Question

For the following matrices verify the distributivity of matrix multiplication over matrix addition i.e. A (B + C) = AB + AC:

`A=[[2    -1],[1        1],[-1         2]]` `B=[[0     1],[1      1]]` C=`[[1      -1],[0                1]]`

Answer 1

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\[\left( ii \right) \]
\[A\left( B + C \right) = AB + AC\]
\[ \Rightarrow \begin{bmatrix}2 & - 1 \\ 1 & 1 \\ - 1 & 2\end{bmatrix}\left( \begin{bmatrix}0 & 1 \\ 1 & 1\end{bmatrix} + \begin{bmatrix}1 & - 1 \\ 0 & 1\end{bmatrix} \right) = \begin{bmatrix}2 & - 1 \\ 1 & 1 \\ - 1 & 2\end{bmatrix}\begin{bmatrix}0 & 1 \\ 1 & 1\end{bmatrix} + \begin{bmatrix}2 & - 1 \\ 1 & 1 \\ - 1 & 2\end{bmatrix}\begin{bmatrix}1 & - 1 \\ 0 & 1\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}2 & - 1 \\ 1 & 1 \\ - 1 & 2\end{bmatrix}\begin{bmatrix}0 + 1 & 1 - 1 \\ 1 + 0 & 1 + 1\end{bmatrix} = \begin{bmatrix}0 - 1 & 2 - 1 \\ 0 + 1 & 1 + 1 \\ 0 + 2 & - 1 + 2\end{bmatrix} + \begin{bmatrix}2 - 0 & - 2 - 1 \\ 1 + 0 & - 1 + 1 \\ - 1 + 0 & 1 + 2\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}2 & - 1 \\ 1 & 1 \\ - 1 & 2\end{bmatrix}\begin{bmatrix}1 & 0 \\ 1 & 2\end{bmatrix} = \begin{bmatrix}- 1 & 1 \\ 1 & 2 \\ 2 & 1\end{bmatrix} + \begin{bmatrix}2 & - 3 \\ 1 & 0 \\ - 1 & 3\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}2 - 1 & 0 - 2 \\ 1 + 1 & 0 + 2 \\ - 1 + 2 & 0 + 4\end{bmatrix} = \begin{bmatrix}- 1 + 2 & 1 - 3 \\ 1 + 1 & 2 + 0 \\ 2 - 1 & 1 + 3\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}1 & - 2 \\ 2 & 2 \\ 1 & 4\end{bmatrix} = \begin{bmatrix}1 & - 2 \\ 2 & 2 \\ 1 & 4\end{bmatrix}\]
\[ \therefore LHS = RHS\]
Hence proved .