CBSE 2026 · Region 4 · Set 1 · Q37 · 4 marks
A carpenter needs to design a wooden box in the shape of a cuboid such that the sum of its length and breadth is $\displaystyle 3$ cm more than its height. Twice of its length, thrice of its breadth and its height add up to $\displaystyle 10$ cm. Its breadth added to $\displaystyle 7$ times its height is $\displaystyle 1$ cm less than $\displaystyle 3$ times its length. On the basis of the above information, answer the following questions :(i)Write the equations representing the various dimensions and express them as the matrix equation $\displaystyle \mathrm{AX}=\mathrm{B}$.(ii)Find if $\displaystyle \mathrm{A}^{-1}$ exists. Justify your answer.(iii)Find $\displaystyle \mathrm{A}^{-1}$.Find $\displaystyle \mathrm{A}^{2}+7 \mathrm{I}$.
A carpenter needs to design a wooden box in the shape of a cuboid such that the sum of its length and breadth is $\displaystyle 3$ cm more than its height. Twice of its length, thrice of its breadth and its height add up to $\displaystyle 10$ cm. Its breadth added to $\displaystyle 7$ times its height is $\displaystyle 1$ cm less than $\displaystyle 3$ times its length. On the basis of the above information, answer the following questions :
(i)
Write the equations representing the various dimensions and express them as the matrix equation $\displaystyle \mathrm{AX}=\mathrm{B}$.
(ii)
Find if $\displaystyle \mathrm{A}^{-1}$ exists. Justify your answer.
(iii)
Find $\displaystyle \mathrm{A}^{-1}$.
Find $\displaystyle \mathrm{A}^{2}+7 \mathrm{I}$.
Marking-scheme solution
(i)
$\displaystyle \begin{pmatrix} 1 & 1 & -1 \\ 2 & 3 & 1 \\ -3 & 1 & 7 \end{pmatrix}\begin{pmatrix} l \\ b \\ h \end{pmatrix}=\begin{pmatrix} 3 \\ 10 \\ -1 \end{pmatrix}$
i.e. $\displaystyle A X=B$
(ii)
$\displaystyle |A|=-8 \neq 0$, so $\displaystyle A^{-1}$ exists.
(iii)
$\displaystyle A^{-1}=\dfrac{1}{|A|}(\operatorname{adj} A)=\dfrac{1}{-8}\begin{pmatrix} 20 & -8 & 4 \\ -17 & 4 & -3 \\ 11 & -4 & 1 \end{pmatrix}$
$\displaystyle A^{2}=\begin{pmatrix} 1 & 1 & -1 \\ 2 & 3 & 1 \\ -3 & 1 & 7 \end{pmatrix}\begin{pmatrix} 1 & 1 & -1 \\ 2 & 3 & 1 \\ -3 & 1 & 7 \end{pmatrix}=\begin{pmatrix} 6 & 3 & -7 \\ 5 & 12 & 8 \\ -22 & 7 & 53 \end{pmatrix}$
Thus, $\displaystyle A^{2}+7 I=\begin{pmatrix} 13 & 3 & -7 \\ 5 & 19 & 8 \\ -22 & 7 & 60 \end{pmatrix}$
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CBSE Class 12 Mathematics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.