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Exercise 4.5 · Q119

Q.If [2a+b3a−bc+2d2c−d]=[234−1]\begin{bmatrix} 2a+b & 3a-b \\ c+2d & 2c-d \end{bmatrix} = \begin{bmatrix} 2 & 3 \\ 4 & -1 \end{bmatrix}, find a,b,ca, b, c and dd.

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Given [2a+b3a−bc+2d2c−d]=[234−1]\begin{bmatrix} 2a+b & 3a-b \\ c+2d & 2c-d \end{bmatrix} = \begin{bmatrix} 2 & 3 \\ 4 & -1 \end{bmatrix}. Equating entries:

2a+b=22a+b=2 ...(1)

3a−b=33a-b=3 ...(2)

c+2d=4c+2d=4 ...(3)

2c−d=−12c-d=-1 ...(4)

Adding (1) and (2): 5a=5⇒a=15a=5 \Rightarrow a=1. Substituting into (1): 2(1)+b=2⇒b=02(1)+b=2 \Rightarrow b=0. …

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