Question

A salesman has the following record of sales during three months for three items A, B and C which have different rates of commission 

Month	Sale of units		Total commission drawn (in Rs)
	A	B	C
Jan	90	100	20	800
Feb	130	50	40	900
March	60	100	30	850

Find out the rates of commission on items A, B and C by using determinant method.

Answer 1

Let x, y and z be the rates of commission on items A, B and C respectively. Based on the given data, we get

\[90x + 100y + 20z = 800\] 
\[130x + 50y + 40z = 900\] 
\[60x + 100y + 30z = 850\]
Dividing all the equations by 10 on both sides, we get
\[9x + 10y + 2z = 80\] 
\[13x + 5y + 4z = 90\] 
\[6x + 10y + 3z = 85\]
\[D = \begin{vmatrix}9 & 10 & 2 \\ 13 & 5 & 4 \\ 6 & 10 & 3\end{vmatrix} \left[\text{ Expressing the equation as a determinant }\right]\] 
\[ = 9(15 - 40) - 10(39 - 24) + 2(130 - 30)\] 
\[ = 9( - 25) - 10(15) + 2(100)\] 
\[ = - 175\] 
\[ D_1 = \begin{vmatrix}80 & 10 & 2 \\ 90 & 5 & 4 \\ 85 & 10 & 3\end{vmatrix}\] 
\[ = 80(15 - 40) - 10(270 - 340) + 2(900 - 425)\] 
\[ = 80( - 25) - 10( - 70) + 2(475)\] 
\[ = - 350\] 
\[ D_2 = \begin{vmatrix}9 & 80 & 2 \\ 13 & 90 & 4 \\ 6 & 85 & 3\end{vmatrix}\] 
\[ = 9(270 - 340) - 80(39 - 24) + 2(1105 - 540)\] 
\[ = 9( - 70) - 80(15) + 2(565)\] 
\[ = - 700\] 
\[ D_3 = \begin{vmatrix}9 & 10 & 80 \\ 13 & 5 & 90 \\ 6 & 10 & 85\end{vmatrix}\] 
\[ = 9(425 - 900) - 10(1105 - 540) + 80(130 - 30)\] 
\[ = 9( - 475) - 10(565) + 80(100)\] 
\[ = - 1925\] 
Thus, 
\[x = \frac{D_1}{D} = \frac{- 350}{- 175} = 2\] 
\[y = \frac{D_2}{D} = \frac{- 700}{- 175} = 4\] 
\[z = \frac{D_3}{D} = \frac{- 1925}{- 175} = 11\]
Therefore, the rates of commission on items A, B and C are 2, 4 and 11, respectively.