# if

Question:

If $A=\left[\begin{array}{rrr}1 & 0 & -2 \\ 3 & -1 & 0 \\ -2 & 1 & 1\end{array}\right], B=\left[\begin{array}{rrr}0 & 5 & -4 \\ -2 & 1 & 3 \\ -1 & 0 & 2\end{array}\right]$ and $C=\left[\begin{array}{rrr}1 & 5 & 2 \\ -1 & 1 & 0 \\ 0 & -1 & 1\end{array}\right]$ , verify that $A(B-C)=A B-A C$.

Solution:

Given : $A(B-C)=A B-A C$

$\Rightarrow\left[\begin{array}{ccc}1 & 0 & -2 \\ 3 & -1 & 0 \\ -2 & 1 & 1\end{array}\right]\left(\left[\begin{array}{ccc}0 & 5 & -4 \\ -2 & 1 & 3 \\ -1 & 0 & 2\end{array}\right]-\left[\begin{array}{ccc}1 & 5 & 2 \\ -1 & 1 & 0 \\ 0 & -1 & 1\end{array}\right]\right)=\left[\begin{array}{ccc}1 & 0 & -2 \\ 3 & -1 & 0 \\ -2 & 1 & 1\end{array}\right]\left[\begin{array}{ccc}0 & 5 & -4 \\ -2 & 1 & 3 \\ -1 & 0 & 2\end{array}\right]-\left[\begin{array}{ccc}1 & 0 & -2 \\ 3 & -1 & 0 \\ -2 & 1 & 1\end{array}\right]\left[\begin{array}{ccc}1 & 5 & 2 \\ -1 & 1 & 0 \\ 0 & -1 & 1\end{array}\right]$

$\Rightarrow\left[\begin{array}{ccc}1 & 0 & -2 \\ 3 & -1 & 0 \\ -2 & 1 & 1\end{array}\right]\left[\begin{array}{cc}0-1 & 5-5 & -4-2 \\ -2+1 & 1-1 & 3-0 \\ -1-0 & 0+1 & 2-1\end{array}\right]=\left[\begin{array}{cc}0-0+2 & 5+0-0 & -4+0-4 \\ 0+2-0 & 15-1+0 & -12-3+0 \\ 0-2-1 & -10+1+0 & 8+3+2\end{array}\right]-\left[\begin{array}{c}1-0-0 & 5+0+2 \\ 2+0-2 \\ -2-1+0\end{array}\right]$

$\Rightarrow\left[\begin{array}{ccc}1 & 0 & -2 \\ 3 & -1 & 0 \\ -2 & 1 & 1\end{array}\right]\left[\begin{array}{ccc}-1 & 0 & -6 \\ -1 & 0 & 3 \\ -1 & 1 & 1\end{array}\right]=\left[\begin{array}{ccc}2 & 5 & -8 \\ 2 & 14 & -15 \\ -3 & -9 & 13\end{array}\right]-\left[\begin{array}{ccc}1 & 7 & 0 \\ 4 & 14 & 6 \\ -3 & -10 & -3\end{array}\right]$

$\Rightarrow\left[\begin{array}{ccc}-1-0+2 & 0+0-2 & -6+0-2 \\ -3+1-0 & 0-0+0 & -18-3+0 \\ 2-1-1 & 0+0+1 & 12+3+1\end{array}\right]=\left[\begin{array}{cc}2-1 & 5-7 & -10+1-1 & 6-0+0 \\ 2-4 & 14-14 & -15-6 & -0+1\end{array}\right]$

$\Rightarrow\left[\begin{array}{ccc}1 & -2 & -8 \\ -2 & 0 & -21 \\ 0 & 1 & 16\end{array}\right]=\left[\begin{array}{ccc}1 & -2 & -8 \\ -2 & 0 & -21 \\ 0 & 1 & 16\end{array}\right]$

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

Hence proved.