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Mathematics

Given A = [2130],B=[1152] and C=[−3−100]\begin{bmatrix}[r] 2 & 1 \ 3 & 0 \end{bmatrix}, B = \begin{bmatrix}[r] 1 & 1 \ 5 & 2 \end{bmatrix} \text{ and } C = \begin{bmatrix}[r] -3 & -1 \ 0 & 0 \end{bmatrix}; find :

(i) 2A - 3B + C

(ii) A + 2C - B

Matrices

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Answer

(i) Given,

2A - 3B + C

Substituting values of A, B and C in above equation we get,

⇒2A−3B+C=2[2130]−3[1152]+[−3−100]=[4260]−[33156]+[−3−100]=[4−3+(−3)2−3+(−1)6−15+00−6+0]=[−2−2−9−6].\Rightarrow 2A - 3B + C = 2\begin{bmatrix}[r] 2 & 1 \ 3 & 0 \end{bmatrix} - 3\begin{bmatrix}[r] 1 & 1 \ 5 & 2 \end{bmatrix} + \begin{bmatrix}[r] -3 & -1 \ 0 & 0 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 4 & 2 \ 6 & 0 \end{bmatrix} - \begin{bmatrix}[r] 3 & 3 \ 15 & 6 \end{bmatrix} + \begin{bmatrix}[r] -3 & -1 \ 0 & 0 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 4 - 3 + (-3) & 2 - 3 + (-1) \ 6 - 15 + 0 & 0 - 6 + 0 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] -2 & -2 \ -9 & -6 \end{bmatrix}.

Hence 2A - 3B + C = [−2−2−9−6].\begin{bmatrix}[r] -2 & -2 \ -9 & -6 \end{bmatrix}.

(ii) Given,

A + 2C - B

Substituting values of A, B and C in above equation we get,

⇒[2130]+2[−3−100]−[1152]=[2130]+[−6−200]−[1152]=[2+(−6)−11+(−2)−13+0−50+0−2]=[−5−2−2−2].\Rightarrow \begin{bmatrix}[r] 2 & 1 \ 3 & 0 \end{bmatrix} + 2\begin{bmatrix}[r] -3 & -1 \ 0 & 0 \end{bmatrix} - \begin{bmatrix}[r] 1 & 1 \ 5 & 2 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 2 & 1 \ 3 & 0 \end{bmatrix} + \begin{bmatrix}[r] -6 & -2 \ 0 & 0 \end{bmatrix} - \begin{bmatrix}[r] 1 & 1 \ 5 & 2 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 2 + (-6) - 1 & 1 + (-2) - 1 \ 3 + 0 - 5 & 0 + 0 - 2 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] -5 & -2 \ -2 & -2 \end{bmatrix}.

Hence, A + 2C - B = [−5−2−2−2].\begin{bmatrix}[r] -5 & -2 \ -2 & -2 \end{bmatrix}.

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