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

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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 ratio5:7.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 and7Y 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.

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