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If A = [3051]\begin{bmatrix} 3 & 0 \ 5 & 1 \end{bmatrix} and B = [−4210]\begin{bmatrix} -4 & 2 \ 1 & 0 \end{bmatrix}, find A2 – 2AB + B2.

Matrices

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Answer

Given,

A = [3051]\begin{bmatrix} 3 & 0 \ 5 & 1 \end{bmatrix} and B = [−4210]\begin{bmatrix} -4 & 2 \ 1 & 0 \end{bmatrix}

Solving for A2:

⇒A2=[3051]×[3051]=[(3)(3)+(0)(5)(3)(0)+(0)(1)(5)(3)+(1)(5)(5)(0)+(1)(1)]=[9+00+015+50+1]=[90201].\Rightarrow A^2 = \begin{bmatrix} 3 & 0 \ 5 & 1 \end{bmatrix} \times \begin{bmatrix} 3 & 0 \ 5 & 1 \end{bmatrix} \\[1em] = \begin{bmatrix} (3)(3) + (0)(5) & (3)(0) + (0)(1) \ (5)(3) + (1)(5) & (5)(0) + (1)(1) \end{bmatrix} \\[1em] = \begin{bmatrix} 9 + 0 & 0 + 0 \ 15 + 5 & 0 + 1 \end{bmatrix} \\[1em] = \begin{bmatrix} 9 & 0 \ 20 & 1 \end{bmatrix}.

Solving for B2:

⇒B2=[−4210]×[−4210]=[(−4)(−4)+(2)(1)(−4)(2)+(2)(0)(1)(−4)+(0)(1)(1)(2)+(0)(0)]=[16+2−8+0−4+02+0]=[18−8−42].\Rightarrow B^2 = \begin{bmatrix} -4 & 2 \ 1 & 0 \end{bmatrix} \times \begin{bmatrix} -4 & 2 \ 1 & 0 \end{bmatrix} \\[1em] = \begin{bmatrix} (-4)(-4) + (2)(1) & (-4)(2) + (2)(0) \ (1)(-4) + (0)(1) & (1)(2) + (0)(0) \end{bmatrix} \\[1em] = \begin{bmatrix} 16 + 2 & -8 + 0 \ -4 + 0 & 2 + 0 \end{bmatrix} \\[1em] = \begin{bmatrix} 18 & -8 \ -4 & 2 \end{bmatrix}.

Solving for 2AB:

⇒2AB=2([3051]×[−4210])=2[(3)(−4)+(0)(1)(3)(2)+(0)(0)(5)(−4)+(1)(1)(5)(2)+(1)(0)]=2[−12+06+0−20+110+0]=2[−126−1910]=[−2412−3820].\Rightarrow 2AB = 2 \Big(\begin{bmatrix} 3 & 0 \ 5 & 1 \end{bmatrix} \times \begin{bmatrix} -4 & 2 \ 1 & 0 \end{bmatrix}\Big) \\[1em] = 2\begin{bmatrix} (3)(-4) + (0)(1) & (3)(2) + (0)(0) \ (5)(-4) + (1)(1) & (5)(2) + (1)(0) \end{bmatrix} \\[1em] = 2\begin{bmatrix} -12 + 0 & 6 + 0 \ -20 + 1 & 10 + 0 \end{bmatrix} \\[1em] = 2\begin{bmatrix} -12 & 6 \ -19 & 10 \end{bmatrix} \\[1em] = \begin{bmatrix} -24 & 12 \ -38 & 20 \end{bmatrix}.

A2 – 2AB + B2

⇒[90201]−[−2412−3820]+[18−8−42]⇒[9−(−24)0−1220−(−38)1−20]+[18−8−42]⇒[33−1258−19]+[18−8−42]⇒[33+18−12−858−4−19+2]⇒[51−2054−17].\Rightarrow \begin{bmatrix} 9 & 0 \ 20 & 1 \end{bmatrix} - \begin{bmatrix} -24 & 12 \ -38 & 20 \end{bmatrix} + \begin{bmatrix} 18 & -8 \ -4 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 9 - (-24) & 0 - 12 \ 20 - (-38) & 1 - 20 \end{bmatrix} + \begin{bmatrix} 18 & -8 \ -4 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 33 & -12 \ 58 & -19 \end{bmatrix} + \begin{bmatrix} 18 & -8 \ -4 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 33 + 18 & -12 - 8 \ 58 - 4 & -19 + 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 51 & -20 \ 54 & -17 \end{bmatrix}.

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

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