#### Please solve RD Sharma class 12 chapter Algebra of matrices exercise 4.3 question 18 maths textbook solution

Hence, verified$A(B-C)=A B-A C$

Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.

Given:

$A=\left[\begin{array}{ccc}1 & 0 & -2 \\ 3 & -1 & 0 \\ -2 & 1 & 1\end{array}\right], B=\left[\begin{array}{ccc}0 & 5 & -4 \\ -2 & 1 & 3 \\ -1 & 0 & 2\end{array}\right], C=\left[\begin{array}{ccc}1 & 5 & 2 \\ -1 & 1 & 0 \\ 0 & -1 & 1\end{array}\right]$

Consider,

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

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

$=\left[\begin{array}{ccc} 1 & 0 & -2 \\ 3 & -1 & 0 \\ -2 & 1 & 1 \end{array}\right]\left[\begin{array}{ccc} -1 & 0 & -6 \\ -1 & 0 & 3 \\ -1 & 1 & 1 \end{array}\right] \\\\ \\ =\left[\begin{array}{ccc} 1(-1)+0(-1)+(-2)(-1) & 1(0)+0(0)+(-2)(1) & 1(-6)+0(3)+(-2)(1) \\ 3(-1)+(-1)(-1)+0(-1) & 3(0)+(-1)(0)+0(1) & 3(-6)+(-1)(3)+0(1) \\ -2(-1)+1(-1)+1(-1) & -2(0)+1(0)+1(1) & -2(-6)+1(3)+1(1) \end{array}\right] \\\\\ \\ =\left[\begin{array}{ccc} -1-0+2 & 0+0-2 & -6+0-2 \\ -3+1-0 & 0+0+0 & -18-3+0 \\ 2-1-1 & -0+0+1 & 12+3+1 \end{array}\right] \\\\ \\ A(B-C) =\left[\begin{array}{ccc} 1 & -2 & -8 \\ -2 & 0 & -21 \\ 0 & 1 & 16 \end{array}\right] ...(i)$

Now consider,

$A B-A C=\left[\begin{array}{ccc} 1 & 0 & -2 \\ 3 & -1 & 0 \\ -2 & 1 & 1 \end{array}\right]\left[\begin{array}{ccc} 0 & 5 & -4 \\ -2 & 1 & 3 \\ -1 & 0 & 2 \end{array}\right]-\left[\begin{array}{ccc} 1 & 0 & -2 \\ 3 & -1 & 0 \\ -2 & 1 & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 5 & 2 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{array}\right] \\\\\\ =\left[\begin{array}{ccc} 0+0+2 & 5+0+0 & -4+0-4 \\ 0+2+0 & 15-1+0 & -12-3+0 \\ 0-2-1 & -10+1+0 & 8+3+2 \end{array}\right]-\left[\begin{array}{ccc} 1+0+0 & 5+0+2 & 2+0-2 \\ 3+1+0 & 15-1+0 & 6+0+0 \\ -2-1+0 & -10+1-1 & -4+0+1 \end{array}\right] \\ \\\\ =\left[\begin{array}{ccc} 2 & 5 & -8 \\ 2 & 14 & -15 \\ -3 & -9 & 13 \end{array}\right]-\left[\begin{array}{ccc} 1 & 7 & 0 \\ 4 & 14 & 6 \\ -3 & -10 & -3 \end{array}\right]$

$A B-A C=\left[\begin{array}{ccc} 2-1 & 5-7 & -8-0 \\ 2-4 & 14-14 & -15-6 \\ -3+3 & -9+10 & 13+3 \end{array}\right]\\\\ \\ A B-A C =\left[\begin{array}{ccc} 1 & -2 & -8 \\ -2 & 0 & -21 \\ 0 & 1 & 16 \end{array}\right] ... (ii)$

From equation i & ii

$A(B-C) = AB - AC$