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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}, find A2 - 2AB + B2.

Matrices

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Answer

Given, 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}

⇒A2=[3051]×[3051]=[3×3+0×53×0+0×15×3+1×55×0+1×1]=[9+00+015+50+1]=[90201]⇒2AB=2×[3051]×[−4210]=[60102]×[−4210]=[6×(−4)+0×16×2+0×010×(−4)+2×110×2+2×0]=[−24+012+0−40+220+0]=[−2412−3820]⇒B2=[−4210]×[−4210]=[−4×(−4)+2×1−4×2+2×01×(−4)+0×11×2+0×0]=[16+2−8+0−4+02+0]=[18−8−42]\Rightarrow A^2 =\begin{bmatrix}[r] 3 & 0 \ 5 & 1 \end{bmatrix} \times \begin{bmatrix}[r] 3 & 0 \ 5 & 1 \end{bmatrix}\\[1em] = \begin{bmatrix}[r] 3 \times 3 + 0 \times 5 & 3 \times 0 + 0 \times 1 \ 5 \times 3 + 1 \times 5 & 5 \times 0 + 1 \times 1 \end{bmatrix}\\[1em] = \begin{bmatrix}[r] 9 + 0 & 0 + 0 \ 15 + 5 & 0 + 1 \end{bmatrix}\\[1em] = \begin{bmatrix}[r] 9 & 0 \ 20 & 1 \end{bmatrix}\\[1em] \Rightarrow \text{2AB} = 2 \times \begin{bmatrix}[r] 3 & 0 \ 5 & 1 \end{bmatrix} \times \begin{bmatrix}[r] -4 & 2 \ 1 & 0 \end{bmatrix}\\[1em] = \begin{bmatrix}[r] 6 & 0 \ 10 & 2 \end{bmatrix} \times \begin{bmatrix}[r] -4 & 2 \ 1 & 0 \end{bmatrix}\\[1em] = \begin{bmatrix}[r] 6 \times (-4) + 0 \times 1 & 6 \times 2 + 0 \times 0 \ 10 \times (-4) + 2 \times 1 & 10 \times 2 + 2 \times 0 \end{bmatrix}\\[1em] = \begin{bmatrix}[r] -24 + 0 & 12 + 0 \ -40 + 2 & 20 + 0 \end{bmatrix}\\[1em] = \begin{bmatrix}[r] -24 & 12 \ -38 & 20 \end{bmatrix}\\[1em] \Rightarrow \text{B}^2 = \begin{bmatrix}[r] -4 & 2 \ 1 & 0 \end{bmatrix} \times \begin{bmatrix}[r] -4 & 2 \ 1 & 0 \end{bmatrix}\\[1em] = \begin{bmatrix}[r] -4 \times (-4) + 2 \times 1 & -4 \times 2 + 2 \times 0 \ 1 \times (-4) + 0 \times 1 & 1 \times 2 + 0 \times 0 \end{bmatrix}\\[1em] = \begin{bmatrix}[r] 16 + 2 & -8 + 0 \ -4 + 0 & 2 + 0 \end{bmatrix}\\[1em] = \begin{bmatrix}[r] 18 & -8 \ -4 & 2 \end{bmatrix}

Substituting values in A2 - 2AB + B2, we get :

A2−2AB+B2=[90201]−[−2412−3820]+[18−8−42]=[9−(−24)+180−12+(−8)20−(−38)+(−4)1−20+2]=[51−2054−17]A^2 - 2AB + B^2 = \begin{bmatrix}[r] 9 & 0 \ 20 & 1 \end{bmatrix} - \begin{bmatrix}[r] -24 & 12 \ -38 & 20 \end{bmatrix} + \begin{bmatrix}[r] 18 & -8 \ -4 & 2 \end{bmatrix}\\[1em] = \begin{bmatrix}[r] 9 - (-24) + 18 & 0 - 12 + (-8) \ 20 - (-38) + (-4) & 1 - 20 + 2 \end{bmatrix}\\[1em] = \begin{bmatrix}[r] 51 & -20 \ 54 & -17 \end{bmatrix}

Hence, the value of A2 - 2AB + B2 = [51−2054−17]\begin{bmatrix}[r] 51 & -20 \ 54 & -17 \end{bmatrix}

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