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If A = [1334]\begin{bmatrix} 1 & 3 \ 3 & 4 \end{bmatrix} and B = [−21−32]\begin{bmatrix} -2 & 1 \ -3 & 2 \end{bmatrix} and A2 – 5B2 = 5C. Find matrix C, where C is a 2 × 2 matrix.

Matrices

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Answer

Given,

A = [1324]\begin{bmatrix} 1 & 3 \ 2 & 4 \end{bmatrix} and B = [−21−32]\begin{bmatrix} -2 & 1 \ -3 & 2 \end{bmatrix}

Solving for A2:

⇒A2=[1324]×[1324]=[(1)(1)+(3)(2)(1)(3)+(3)(4)(2)(1)+(4)(2)(2)(3)+(4)(4)]=[1+63+122+86+16]=[7151022].\Rightarrow A^2 = \begin{bmatrix} 1 & 3 \ 2 & 4 \end{bmatrix} \times \begin{bmatrix} 1 & 3 \ 2 & 4 \end{bmatrix} \\[1em] = \begin{bmatrix} (1)(1) + (3)(2) & (1)(3) + (3)(4) \ (2)(1) + (4)(2) & (2)(3) + (4)(4) \end{bmatrix} \\[1em] = \begin{bmatrix} 1 + 6 & 3 + 12 \ 2 + 8 & 6 + 16 \end{bmatrix} \\[1em] = \begin{bmatrix} 7 & 15 \ 10 & 22 \end{bmatrix}.

Solving for 5B2:

⇒5B2=5([−21−32]×[−21−32])=5([(−2)(−2)+(1)(−3)(−2)(1)+(1)(2)(−3)(−2)+(2)(−3)(−3)(1)+(2)(2)])=5[4−3−2+26−6−3+4]=5[1001]=[5005].\Rightarrow 5B^2 = 5\Big(\begin{bmatrix} -2 & 1 \ -3 & 2 \end{bmatrix} \times \begin{bmatrix} -2 & 1 \ -3 & 2 \end{bmatrix}\Big) \\[1em] = 5\Big(\begin{bmatrix} (-2)(-2) + (1)(-3) & (-2)(1) + (1)(2) \ (-3)(-2) + (2)(-3) & (-3)(1) + (2)(2) \end{bmatrix}\Big) \\[1em] = 5\begin{bmatrix} 4 - 3 & -2 + 2 \ 6 - 6 & -3 + 4 \end{bmatrix} \\[1em] = 5\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \\[1em] = \begin{bmatrix} 5 & 0 \ 0 & 5 \end{bmatrix}.

Solving for A2 – 5B2 = 5C:

⇒[7151022]−[5005]=5C⇒[7−515−010−022−5]=5C⇒15[2151517]=C⇒[2533175]=C\Rightarrow \begin{bmatrix} 7 & 15 \ 10 & 22 \end{bmatrix} - \begin{bmatrix} 5 & 0 \ 0 & 5 \end{bmatrix} = 5C \\[1em] \Rightarrow \begin{bmatrix} 7 - 5 & 15 - 0 \ 10 - 0 & 22 - 5 \end{bmatrix} = 5C \\[1em] \Rightarrow \dfrac{1}{5} \begin{bmatrix} 2 & 15 \ 15 & 17 \end{bmatrix} = C \\[1em] \Rightarrow \begin{bmatrix} \dfrac{2}{5} & 3 \ 3 & \dfrac{17}{5} \end{bmatrix} = C

Hence, C = [2533175]\begin{bmatrix} \dfrac{2}{5} & 3 \ 3 & \dfrac{17}{5} \end{bmatrix}

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