \begin{bmatrix} 1 \times 2 + 0 \times (-1) & 1 \times 3 + 0 \times 0 \ 2 \times 2 + 1 \times (-1) & 2 \times 3 + 1 \times 0 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 2 + 0 & 3 + 0 \ 4 - 1 & 6 + 0 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 2 & 3 \ 3 & 6 \end{bmatrix} \\[1em] B^2 = \begin{bmatrix}[r] 2 & 3 \ -1 & 0 \end{bmatrix} \begin{bmatrix}[r] 2 & 3 \ -1 & 0 \end{bmatrix} \\[1em] = \begin{bmatrix} 2 \times 2 + 3 \times (-1) & 2 \times 3 + 3 \times 0 \ (-1) \times 2 + 0 \times (-1) & (-1) \times 3 + 0 \times 0 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 4 - 3 & 6 + 0 \ -2 + 0 & -3 + 0 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 1 & 6 \ -2 & -3 \end{bmatrix}. \\[1em] \therefore A^2 + AB + B^2 = \begin{bmatrix} 1 & 0 \ 4 & 1 \end{bmatrix} + \begin{bmatrix}[r] 2 & 3 \ 3 & 6 \end{bmatrix} + \begin{bmatrix}[r] 1 & 6 \ -2 & -3 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 1 + 2 + 1 & 0 + 3 + 6 \ 4 + 3 + (-2) & 1 + 6 + (-3) \end{bmatrix} \\[1em] = \begin{bmatrix} 4 & 9 \ 5 & 4 \end{bmatrix}.A2=[1201][1201]=[1×1+0×22×1+1×21×0+0×12×0+1×1]=[1+02+20+00+1]=[1401]AB=[1201][2−130]=[1×2+0×(−1)2×2+1×(−1)1×3+0×02×3+1×0]=[2+04−13+06+0]=[2336]B2=[2−130][2−130]=[2×2+3×(−1)(−1)×2+0×(−1)2×3+3×0(−1)×3+0×0]=[4−3−2+06+0−3+0]=[1−26−3].∴A2+AB+B2=[1401]+[2336]+[1−26−3]=[1+2+14+3+(−2)0+3+61+6+(−3)]=[4594].
Hence, the matrix A2 + AB + B2 = [4954]\begin{bmatrix} 4 & 9 \ 5 & 4 \end{bmatrix}[4594].
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If A = [1234], B =[2142] and C =[5174],\begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix}, \text{ B } = \begin{bmatrix} 2 & 1 \ 4 & 2 \end{bmatrix} \text{ and C } = \begin{bmatrix} 5 & 1 \ 7 & 4 \end{bmatrix},[1324], B =[2412] and C =[5714], compute
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If A = [1223], B =[2132] and C =[1331],\begin{bmatrix} 1 & 2 \ 2 & 3 \end{bmatrix}, \text{ B } = \begin{bmatrix} 2 & 1 \ 3 & 2 \end{bmatrix} \text{ and C } = \begin{bmatrix} 1 & 3 \ 3 & 1 \end{bmatrix},[1223], B =[2312] and C =[1331], find the matrix C(B - A).
If A = [3051] and B[−4210]\begin{bmatrix}[r] 3 & 0 \ 5 & 1 \end{bmatrix} \text{ and B} \begin{bmatrix}[r] -4 & 2 \ 1 & 0 \end{bmatrix}[3501] and B[−4120], find A2 - 2AB + B2.
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.
