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Question
If A= `[[1 0 -2],[3 -1 0],[-2 1 1]]` B=,`[[0 5 -4],[-2 1 3],[-1 0 2]] and C=[[1 5 2],[-1 1 0],[0 -1 1]]` verify that A (B − C) = AB − AC.
Sum
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Solution
LaTeX
\[Given: A\left( B - C \right) = AB - AC\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & - 2 \\ 3 & - 1 & 0 \\ - 2 & 1 & 1\end{bmatrix}\left( \begin{bmatrix}0 & 5 & - 4 \\ - 2 & 1 & 3 \\ - 1 & 0 & 2\end{bmatrix} - \begin{bmatrix}1 & 5 & 2 \\ - 1 & 1 & 0 \\ 0 & - 1 & 1\end{bmatrix} \right) = \begin{bmatrix}1 & 0 & - 2 \\ 3 & - 1 & 0 \\ - 2 & 1 & 1\end{bmatrix}\begin{bmatrix}0 & 5 & - 4 \\ - 2 & 1 & 3 \\ - 1 & 0 & 2\end{bmatrix} - \begin{bmatrix}1 & 0 & - 2 \\ 3 & - 1 & 0 \\ - 2 & 1 & 1\end{bmatrix}\begin{bmatrix}1 & 5 & 2 \\ - 1 & 1 & 0 \\ 0 & - 1 & 1\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & - 2 \\ 3 & - 1 & 0 \\ - 2 & 1 & 1\end{bmatrix}\begin{bmatrix}0 - 1 & 5 - 5 & - 4 - 2 \\ - 2 + 1 & 1 - 1 & 3 - 0 \\ - 1 - 0 & 0 + 1 & 2 - 1\end{bmatrix} = \begin{bmatrix}0 - 0 + 2 & 5 + 0 - 0 & - 4 + 0 - 4 \\ 0 + 2 - 0 & 15 - 1 + 0 & - 12 - 3 + 0 \\ 0 - 2 - 1 & - 10 + 1 + 0 & 8 + 3 + 2\end{bmatrix} - \begin{bmatrix}1 - 0 - 0 & 5 + 0 + 2 & 2 + 0 - 2 \\ 3 + 1 + 0 & 15 - 1 - 0 & 6 - 0 + 0 \\ - 2 - 1 + 0 & - 10 + 1 - 1 & - 4 + 0 + 1\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & - 2 \\ 3 & - 1 & 0 \\ - 2 & 1 & 1\end{bmatrix}\begin{bmatrix}- 1 & 0 & - 6 \\ - 1 & 0 & 3 \\ - 1 & 1 & 1\end{bmatrix} = \begin{bmatrix}2 & 5 & - 8 \\ 2 & 14 & - 15 \\ - 3 & - 9 & 13\end{bmatrix} - \begin{bmatrix}1 & 7 & 0 \\ 4 & 14 & 6 \\ - 3 & - 10 & - 3\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}- 1 - 0 + 2 & 0 + 0 - 2 & - 6 + 0 - 2 \\ - 3 + 1 - 0 & 0 - 0 + 0 & - 18 - 3 + 0 \\ 2 - 1 - 1 & 0 + 0 + 1 & 12 + 3 + 1\end{bmatrix} = \begin{bmatrix}2 - 1 & 5 - 7 & - 8 - 0 \\ 2 - 4 & 14 - 14 & - 15 - 6 \\ - 3 + 3 & - 9 + 10 & 13 + 3\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}1 & - 2 & - 8 \\ - 2 & 0 & - 21 \\ 0 & 1 & 16\end{bmatrix} = \begin{bmatrix}1 & - 2 & - 8 \\ - 2 & 0 & - 21 \\ 0 & 1 & 16\end{bmatrix}\]
\[ \therefore LHS = RHS\]
Hence proved .
\[ \Rightarrow \begin{bmatrix}1 & 0 & - 2 \\ 3 & - 1 & 0 \\ - 2 & 1 & 1\end{bmatrix}\left( \begin{bmatrix}0 & 5 & - 4 \\ - 2 & 1 & 3 \\ - 1 & 0 & 2\end{bmatrix} - \begin{bmatrix}1 & 5 & 2 \\ - 1 & 1 & 0 \\ 0 & - 1 & 1\end{bmatrix} \right) = \begin{bmatrix}1 & 0 & - 2 \\ 3 & - 1 & 0 \\ - 2 & 1 & 1\end{bmatrix}\begin{bmatrix}0 & 5 & - 4 \\ - 2 & 1 & 3 \\ - 1 & 0 & 2\end{bmatrix} - \begin{bmatrix}1 & 0 & - 2 \\ 3 & - 1 & 0 \\ - 2 & 1 & 1\end{bmatrix}\begin{bmatrix}1 & 5 & 2 \\ - 1 & 1 & 0 \\ 0 & - 1 & 1\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & - 2 \\ 3 & - 1 & 0 \\ - 2 & 1 & 1\end{bmatrix}\begin{bmatrix}0 - 1 & 5 - 5 & - 4 - 2 \\ - 2 + 1 & 1 - 1 & 3 - 0 \\ - 1 - 0 & 0 + 1 & 2 - 1\end{bmatrix} = \begin{bmatrix}0 - 0 + 2 & 5 + 0 - 0 & - 4 + 0 - 4 \\ 0 + 2 - 0 & 15 - 1 + 0 & - 12 - 3 + 0 \\ 0 - 2 - 1 & - 10 + 1 + 0 & 8 + 3 + 2\end{bmatrix} - \begin{bmatrix}1 - 0 - 0 & 5 + 0 + 2 & 2 + 0 - 2 \\ 3 + 1 + 0 & 15 - 1 - 0 & 6 - 0 + 0 \\ - 2 - 1 + 0 & - 10 + 1 - 1 & - 4 + 0 + 1\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}1 & 0 & - 2 \\ 3 & - 1 & 0 \\ - 2 & 1 & 1\end{bmatrix}\begin{bmatrix}- 1 & 0 & - 6 \\ - 1 & 0 & 3 \\ - 1 & 1 & 1\end{bmatrix} = \begin{bmatrix}2 & 5 & - 8 \\ 2 & 14 & - 15 \\ - 3 & - 9 & 13\end{bmatrix} - \begin{bmatrix}1 & 7 & 0 \\ 4 & 14 & 6 \\ - 3 & - 10 & - 3\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}- 1 - 0 + 2 & 0 + 0 - 2 & - 6 + 0 - 2 \\ - 3 + 1 - 0 & 0 - 0 + 0 & - 18 - 3 + 0 \\ 2 - 1 - 1 & 0 + 0 + 1 & 12 + 3 + 1\end{bmatrix} = \begin{bmatrix}2 - 1 & 5 - 7 & - 8 - 0 \\ 2 - 4 & 14 - 14 & - 15 - 6 \\ - 3 + 3 & - 9 + 10 & 13 + 3\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}1 & - 2 & - 8 \\ - 2 & 0 & - 21 \\ 0 & 1 & 16\end{bmatrix} = \begin{bmatrix}1 & - 2 & - 8 \\ - 2 & 0 & - 21 \\ 0 & 1 & 16\end{bmatrix}\]
\[ \therefore LHS = RHS\]
Hence proved .
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