Case Study - 3 A scholarship is a sum of money provided — CBSE Class 12 Mathematics 2024 PYQ

Question

Q. Case Study - 3 A scholarship is a sum of money provided to a student to help him or her pay for education. Some students are granted scholarships based on their academic achievements, while others are rewarded based on their financial needs. Every year a school offers scholarships to girl children and meritorious achievers based on certain criteria. In the session 2022-23, the school offered monthly scholarship of ₹ 3,000 each to some girl students and ₹ 4,000 each to meritorious achievers in academics as well as sports. In all, 50 students were given the scholarships and monthly expenditure incurred by the school on scholarships was ₹ 1,80,000. Based on the above information, answer the following questions:

(i) Express the given information algebraically using matrices. [1]

(ii) Check whether the system of matrix equations so obtained is consistent or not. [1]

(iii)

(a) Find the number of scholarships of each kind given by the school, using matrices. [2]

(OR)

(iii)

(b) Had the amount of scholarship given to each girl child and meritorious student been interchanged, what would be the monthly expenditure incurred by the school? [2]

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Part (a)Concept understanding — Inverse Matrix Method

The Inverse Matrix Method

Many problems reduce to a system of linear equations, for example

\begin{aligned}2x+3y&=8\\x-\ \ y&=-1.\end{aligned}

The inverse matrix method solves such a system by writing it as a single matrix equation and then undoing the coefficient matrix with its inverse — the matrix analogue of dividing.

Writing the system as $AX=B$

Collect the coefficients, the unknowns and the constants:

A=\begin{pmatrix}2&3\\1&-1\end{pmatrix},\quad X=\begin{pmatrix}x\\y\end{pmatrix},\quad B=\begin{pmatrix}8\\-1\end{pmatrix},

so the whole system becomes $AX=B$ .

The idea

For numbers, $ax=b$ gives $x=a^{-1}b$ provided $a\neq0$ . The same works for matrices: if $A$ is invertible, multiply $AX=B$ on the left by $A^{-1}$ :

A^{-1}(AX)=A^{-1}B\;\Rightarrow\;IX=A^{-1}B\;\Rightarrow\;X=A^{-1}B.

X=A^{-1}B

Multiplying on the left matters — matrix products do not commute, so $BA^{-1}$ would be wrong.

When it works

The inverse $A^{-1}$ exists only when $\det A\neq0$ , so:

$\det A\neq0$ : the system is consistent with the unique solution $X=A^{-1}B$ .
$\det A=0$ : no inverse; the system is either inconsistent (no solution) or has infinitely many — handle it by another method.

Worked steps

For the system above, $\det A=(2)(-1)-(3)(1)=-5\neq0$ , and

A^{-1}=\frac{1}{-5}\begin{pmatrix}-1&-3\\-1&2\end{pmatrix}.

Then …

Part (b)Concept understanding — Inverse Matrix Method

The Inverse Matrix Method

Many problems reduce to a system of linear equations, for example

\begin{aligned}2x+3y&=8\\x-\ \ y&=-1.\end{aligned}

The inverse matrix method solves such a system by writing it as a single matrix equation and then undoing the coefficient matrix with its inverse — the matrix analogue of dividing.

Writing the system as $AX=B$

Collect the coefficients, the unknowns and the constants:

A=\begin{pmatrix}2&3\\1&-1\end{pmatrix},\quad X=\begin{pmatrix}x\\y\end{pmatrix},\quad B=\begin{pmatrix}8\\-1\end{pmatrix},

so the whole system becomes $AX=B$ .

The idea

For numbers, $ax=b$ gives $x=a^{-1}b$ provided $a\neq0$ . The same works for matrices: if $A$ is invertible, multiply $AX=B$ on the left by $A^{-1}$ :

A^{-1}(AX)=A^{-1}B\;\Rightarrow\;IX=A^{-1}B\;\Rightarrow\;X=A^{-1}B.

X=A^{-1}B

Multiplying on the left matters — matrix products do not commute, so $BA^{-1}$ would be wrong.

When it works

The inverse $A^{-1}$ exists only when $\det A\neq0$ , so:

$\det A\neq0$ : the system is consistent with the unique solution $X=A^{-1}B$ .
$\det A=0$ : no inverse; the system is either inconsistent (no solution) or has infinitely many — handle it by another method.

Worked steps

For the system above, $\det A=(2)(-1)-(3)(1)=-5\neq0$ , and

A^{-1}=\frac{1}{-5}\begin{pmatrix}-1&-3\\-1&2\end{pmatrix}.

Then …

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Modelling gives $x+y=50$ , $3x+4y=180$ . The coefficient determinant is $1\ne0$ , so the system is consistent; $A^{-1}B$ gives $x=20$ girl-student and $y=30$ meritorious scholarships.

Part (b): with amounts swapped, expenditure $=4000(20)+3000(30)=$ Rs $1{,}70{,}000$ .

Let $x$ be the number of girl-student scholarships and $y$ the number of meritorious-achiever scholarships. "In all $50$ students" gives $x+y=50$ ; the total spend Rs 1,80,000 at Rs 3000 and Rs 4000 gives $3000x+4000y=180000$ , i.e. $3x+4y=180$ .

Part (a)

(i) Matrix form $AX=B$ :

\begin{pmatrix}1&1\\3&4\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}50\\180\end{pmatrix}.

(ii) Consistency: $\det A=(1)(4)-(1)(3)=1$ . Since $\det A=1\ne0$ , $A$ is invertible and the system has a unique solution — it is consistent.

(iii)(a) Solve by $X=A^{-1}B$ . With $\det A=1$ ,

A^{-1}=\frac1{\det A}\begin{pmatrix}4&-1\\-3&1\end{pmatrix}=\begin{pmatrix}4&-1\\-3&1\end{pmatrix},

$X=\begin{pmatrix}4&-1\\-3&1\end{pmatrix}\begin{pmatrix}50\\180\end{pmatrix}=\begin{pmatrix}200-180\\-150+180\end{pmatrix}=\begin{pmatrix}20\\30\end{pmatrix}.$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 markMCQ
$A$
(a) $\det(A)$
(b) $\frac{1}{\det(A)}$
(c) $1$
(d) $0$
›Reveal solutionSolution

$\det(A^{-1})=\dfrac{1}{\det A}$ .

From $AA^{-1}=I$ , $\det(A)\det(A^{-1})=\det(I)=1$ , so $\det(A^{-1})=\dfrac{1}{\det(A)}$ . (This holds for a …

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CBSE 2026Set ANNUAL1 mark
$AX = B$
›Reveal solutionSolution

The unique solution of $AX=B$ (when $|A| e0$ ) is $X=A^{-1}B$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$A = \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}$
(a) $19$
(b) $\dfrac{1}{19}$
(c) $-19$
(d) $-\dfrac{1}{19}$
›Reveal solutionSolution

Compute |A| and adj(A), form A⁻¹, and compare it entry-by-entry with kA.

$A = \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}$ , $|A| = 2(-2) - 3(5) = -4-15 = -19$

$\text{adj}(A) = \begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix}$

…

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CBSE 2024Set D1 markMCQ
$A = \begin{bmatrix} 2 & -3 \\ 4 & 6 \end{bmatrix}$
(a) $\begin{bmatrix} \frac{1}{4} & \frac{1}{8} \\ \frac{1}{6} & \frac{1}{12} \end{bmatrix}$
(b) $\begin{bmatrix} \frac{1}{4} & \frac{1}{8} \\ -\frac{1}{6} & \frac{1}{12} \end{bmatrix}$
(c) $\begin{bmatrix} 4 & 8 \\ 6 & 12 \end{bmatrix}$
(d) $\begin{bmatrix} 4 & 8 \\ -6 & 12 \end{bmatrix}$
›Reveal solutionSolution

For a $2\times2$ matrix $A^{-1} = \frac{1}{\det A}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ .

Here $A = \begin{bmatrix} 2 & -3 \\ 4 & 6 \end{bmatrix}$ , so $\det A = (2)(6) - (-3)(4) = 12 + 12 = 24.$

The adjoint (swap diagonal, negate off-diagonal) is $\begin{bmatrix} 6 & 3 \\ -4 & 2 \end{bmatrix}$ . Hence …

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Part (a)

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