#### Need solution for RD Sharma maths class 12 chapter Algebra of matrices exercise 4.3 question 5 sub question (i)

$\begin{bmatrix} 6 &16 &26 \\ -8&- 18 &-28 \end{bmatrix}$

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: 1$\left(\left[\begin{array}{cc} 1 & 3 \\ -1 & -4 \end{array}\right]+\left[\begin{array}{cc} 3 & -2 \\ -1 & 1 \end{array}\right]\right)\left[\begin{array}{ccc} 1 & 3 & 5 \\ 2 & 4 & 6 \end{array}\right]$

Firstly, we have to add first two matrix,

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

Then, we multiply two matrices

$\begin{bmatrix} 4 &1 \\ -2 &-3 \end{bmatrix} \begin{bmatrix} 1 &3 &5 \\ 2& 4 &6 \end{bmatrix}$

\begin{aligned} &=\left[\begin{array}{ccc} 4 \times 1+1 \times 2 & 4 \times 3+1 \times 4 & 4 \times 5+1 \times 6 \\ -2 \times 1+(-3) \times 2 & -2 \times 3+(-3) \times 4 & -2 \times 5+(-3) \times 6 \end{array}\right] \\ &=\left[\begin{array}{ccc} 4+2 & 12+4 & 20+6 \\ -2-6 & -6-12 & -10-18 \end{array}\right] \end{aligned}

On simplification we get

$=\begin{bmatrix} 6 &16 &26 \\ -8&- 18 &-28 \end{bmatrix}$