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

Question:

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

(i) $A=\left[\begin{array}{rr}1 & -1 \\ 0 & 2\end{array}\right], B=\left[\begin{array}{rr}-1 & 0 \\ 2 & 1\end{array}\right]$ and $C=\left[\begin{array}{rr}0 & 1 \\ 1 & -1\end{array}\right]$

(ii) $A=\left[\begin{array}{rr}2 & -1 \\ 1 & 1 \\ -1 & 2\end{array}\right], B=\left[\begin{array}{ll}0 & 1 \\ 1 & 1\end{array}\right]$ and $C=\left[\begin{array}{rr}1 & -1 \\ 0 & 1\end{array}\right]$.

Solution:

(i)

$A(B+C)=A B+A C$

$\Rightarrow\left[\begin{array}{cc}1 & -1 \\ 0 & 2\end{array}\right]\left(\left[\begin{array}{cc}-1 & 0 \\ 2 & 1\end{array}\right]+\left[\begin{array}{cc}0 & 1 \\ 1 & -1\end{array}\right]\right)=\left[\begin{array}{cc}1 & -1 \\ 0 & 2\end{array}\right]\left[\begin{array}{cc}-1 & 0 \\ 2 & 1\end{array}\right]+\left[\begin{array}{cc}1 & -1 \\ 0 & 2\end{array}\right]\left[\begin{array}{cc}0 & 1 \\ 1 & -1\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}1 & -1 \\ 0 & 2\end{array}\right]\left[\begin{array}{cc}-1+0 & 0+1 \\ 2+1 & 1-1\end{array}\right]=\left[\begin{array}{cc}-1-2 & 0-1 \\ 0+4 & 0+2\end{array}\right]+\left[\begin{array}{cc}0-1 & 1+1 \\ 0+2 & 0-2\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}1 & -1 \\ 0 & 2\end{array}\right]\left[\begin{array}{cc}-1 & 1 \\ 3 & 0\end{array}\right]=\left[\begin{array}{cc}-3 & -1 \\ 4 & 2\end{array}\right]+\left[\begin{array}{cc}-1 & 2 \\ 2 & -2\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}-1-3 & 1-0 \\ 0+6 & 0+0\end{array}\right]=\left[\begin{array}{cc}-3-1 & -1+2 \\ 4+2 & 2-2\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}-4 & 1 \\ 6 & 0\end{array}\right]=\left[\begin{array}{cc}-4 & 1 \\ 6 & 0\end{array}\right]$

$\therefore \mathrm{LHS}=\mathrm{RHS}$

Hence proved.

(ii)

$A(B+C)=A B+A C$

$\Rightarrow\left[\begin{array}{cc}2 & -1 \\ 1 & 1 \\ -1 & 2\end{array}\right]\left(\left[\begin{array}{ll}0 & 1 \\ 1 & 1\end{array}\right]+\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right]\right)=\left[\begin{array}{cc}2 & -1 \\ 1 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{cc}0 & 1 \\ 1 & 1\end{array}\right]+\left[\begin{array}{cc}2 & -1 \\ 1 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}2 & -1 \\ 1 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{ll}0+1 & 1-1 \\ 1+0 & 1+1\end{array}\right]=\left[\begin{array}{cc}0-1 & 2-1 \\ 0+1 & 1+1 \\ 0+2 & -1+2\end{array}\right]+\left[\begin{array}{c}2-0 & -2-1 \\ 1+0 & -1+1 \\ -1+0 & 1+2\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}2 & -1 \\ 1 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{ll}1 & 0 \\ 1 & 2\end{array}\right]=\left[\begin{array}{cc}-1 & 1 \\ 1 & 2 \\ 2 & 1\end{array}\right]+\left[\begin{array}{cc}2 & -3 \\ 1 & 0 \\ -1 & 3\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}2-1 & 0-2 \\ 1+1 & 0+2 \\ -1+2 & 0+4\end{array}\right]=\left[\begin{array}{cc}-1+2 & 1-3 \\ 1+1 & 2+0 \\ 2-1 & 1+3\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}1 & -2 \\ 2 & 2 \\ 1 & 4\end{array}\right]=\left[\begin{array}{cc}1 & -2 \\ 2 & 2 \\ 1 & 4\end{array}\right]$

$\therefore$ LHS $=$ RHS

Hence proved.

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