#### Explain solution for rd sharma class 12 chapter Algebra of matrices exercise 4.3 question 17 sub question (ii) math

: Hence, verify the distribution of matrix multiplication over matrix addition $A(B+C)=A B +A C$

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:

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

Consider,

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

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

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

Now consider

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

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

$A B+A C=\left[\begin{array}{cc} 1 & -2 \\ 2 & 2 \\ 1 & 4 \end{array}\right] .... (ii)$

From equation i & ii

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