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Mathematics

Let A = [1021],B=[23−10].\begin{bmatrix}[r] 1 & 0 \ 2 & 1 \end{bmatrix}, B = \begin{bmatrix}[r] 2 & 3 \ -1 & 0 \end{bmatrix}. Find A2 + AB + B2.

Matrices

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Answer

A2=[1021][1021]=[1×1+0×21×0+0×12×1+1×22×0+1×1]=[1+00+02+20+1]=[1041]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].A^2 = \begin{bmatrix}[r] 1 & 0 \ 2 & 1 \end{bmatrix}\begin{bmatrix}[r] 1 & 0 \ 2 & 1 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 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}[r] 1 + 0 & 0 + 0 \ 2 + 2 & 0 + 1 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 1 & 0 \ 4 & 1 \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}[r] 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}.

Substituting value of A2 and B2 in A2 + AB + B2 we get,

A2+AB+B2=[1041]+[1021][23−10]+[16−2−3]=[1041]+[1×2+0×(−1)1×3+0×02×2+1×(−1)2×3+1×0]+[16−2−3]=[1041]+[2+03+04−16+0]+[16−2−3]=[1041]+[2336]+[16−2−3]=[1+2+10+3+64+3+(−2)1+6+(−3)]=[4954].A^2 + AB + B^2 = \begin{bmatrix}[r] 1 & 0 \ 4 & 1 \end{bmatrix} + \begin{bmatrix}[r] 1 & 0 \ 2 & 1 \end{bmatrix}\begin{bmatrix}[r] 2 & 3 \ -1 & 0 \end{bmatrix} + \begin{bmatrix}[r] 1 & 6 \ -2 & -3 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 1 & 0 \ 4 & 1 \end{bmatrix} + \begin{bmatrix}[r] 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} + \begin{bmatrix}[r] 1 & 6 \ -2 & -3 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 1 & 0 \ 4 & 1 \end{bmatrix} + \begin{bmatrix}[r] 2 + 0 & 3 + 0 \ 4 - 1 & 6 + 0 \end{bmatrix} + \begin{bmatrix}[r] 1 & 6 \ -2 & -3 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 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}[r] 4 & 9 \ 5 & 4 \end{bmatrix}.

Hence, A2 + AB + B2 = [4954].\begin{bmatrix}[r] 4 & 9 \ 5 & 4 \end{bmatrix}.

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