CBSE 2023 · Region 5 · Set 1 · Q37 · 4 marks
Gautam buys $\displaystyle 5$ pens, $\displaystyle 3$ bags and $\displaystyle 1$ instrument box and pays a sum of ₹ $\displaystyle 160$ . From the same shop, Vikram buys $\displaystyle 2$ pens, $\displaystyle 1$ bag and $\displaystyle 3$ instrument boxes and pays a sum of ₹ $\displaystyle 190$ . Also Ankur buys $\displaystyle 1$ pen, $\displaystyle 2$ bags and $\displaystyle 4$ instrument boxes and pays a sum of ₹ $\displaystyle 250$ . Based on the above information, answer the following questions :(I)Convert the given above situation into a matrix equation of the form $\displaystyle \mathrm{AX}=\mathrm{B}$.(II)Find $\displaystyle |\mathrm{A}|$.Find $\displaystyle \mathrm{A}^{-1}$.Determine $\displaystyle \mathrm{P}=\mathrm{A}^{2}-5 \mathrm{~A}$. Case Study-III
Gautam buys $\displaystyle 5$ pens, $\displaystyle 3$ bags and $\displaystyle 1$ instrument box and pays a sum of ₹ $\displaystyle 160$ . From the same shop, Vikram buys $\displaystyle 2$ pens, $\displaystyle 1$ bag and $\displaystyle 3$ instrument boxes and pays a sum of ₹ $\displaystyle 190$ . Also Ankur buys $\displaystyle 1$ pen, $\displaystyle 2$ bags and $\displaystyle 4$ instrument boxes and pays a sum of ₹ $\displaystyle 250$ . Based on the above information, answer the following questions :
(I)
Convert the given above situation into a matrix equation of the form $\displaystyle \mathrm{AX}=\mathrm{B}$.
(II)
Find $\displaystyle |\mathrm{A}|$.
Find $\displaystyle \mathrm{A}^{-1}$.
Determine $\displaystyle \mathrm{P}=\mathrm{A}^{2}-5 \mathrm{~A}$. Case Study-III
Marking-scheme solution
(I)
Matrix equation is $\displaystyle \mathrm{AX}=\mathrm{B}$, where
$$\mathrm{A}=\left[\begin{array}{lll}
5 & 3 & 1
2 & 1 & 3
1 & 2 & 4
\end{array}\right], X=\left[\begin{array}{l}
x
y
z
\end{array}\right], \mathrm{B}=\left[\begin{array}{l}
160
190
250
\end{array}\right]
$$where x is the number of pens bought, y the number of bags and z the number of instrument boxes.
(II)
$\displaystyle |\mathrm{A}|=5(4-6)-3(8-3)+1(4-1)=-22$
(III)
$\displaystyle \operatorname{adj}(\mathrm{A})=\left[\begin{array}{ccc} -2 & -5 & 3 \\ -10 & 19 & -7 \\ 8 & -13 & -1 \end{array}\right]^{\prime}=\left[\begin{array}{ccc} -2 & -10 & 8 \\ -5 & 19 & -13 \\ 3 & -7 & -1 \end{array}\right]$
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CBSE Class 12 Mathematics past-paper question from the 2023board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.