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Mathematics

If A = [sec 60°cos 90°-3 tan 45°sin 90°] and B =[0cot 45°−23 sin 90°],\begin{bmatrix}[r] \text{sec 60°} & \text{cos 90°} \ \text{-3 tan 45°} & \text{sin 90°} \end{bmatrix} \text{ and B } = \begin{bmatrix}[r] 0 & \text{cot 45°} \ -2 & \text{3 sin 90°} \end{bmatrix}, find

(i) 2A - 3B

(ii) A2

(iii) BA

Matrices

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Answer

(i) Given,

A=[sec 60°cos 90°-3 tan 45°sin 90°]=[20−31]B=[0cot 45°−23 sin 90°]=[01−23]2A - 3B=2[20−31]−3[01−23]=[40−62]−[03−69]=[4−00−3−6−(−6)2−9]=[4−30−7].\text{A} = \begin{bmatrix}[r] \text{sec 60°} & \text{cos 90°} \ \text{-3 tan 45°} & \text{sin 90°} \end{bmatrix} = \begin{bmatrix}[r] 2 & 0 \ -3 & 1 \end{bmatrix} \\[1em] \text{B} = \begin{bmatrix}[r] 0 & \text{cot 45°} \ -2 & \text{3 sin 90°} \end{bmatrix} = \begin{bmatrix}[r] 0 & 1 \ -2 & 3 \end{bmatrix} \\[1em] \text{2A - 3B} = 2\begin{bmatrix}[r] 2 & 0 \ -3 & 1 \end{bmatrix} - 3 \begin{bmatrix}[r] 0 & 1 \ -2 & 3 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 4 & 0 \ -6 & 2 \end{bmatrix} - \begin{bmatrix}[r] 0 & 3 \ -6 & 9 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 4 - 0 & 0 - 3 \ -6 - (-6) & 2 - 9 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 4 & -3 \ 0 & -7 \end{bmatrix}.

Hence, the value of 2A - 3B =[4−30−7].= \begin{bmatrix}[r] 4 & -3 \ 0 & -7 \end{bmatrix}.

(ii) Given,

A2=[20−31][20−31]=[2×2+0×(−3)2×0+0×1(−3)×2+1×(−3)(−3)×0+1×1]=[4+00+0−6−30+1]=[40−91].\text{A}^2 = \begin{bmatrix}[r] 2 & 0 \ -3 & 1 \end{bmatrix} \begin{bmatrix}[r] 2 & 0 \ -3 & 1 \end{bmatrix} \\[0.5em] = \begin{bmatrix}[r] 2 \times 2 + 0 \times (-3) & 2 \times 0 + 0 \times 1 \ (-3) \times 2 + 1 \times (-3) & (-3) \times 0 + 1 \times 1 \end{bmatrix} \\[0.5em] = \begin{bmatrix}[r] 4 + 0 & 0 + 0 \ -6 - 3 & 0 + 1 \end{bmatrix} \\[0.5em] = \begin{bmatrix}[r] 4 & 0 \ -9 & 1 \end{bmatrix}.

Hence, the value of A2=[40−91].\text{A}^2 = \begin{bmatrix}[r] 4 & 0 \ -9 & 1 \end{bmatrix}.

(iii)

BA =[01−23][20−31]=[0×2+1×(−3)0×0+1×1−2×2+3×(−3)−2×0+3×1]=[−31−133].\text{BA } = \begin{bmatrix}[r] 0 & 1 \ -2 & 3 \end{bmatrix} \begin{bmatrix}[r] 2 & 0 \ -3 & 1 \end{bmatrix} \\[0.5em] = \begin{bmatrix}[r] 0 \times 2 + 1 \times (-3) & 0 \times 0 + 1 \times 1 \ -2 \times 2 + 3 \times (-3) & -2 \times 0 + 3 \times 1 \end{bmatrix} \\[0.5em] = \begin{bmatrix}[r] -3 & 1 \ -13 & 3 \end{bmatrix}.

Hence, the value of matrix BA = [−31−133].\begin{bmatrix}[r] -3 & 1 \ -13 & 3 \end{bmatrix}.

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