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Mathematics

Find the matrices A and B, if 2A + B = [3−427]\begin{bmatrix}[r] 3 & -4 \ 2 & 7 \end{bmatrix} and A - 2B = [4311]\begin{bmatrix}[r] 4 & 3 \ 1 & 1 \end{bmatrix}

Matrices

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Answer

Given,

⇒ 2A + B = [3−427]\begin{bmatrix}[r] 3 & -4 \ 2 & 7 \end{bmatrix} ………(1)

⇒ A - 2B = [4311]\begin{bmatrix}[r] 4 & 3 \ 1 & 1 \end{bmatrix} ………(2)

Multiplying equation (1) by 2 and adding in equation (2), we get :

⇒2(2A+B)+A−2B=2[3−427]+[4311]⇒4A+2B+A−2B=[6−8414]+[4311]⇒5A=[6+4−8+34+114+1]⇒5A=[10−5515]⇒A=15[10−5515]⇒A=[2−113].\Rightarrow 2(2A + B) + A - 2B = 2\begin{bmatrix}[r] 3 & -4 \ 2 & 7 \end{bmatrix} + \begin{bmatrix}[r] 4 & 3 \ 1 & 1 \end{bmatrix} \\[1em] \Rightarrow 4A + 2B + A - 2B = \begin{bmatrix}[r] 6 & -8 \ 4 & 14 \end{bmatrix} + \begin{bmatrix}[r] 4 & 3 \ 1 & 1 \end{bmatrix} \\[1em] \Rightarrow 5A = \begin{bmatrix}[r] 6 + 4 & -8 + 3 \ 4 + 1 & 14 + 1 \end{bmatrix} \\[1em] \Rightarrow 5A = \begin{bmatrix}[r] 10 & -5 \ 5 & 15 \end{bmatrix} \\[1em] \Rightarrow A = \dfrac{1}{5}\begin{bmatrix}[r] 10 & -5 \ 5 & 15 \end{bmatrix} \\[1em] \Rightarrow A = \begin{bmatrix}[r] 2 & -1 \ 1 & 3 \end{bmatrix}.

Substituting value of A in equation (1), we get :

⇒2[2−113]+B=[3−427]⇒[4−226]+B=[3−427]⇒B=[3−427]−[4−226]⇒B=[3−4−4−(−2)2−27−6]⇒B=[−1−201].\Rightarrow 2\begin{bmatrix}[r] 2 & -1 \ 1 & 3 \end{bmatrix} + B = \begin{bmatrix}[r] 3 & -4 \ 2 & 7 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 4 & -2 \ 2 & 6 \end{bmatrix} + B = \begin{bmatrix}[r] 3 & -4 \ 2 & 7 \end{bmatrix} \\[1em] \Rightarrow B = \begin{bmatrix}[r] 3 & -4 \ 2 & 7 \end{bmatrix} - \begin{bmatrix}[r] 4 & -2 \ 2 & 6 \end{bmatrix} \\[1em] \Rightarrow B = \begin{bmatrix}[r] 3 - 4 & -4 - (-2) \ 2 - 2 & 7 - 6 \end{bmatrix} \\[1em] \Rightarrow B = \begin{bmatrix}[r] -1 & -2 \ 0 & 1 \end{bmatrix}.

Hence, A=[2−113],B=[−1−201].A = \begin{bmatrix}[r] 2 & -1 \ 1 & 3 \end{bmatrix}, B = \begin{bmatrix}[r] -1 & -2 \ 0 & 1 \end{bmatrix}.

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