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Mathematics

If A=[1324],B=[1224],C=[4115] and I=[1001]A = \begin{bmatrix}[r] 1 & 3 \ 2 & 4 \end{bmatrix}, B = \begin{bmatrix}[r] 1 & 2 \ 2 & 4 \end{bmatrix}, C = \begin{bmatrix}[r] 4 & 1 \ 1 & 5 \end{bmatrix} \text{ and } I = \begin{bmatrix}[r] 1 & 0 \ 0 & 1 \end{bmatrix}. Find A(B + C) - 14I.

Matrices

ICSE 2023

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Answer

B + C = [1224]+[4115]=[5339]\begin{bmatrix}[r] 1 & 2 \ 2 & 4 \end{bmatrix} + \begin{bmatrix}[r] 4 & 1 \ 1 & 5 \end{bmatrix} = \begin{bmatrix}[r] 5 & 3 \ 3 & 9 \end{bmatrix}

Substituting values we get :

A(B+C)−14I=[1324][5339]−14[1001]=[1×5+3×31×3+3×92×5+4×32×3+4×9]−[140014]=[5+93+2710+126+36]−[140014]=[14302242]−[140014]=[14−1430−022−042−14]=[0302228].A(B + C) - 14I = \begin{bmatrix}[r] 1 & 3 \ 2 & 4 \end{bmatrix}\begin{bmatrix}[r] 5 & 3 \ 3 & 9 \end{bmatrix} - 14\begin{bmatrix}[r] 1 & 0 \ 0 & 1 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 1 \times 5 + 3 \times 3 & 1 \times 3 + 3 \times 9 \ 2 \times 5 + 4 \times 3 & 2 \times 3 + 4 \times 9 \end{bmatrix} - \begin{bmatrix}[r] 14 & 0 \ 0 & 14 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 5 + 9 & 3 + 27 \ 10 + 12 & 6 + 36 \end{bmatrix} - \begin{bmatrix}[r] 14 & 0 \ 0 & 14 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 14 & 30 \ 22 & 42 \end{bmatrix} - \begin{bmatrix}[r] 14 & 0 \ 0 & 14 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 14 - 14 & 30 - 0 \ 22 - 0 & 42 - 14 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 0 & 30 \ 22 & 28 \end{bmatrix}.

Hence, A(B + C) - 14I = [0302228].\begin{bmatrix}[r] 0 & 30 \ 22 & 28 \end{bmatrix}.

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