A standard Class 12 matrix-method application problem — not a JEE PYQ.
Problem
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 the matrix method.
Key insight. “Ratio 3:4” and “ratio 5:7” don’t mean the incomes and expenditures are 3, 4, 5 and 7 rupees — they mean income $=3x$ and $4x$ for some common scale $x$, and expenditure $=5y$ and $7y$ for some (different) common scale $y$. Setting it up this way turns two vague ratios into a clean pair of linear equations in $x$ and $y$.
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Approach
Since saving = income − expenditure, and both people save exactly ₹15,000, writing income as $3x, 4x$ and expenditure as $5y, 7y$ turns “savings = 15,000” into two linear equations. Those equations are then solved using the matrix inverse method: $AX = B \implies X = A^{-1}B$.
Solution
Step 1 — Set up the equations
Let Aryan’s income be $3x$ and Babban’s be $4x$ (common ratio $3:4$). Let Aryan’s expenditure be $5y$ and Babban’s be $7y$ (common ratio $5:7$).
Since savings $=$ income $-$ expenditure, and both save ₹15,000:
$$3x – 5y = 15000$$ $$4x – 7y = 15000$$
Step 2 — Write in matrix form AX = B
$$\begin{bmatrix} 3 & -5 \\ 4 & -7 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 15000 \\ 15000 \end{bmatrix}$$
Step 3 — Find A⁻¹
$$\det(A) = 3(-7) – (-5)(4) = -21 + 20 = -1$$
$$\text{adj}(A) = \begin{bmatrix} -7 & 5 \\ -4 & 3 \end{bmatrix}$$
$$A^{-1} = \frac{1}{\det(A)}\,\text{adj}(A) = \frac{1}{-1}\begin{bmatrix} -7 & 5 \\ -4 & 3 \end{bmatrix} = \begin{bmatrix} 7 & -5 \\ 4 & -3 \end{bmatrix}$$
Step 4 — Solve X = A⁻¹B
$$\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 7 & -5 \\ 4 & -3 \end{bmatrix}\begin{bmatrix} 15000 \\ 15000 \end{bmatrix} = \begin{bmatrix} 7(15000) – 5(15000) \\ 4(15000) – 3(15000) \end{bmatrix} = \begin{bmatrix} 30000 \\ 15000 \end{bmatrix}$$
So $x = 30000$ and $y = 15000$.
Step 5 — Recover the actual incomes
$$\text{Aryan’s income} = 3x = 3(30000) = ₹90{,}000$$ $$\text{Babban’s income} = 4x = 4(30000) = ₹120{,}000$$
Answer
Aryan’s monthly income is $₹90{,}000$ and Babban’s is $₹120{,}000$.
Common mistakes
- Using the same scale variable for both income and expenditure. Income and expenditure follow different ratios (3:4 versus 5:7), so they need two independent variables, $x$ and $y$ — not a single shared one.
- Sign errors in the adjoint. For a $2\times2$ matrix $\begin{bmatrix}a&b\\c&d\end{bmatrix}$, the adjoint is $\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$ — it’s easy to forget to negate the off-diagonal entries.
Practise next
- Two friends’ monthly incomes are in the ratio $5:6$ and expenditures in the ratio $3:4$. If each saves ₹6,000 per month, find their incomes using the same matrix method.
Show answer
₹15,000 and ₹18,000. Let the incomes be $5x$ and $6x$, the expenditures $3y$ and $4y$. Saving is income minus expenditure, so
$$5x-3y=6000,\qquad 6x-4y=6000.$$
In matrix form $\begin{pmatrix}5&-3\\6&-4\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}6000\\6000\end{pmatrix}$, and the determinant is $-20+18=-2$, so the inverse exists and $x=y=3000$.
The incomes are $5(3000)=₹15{,}000$ and $6(3000)=₹18{,}000$, with expenditures ₹9,000 and ₹12,000 — each saving ₹6,000, as required.
Doubts are answered by Shiwam, usually within a day. Ask about this question specifically — a general question about the chapter is better asked in class.