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The Monthly Incomes of Aryan and Babban Are in the Ratio 3 : 4 and Their Monthly Expenditures Are in the Ratio 5 : 7. If Each Saves ₹ 15,000 per Month, Find Their Monthly Incomes Using Matrix Method. - Mathematics

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Question

The monthly incomes of Aryan and Babban are in the ratio 3 : 4 and their monthly expenditures are in the ratio 5 : 7. If each saves ₹ 15,000 per month, find their monthly incomes using matrix method. This problem reflects which value?

Sum

Solution

Let the monthly incomes of Aryan and Babban be 3x and 4x, respectively.

Suppose their monthly expenditures are 5y and 7y, respectively.

Since each saves Rs 15,000 per month, 

Monthly saving of Aryan:  \[ 3x - 5y = 15, 000\]

Monthly saving of Babban: \[4x - 7y = 15, 000\]
The above system of equations can be written in the matrix form as follows:
`= [[3   -5 ],[4  -7   ]]``[[x] ,[ y ]]` =`[[15000],[15000]]`
or,
AX = B, where
`A= [[3   -5 ],[4  -7   ]]`,` X=[[x] ,[ y ]]` and `B = [[15000],[15000]]`

Now,

\[\left| A \right| = \begin{vmatrix}3 & - 5 \\ 4 & - 7\end{vmatrix} = - 21 - \left( - 20 \right) = - 1\]

Adj A=

\[\begin{bmatrix}- 7 & - 4 \\ 5 & 3\end{bmatrix}^T = \begin{bmatrix}- 7 & 5 \\ - 4 & 3\end{bmatrix}\]

So,

\[A^{- 1} = \frac{1}{\left| A \right|}adjA = - 1\begin{bmatrix}- 7 & 5 \\ - 4 & 3\end{bmatrix} = \begin{bmatrix}7 & - 5 \\ 4 & - 3\end{bmatrix}\]

\[\therefore X = A^{- 1} B\]

⇒  `[[x] ,[ y ]]` = `[[7      5],[4    -3]]` `[[15000 ] , [15000]]`

⇒  `[[x] ,[ y ]]` = `[[105000-75000],[60000-45000]]`

⇒  `[[x] ,[ y ]]` =`[[30000],[15000]]`

⇒ x = 30, 000 and y= 15, 000

Therefore,
Monthly income of Aryan =

\[3 \times \text{Rs 30, 000} = \text{Rs 90, 000}\]

Monthly income of Babban =

\[4 \times Rs 30, 000 = Rs 1, 20, 000\]

From this problem, we are encouraged to understand the power of savings. We should save certain part of our monthly income for the future.

 
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Chapter 5: Algebra of Matrices - Exercise 5.3 [Page 48]

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RD Sharma Mathematics [English] Class 12
Chapter 5 Algebra of Matrices
Exercise 5.3 | Q 78 | Page 48

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