Given, A = 12-35021-11 and B= 3-12425203. Find the matrix C such that A + 2C = B

2C = B-A = 3-12425203 - 12-35021-11 = 3-1-1-22+34-52-05-22-10+13-1 ∴ 2C = 2-35-123112 ⇒ C = 122-35-123112 = 1-3252-1213212121

Given, A = 12-35021-11 and B= 3-12425203. Find the matrix C such that A + 2C = B

2C = B-A = 3-12425203 - 12-35021-11 = 3-1-1-22+34-52-05-22-10+13-1 ∴ 2C = 2-35-123112 ⇒ C = 122-35-123112 = 1-3252-1213212121

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Given, $A = [\begin{matrix} 1 & 2 & - 3 \\ 5 & 0 & 2 \\ 1 & - 1 & 1 \end{matrix}]$ and $B = [\begin{matrix} 3 & - 1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3 \end{matrix}]$ . Find the matrix $C$ such that $A + 2 C = B$

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$[\begin{matrix} 1 & \frac{3}{2} & \frac{5}{2} \\ \frac{1}{2} & 1 & - \frac{3}{2} \\ - \frac{1}{2} & \frac{1}{2} & 1 \end{matrix}]$

$[\begin{matrix} 1 & \frac{- 3}{2} & \frac{5}{2} \\ \frac{- 1}{2} & 1 & \frac{3}{2} \\ \frac{1}{2} & \frac{1}{2} & 1 \end{matrix}]$

$[\begin{matrix} 1 & 1 & \frac{5}{2} \\ \frac{- 1}{2} & 1 & \frac{3}{2} \\ \frac{1}{2} & \frac{1}{2} & \frac{- 3}{2} \end{matrix}]$

$[\begin{matrix} 2 & 1 & \frac{5}{2} \\ 1 & 1 & \frac{3}{2} \\ \frac{1}{2} & \frac{1}{2} & \frac{- 3}{2} \end{matrix}]$

answer is B.

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Detailed Solution

$2 C = B - A = [\begin{matrix} 3 & - 1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3 \end{matrix}] - [\begin{matrix} 1 & 2 & - 3 \\ 5 & 0 & 2 \\ 1 & - 1 & 1 \end{matrix}] = [\begin{matrix} 3 - 1 & - 1 - 2 & 2 + 3 \\ 4 - 5 & 2 - 0 & 5 - 2 \\ 2 - 1 & 0 + 1 & 3 - 1 \end{matrix}] ∴ 2 C = [\begin{matrix} 2 & - 3 & 5 \\ - 1 & 2 & 3 \\ 1 & 1 & 2 \end{matrix}] \Rightarrow C = \frac{1}{2} [\begin{matrix} 2 & - 3 & 5 \\ - 1 & 2 & 3 \\ 1 & 1 & 2 \end{matrix}] = [\begin{matrix} 1 & \frac{- 3}{2} & \frac{5}{2} \\ \frac{- 1}{2} & 1 & \frac{3}{2} \\ \frac{1}{2} & \frac{1}{2} & 1 \end{matrix}]$