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#### Please Solve RD Sharma Class 12 Chapter Algebra of Matrices Exercise 4.2 Question 22 Maths Textbook Solution.

Answer:$X = 30000\: and \: Y = 15000$
Hint: Try to solve monthly saving of Aryan and Babban using the ratio.
Recall that the solution to the system of equations that can be written as $AX = B.$$AX = B.$
Given: The monthly income of Aryan and Babban are in the ratio $3:4.$The monthly expenditures are in the ratio$5:7.$Saving of each month is $Rs. 15000.$
Solution:
Let the monthly income of Aryan and Babban be $3X$ and $4X$ respectively. Suppose that monthly expenditures are $5Y$ and$7Y$ respectively.
Since, each save $Rs. 15000$ per month,

Monthly saving of Aryan:$3X - 5Y = 15000$

Monthly saving of Babban:$4X - 7Y = 15000$
The above system of equations can be written in the matrix form as follows:

$\begin{bmatrix} 3 &-5 \\ 4 & -7 \end{bmatrix}\begin{bmatrix} X\\Y \end{bmatrix}= \begin{bmatrix} 15000\\15000 \end{bmatrix}$
Now,
$A = \begin{bmatrix} 3 & -5\\ 4 & -7 \end{bmatrix} = 3\times \left ( -7 \right )-4\times \left ( -5 \right )$

$= -21 - (-20)$

$= -1$

Adj$(A) = \begin{bmatrix} -7 &-4 \\ 5 & 3 \end{bmatrix}^{T} = \begin{bmatrix} -7 &5 \\ -4& 3 \end{bmatrix}$
So,
$A^{-1} = \frac{1}{\left | A \right |} Adj (A)$

$= (-1) \begin{bmatrix} - 7& 5\\ -4& 3 \end{bmatrix}$

$= \begin{bmatrix} 7& -5\\ 4& -3 \end{bmatrix}$

$\therefore X = A^{-1}B$

$\begin{bmatrix} X\\Y \end{bmatrix} = \begin{bmatrix} 7 & -5\\ 4 & -3 \end{bmatrix}\begin{bmatrix} 15000\\15000 \end{bmatrix}$

$\begin{bmatrix} X\\Y \end{bmatrix} = \begin{bmatrix} 105000 &-75000 \\ 60000 & -45000 \end{bmatrix}$

$\begin{bmatrix} X\\Y \end{bmatrix} = \begin{bmatrix} 30000\\15000 \end{bmatrix}$

$X = 30000, Y = 15000$
Therefore,
Monthly income of Aryan $= 3X$
$\! \! \! \! \! \! \! \! \! = 3 (30000)\\ = Rs. 90,000$
Monthly income of Aryan$= 4X$
$\! \! \! \! \! \! \! \! = 4 (30000)\\ = Rs. 1,20,000$
Note:
From this problem, we are encouraged to understand the power of savings. We should save certain part of our monthly income for the future.