If [a342]+[2b1−2]−[11−2c]=[5073],\begin{bmatrix}[r] a & 3 \ 4 & 2 \end{bmatrix} + \begin{bmatrix}[r] 2 & b \ 1 & -2 \end{bmatrix} - \begin{bmatrix}[r] 1 & 1 \ -2 & c \end{bmatrix} = \begin{bmatrix}[r] 5 & 0 \ 7 & 3 \end{bmatrix},[a432]+[21b−2]−[1−21c]=[5703], find the values of a, b and c.
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Given,
⇒[a342]+[2b1−2]−[11−2c]=[5073]⇒[a+2−13+b−14+1−(−2)2+(−2)−c]=[5073]⇒[a+1b+27−c]=[5073]⇒a+1=5,b+2=0 and −c=3\Rightarrow \begin{bmatrix}[r] a & 3 \ 4 & 2 \end{bmatrix} + \begin{bmatrix}[r] 2 & b \ 1 & -2 \end{bmatrix} - \begin{bmatrix}[r] 1 & 1 \ -2 & c \end{bmatrix} = \begin{bmatrix}[r] 5 & 0 \ 7 & 3 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] a + 2 - 1 & 3 + b - 1 \ 4 + 1 -(-2) & 2 + (-2) - c \end{bmatrix} = \begin{bmatrix}[r] 5 & 0 \ 7 & 3 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] a + 1 & b + 2 \ 7 & -c \end{bmatrix} = \begin{bmatrix}[r] 5 & 0 \ 7 & 3 \end{bmatrix} \\[1em] \Rightarrow a + 1 = 5, b + 2 = 0 \text{ and } -c = 3 \\[1em]⇒[a432]+[21b−2]−[1−21c]=[5703]⇒[a+2−14+1−(−2)3+b−12+(−2)−c]=[5703]⇒[a+17b+2−c]=[5703]⇒a+1=5,b+2=0 and −c=3
∴ a = 4, b = -2 and c = -3.
Hence, the values are a = 4, b = -2 and c = -3.
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