#### Explain solution for rd sharma class 12 chapter Algebra of matrices exercise 4.3 question 75 math

Answer: Total cost incurred by the organization for three villages X, Y and Z are: 30000, 23000 and 39000.

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: The number of attempts made in 3 different villages X, Y and Z are

$\begin{array}{c} X \\ Y \\ Z \end{array}\left[\begin{array}{ccc} 400 & 300 & 100 \\ 300 & 250 & 75 \\ 500 & 400 & 150 \end{array}\right]$

An organization tried to generate awareness through (i) house calls, (ii) letters and (iii)announcements

Cost for mode per attempt in house calls Rs 50, letters Rs 20 and announcements Rs 40

The cost for each mode per attempt is represented by $3 \times 1$ matrix

$A=\left[\begin{array}{l}50 \\ 20 \\ 40\end{array}\right]$

The number of attempts made in the three villages X, Y and Z are represented by a  $3 \times 3$ matrix.

$B=\left[\begin{array}{ccc}400 & 300 & 100 \\ 300 & 250 & 75 \\ 500 & 400 & 150\end{array}\right]$

The total cost incurred by the organization for the three villages separately is given by matrix multiplication.$\\B A=\left[\begin{array}{ccc}400 & 300 & 100 \\ 300 & 250 & 75 \\ 500 & 400 & 150\end{array}\right]\left[\begin{array}{c}50 \\ 20 \\ 40\end{array}\right]\\\\\\ B A=\left[\begin{array}{c}400 \times 50+300 \times 20+100 \times 40 \\ 300 \times 50+250 \times 20+75 \times 40 \\ 500 \times 50+400 \times 20+150 \times 40\end{array}\right]\\\\\\ =\left[\begin{array}{l}20000+6000+4000 \\ 15000+5000+3000 \\ 25000+8000+6000\end{array}\right]\\\\\\ =\left[\begin{array}{l}30000 \\ 23000 \\ 39000\end{array}\right]$

Therefore, cost incurred by the organization for the three villages X,Y and Z is 30000, 23000 and 39000.