CBSE Class 12 Mathematics 2026: question 32

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Step-by-step solution

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1. Let the costs of one apple, one orange and one banana be ₹$x$, ₹$y$ and ₹$z$. Then:

   $4x+3y+2z=60$

   $2x+4y+6z=90$

   $6x+2y+3z=70$

   1 mark

2. In matrix form $AX=B$ with $A=\left[\begin{array}{ccc}4& 3& 2\\ 2& 4& 6\\ 6& 2& 3\end{array}\right]$, $X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$, $B=\left[\begin{array}{c}60\\ 90\\ 70\end{array}\right]$.

   $|A|=4(12-12)-3(6-36)+2(4-24)=0+90-40=50\ne 0$, so $A^{-1}$ exists and $X=A^{-1}B$.

   1 mark

3. Cofactors:

   • $A_{11}=0$, $A_{12}=30$, $A_{13}=-20$
   • $A_{21}=-5$, $A_{22}=0$, $A_{23}=10$
   • $A_{31}=10$, $A_{32}=-20$, $A_{33}=10$

   So $\text{adj}A=\left[\begin{array}{ccc}0& -5& 10\\ 30& 0& -20\\ -20& 10& 10\end{array}\right]$

   1½ marks

4. $A^{-1}=\frac{1}{50}\left[\begin{array}{ccc}0& -5& 10\\ 30& 0& -20\\ -20& 10& 10\end{array}\right]$

   ½ mark

5. $X=\frac{1}{50}\left[\begin{array}{c}0-450+700\\ 1800+0-1400\\ -1200+900+700\end{array}\right]=\frac{1}{50}\left[\begin{array}{c}250\\ 400\\ 400\end{array}\right]=\left[\begin{array}{c}5\\ 8\\ 8\end{array}\right]$

   So one apple costs ₹5, one orange costs ₹8 and one banana costs ₹8.

   1 mark

Answer: Apple ₹5, orange ₹8, banana ₹8

Revise the chapter

• Chapter: Matrices (notes, formulas, where marks are lost and every past question)