CBSE Class 12 Mathematics 2023: question 37

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Step-by-step solution

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1. (I) Let the prices (in ₹) of a pen, a bag and an instrument box be $x,y,z$. Then

   • $5x+3y+z=160$
   • $2x+y+3z=190$
   • $x+2y+4z=250$

   In matrix form, $AX=B$ with
   $$A=\left[\begin{array}{ccc}5& 3& 1\\ 2& 1& 3\\ 1& 2& 4\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right],B=\left[\begin{array}{c}160\\ 190\\ 250\end{array}\right]$$

   1 mark

2. (II) Expanding along the first row:
   $$|A|=5(4-6)-3(8-3)+1(4-1)=-10-15+3=-22$$

   1 mark

3. (III)(a) Cofactors:

   • $C_{11}=-2,C_{12}=-5,C_{13}=3$
   • $C_{21}=-10,C_{22}=19,C_{23}=-7$
   • $C_{31}=8,C_{32}=-13,C_{33}=-1$

   The adjoint is the transpose of the cofactor matrix:
   $$\text{adj}A=\left[\begin{array}{ccc}-2& -10& 8\\ -5& 19& -13\\ 3& -7& -1\end{array}\right]$$

   1 mark

4. Since $|A|=-22\ne 0$, $A^{-1}$ exists and
   $$A^{-1}=\frac{1}{|A|}\text{adj}A=-\frac{1}{22}\left[\begin{array}{ccc}-2& -10& 8\\ -5& 19& -13\\ 3& -7& -1\end{array}\right]$$

   1 mark

5. (III)(b) Compute $A^2$ and $5A$:
   $$A^2=\left[\begin{array}{ccc}32& 20& 18\\ 15& 13& 17\\ 13& 13& 23\end{array}\right],5A=\left[\begin{array}{ccc}25& 15& 5\\ 10& 5& 15\\ 5& 10& 20\end{array}\right]$$

   1½ marks

6. Subtract:
   $$P=A^2-5A=\left[\begin{array}{ccc}7& 5& 13\\ 5& 8& 2\\ 8& 3& 3\end{array}\right]$$

   ½ mark

Answer: (I) A = [[5,3,1],[2,1,3],[1,2,4]], X = [x,y,z]ᵀ, B = [160,190,250]ᵀ; (II) |A| = −22; (III)(a) A⁻¹ = −(1/22)[[−2,−10,8],[−5,19,−13],[3,−7,−1]]; (III)(b) P = [[7,5,13],[5,8,2],[8,3,3]]

Revise the chapter

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