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Mathematics · 2026 · 5 marks
CBSE 2026 · Region 3 · Set 1 · Q34
(a) On the inauguration day of a new showroom, a lucky draw was organized and some vouchers of ₹ 1,000 and ₹ 500 were given to the lucky draw…
On the inauguration day of a new showroom, a lucky draw was organized and some vouchers of ₹ $\displaystyle 1,000$ and ₹ $\displaystyle 500$ were given to the lucky draw winners.
A total of $\displaystyle 60$ vouchers were given on the day. The number of ₹ $\displaystyle 1,000$ vouchers added to $\displaystyle 3$ times the number of ₹ $\displaystyle 500$ vouchers, gives 100. Express the given information as a system of linear equations in two variables. Hence, find the number of vouchers of each type by matrix method.
Given that $\displaystyle \mathrm{P}=\left[\begin{array}{cc}2 & -1 \\ 3 & 4\end{array}\right], \mathrm{Q}=\left[\begin{array}{cc}5 & 2 \\ 7 & 4\end{array}\right]$ and $\displaystyle \mathrm{R}=\left[\begin{array}{cc}2 & 5 \\ 3 & 8\end{array}\right]$, find a matrix S such that $\displaystyle \mathrm{PQ}-\mathrm{RS}$ is a null matrix.
Marking-scheme solution
Let, number of ₹ $\displaystyle 1000$ vouchers $\displaystyle =x$ and number of ₹ $\displaystyle 500$ vouchers $\displaystyle =y$
Hence, $\displaystyle x+y=60$ and $\displaystyle x+3 y=100$
In matrix form, $\displaystyle \begin{bmatrix}1 & 1 \\ 1 & 3\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}=\begin{bmatrix}60 \\ 100\end{bmatrix}$
$\displaystyle \Rightarrow AX=B$ where $\displaystyle A=\begin{bmatrix}1 & 1 \\ 1 & 3\end{bmatrix}, B=\begin{bmatrix}60 \\ 100\end{bmatrix}$ and $\displaystyle X=\begin{bmatrix}x \\ y\end{bmatrix}$
$\displaystyle \Rightarrow X=A^{-1} B \quad \ldots(1)$
$\displaystyle |A|=2 \quad \because|A| \neq 0 \Rightarrow A$ is an invertible matrix.
adj $\displaystyle A=\begin{bmatrix}3 & -1 \\ -1 & 1\end{bmatrix}$ and $\displaystyle A^{-1}=\dfrac{1}{2}\begin{bmatrix}3 & -1 \\ -1 & 1\end{bmatrix}$
By eq. ($\displaystyle 1$), $\displaystyle \begin{bmatrix}x \\ y\end{bmatrix}=\dfrac{1}{2}\begin{bmatrix}3 & -1 \\ -1 & 1\end{bmatrix}\begin{bmatrix}60 \\ 100\end{bmatrix}=\begin{bmatrix}40 \\ 20\end{bmatrix}$
On comparing, $\displaystyle x=40$ and $\displaystyle y=20$
Hence, number of ₹ $\displaystyle 1000$ vouchers $\displaystyle =40$ and number of ₹ $\displaystyle 500$ vouchers $\displaystyle =20$
Let $\displaystyle S=\begin{bmatrix}a & b \\ c & d\end{bmatrix}$
$\displaystyle PQ=\begin{bmatrix}3 & 0 \\ 43 & 22\end{bmatrix}$
$\displaystyle RS=\begin{bmatrix}2 a+5 c & 2 b+5 d \\ 3 a+8 c & 3 b+8 d\end{bmatrix}$
Given that $\displaystyle PQ-RS=$ null matrix $\displaystyle \Rightarrow PQ=RS$
On comparing corresponding elements,
$\displaystyle 2 a+5 c=3, \quad 2 b+5 d=0, \quad 3 a+8 c=43, \quad 3 b+8 d=22$
By solving, $\displaystyle a=-191, b=-110, c=77, d=44$
Hence, $\displaystyle S=\begin{bmatrix}-191 & -110 \\ 77 & 44\end{bmatrix}$
Alternative Method:
Given that $\displaystyle PQ-RS=$ null matrix $\displaystyle \Rightarrow RS=PQ$
$\displaystyle \begin{bmatrix}2 & 5 \\ 3 & 8\end{bmatrix} S=\begin{bmatrix}3 & 0 \\ 43 & 22\end{bmatrix}$
$\displaystyle \Rightarrow S=\begin{bmatrix}2 & 5 \\ 3 & 8\end{bmatrix}^{-1}\begin{bmatrix}3 & 0 \\ 43 & 22\end{bmatrix}$
$\displaystyle \Rightarrow S=\begin{bmatrix}8 & -5 \\ -3 & 2\end{bmatrix}\begin{bmatrix}3 & 0 \\ 43 & 22\end{bmatrix}$
$\displaystyle \Rightarrow S=\begin{bmatrix}-191 & -110 \\ 77 & 44\end{bmatrix}$
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CBSE Class 12 Mathematics past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.