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Mathematics · 2025 · 4 marks
CBSE 2025 · Region 6 · Set 1 · Q36
Three students, Neha, Rani and Sam go to a market to purchase stationery items. Neha buys 4 pens, 3 notepads and 2 erasers and pays ₹ 60. Rani buys 2…
Three students, Neha, Rani and Sam go to a market to purchase stationery items. Neha buys $\displaystyle 4$ pens, $\displaystyle 3$ notepads and $\displaystyle 2$ erasers and pays ₹ $\displaystyle 60$ . Rani buys $\displaystyle 2$ pens, $\displaystyle 4$ notepads and $\displaystyle 6$ erasers for ₹ $\displaystyle 90$ . Sam pays ₹ $\displaystyle 70$ for $\displaystyle 6$ pens, $\displaystyle 2$ notepads and $\displaystyle 3$ erasers. Based upon the above information, answer the following questions :
(i)
Form the equations required to solve the problem of finding the price of each item, and express it in the matrix form $\displaystyle \mathrm{AX}=\mathrm{B}$.
(ii)
Find $\displaystyle |\mathrm{A}|$ and confirm if it is possible to find $\displaystyle \mathrm{A}^{-1}$.
(iii)
(a)
Find $\displaystyle \mathrm{A}^{-1}$, if possible, and write the formula to find X .
OR
(b)
Find $\displaystyle \mathrm{A}^{2}-8 \mathrm{I}$, where I is an identity matrix. Case Study - $\displaystyle 2$
Marking-scheme solution
(i)
Let the price of each pen, notepad, eraser be ₹x, ₹y and ₹z respectively
Given system in the form $\displaystyle \mathrm{AX}=\mathrm{B}$ is $\displaystyle \left(\begin{array}{lll}4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3\end{array}\right)\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{l}60 \\ 90 \\ 70\end{array}\right)$
(ii)
$\displaystyle |\mathrm{A}|=50 \neq 0$, hence $\displaystyle \mathrm{A}^{-1}$ exists
(iii)
$\displaystyle \mathrm{A}^{-1}=\frac{\operatorname{adj} \mathrm{A}}{|\mathrm{A}|}=\frac{1}{50}\left(\begin{array}{ccc}0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10\end{array}\right)$
$\displaystyle \mathrm{X}=\mathrm{A}^{-1} \mathrm{B}$
$\displaystyle \mathrm{A}^{2}=\left(\begin{array}{lll}4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3\end{array}\right)\left(\begin{array}{lll}4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3\end{array}\right)=\left(\begin{array}{lll}34 & 28 & 32 \\ 52 & 34 & 46 \\ 46 & 32 & 33\end{array}\right)$
$\displaystyle \mathrm{A}^{2}-8 \mathrm{I}=\left(\begin{array}{lll}26 & 28 & 32 \\ 52 & 26 & 46 \\ 46 & 32 & 25\end{array}\right)$
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CBSE Class 12 Mathematics past-paper question from the 2025 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.