Class 12 Determinants · Case Study4 Marksmatrix_method_applications
Class 12 Maths · Determinants · Case Study #32
Rahul buys 1 kg each of potato,onion and brinjal for ₹21. Ravi buys 4 kg potato,3 kg onion and 2 kg brinjal for ₹60. Rakesh buys 6 kg potato,2 kg onion and 3 kg brinjal for ₹70. Let the respective prices per kilogram be x,y,z rupees.
Part (1) 1 Marks
Write a system of linear equations.
Answer: x+y+z=21; 4x+3y+2z=60; 6x+2y+3z=70.
- For each purchase, multiply each quantity by its unit price and add the three costs.
- The system is x+y+z=21, 4x+3y+2z=60 and 6x+2y+3z=70.
Part (2) 1 Marks
Express the system as AX=B.
Answer: A=\(\begin{bmatrix}1 & 1 & 1\\4 & 3 & 2\\6 & 2 & 3\end{bmatrix}\), X=[x,y,z]ᵀ, B=[21,60,70]ᵀ.
- The coefficients form the rows [1,1,1],[4,3,2],[6,2,3].
- Thus \(\begin{bmatrix}1 & 1 & 1\\4 & 3 & 2\\6 & 2 & 3\end{bmatrix}\)[x,y,z]ᵀ=[21,60,70]ᵀ, which is AX=B.
Part (3) 2 Marks
Find A⁻¹.
Answer: \(\begin{bmatrix}-1 & 1/5 & 1/5\\0 & 3/5 & -2/5\\2 & -4/5 & 1/5\end{bmatrix}\)
- Expand det(A)=1(9−4)−1(12−12)+1(8−18)=−5≠0.
- Signed minors and transposition give adj(A)=\(\begin{bmatrix}5 & -1 & -1\\0 & -3 & 2\\-10 & 4 & -1\end{bmatrix}\).
- Therefore A⁻¹=adj(A)/(−5)=\(\begin{bmatrix}-1 & 1/5 & 1/5\\0 & 3/5 & -2/5\\2 & -4/5 & 1/5\end{bmatrix}\).
Part (4) 2 Marks
Alternatively, find all three prices.
Answer: Potato ₹5/kg; onion ₹8/kg; brinjal ₹8/kg.
- The coefficient determinant is −5≠0, so X=A⁻¹B.
- Using A⁻¹=\(\begin{bmatrix}-1 & 1/5 & 1/5\\0 & 3/5 & -2/5\\2 & -4/5 & 1/5\end{bmatrix}\), multiply by [21,60,70]ᵀ to obtain [5,8,8]ᵀ.
- Thus potato,onion,brinjal cost ₹5,₹8,₹8 per kg respectively; these satisfy all purchase totals.