Class 12 Determinants · Case Study4 Marksmatrix_method_applications
Class 12 Maths · Determinants · Case Study #31
Gautam buys 5 pens,3 bags and 1 instrument box for ₹160. Vikram buys 2 pens,1 bag and 3 instrument boxes for ₹190. Ankur buys 1 pen,2 bags and 4 instrument boxes for ₹250. Let unit costs be x,y,z rupees respectively.
Part (1) 1 Marks
Write the situation as AX=B.
Answer: A=\(\begin{bmatrix}5 & 3 & 1\\2 & 1 & 3\\1 & 2 & 4\end{bmatrix}\), X=[x,y,z]ᵀ, B=[160,190,250]ᵀ.
- The three purchase equations are 5x+3y+z=160, 2x+y+3z=190 and x+2y+4z=250.
- Thus AX=B, with A=\(\begin{bmatrix}5 & 3 & 1\\2 & 1 & 3\\1 & 2 & 4\end{bmatrix}\), X=[x,y,z]ᵀ and B=[160,190,250]ᵀ.
Part (2) 1 Marks
Find det(A).
Answer: −22
- Expand along row 1: 5(4−6)−3(8−3)+(4−1)=−10−15+3.
- Therefore det(A)=−22.
Part (3) 2 Marks
Find A⁻¹.
Answer: \(\begin{bmatrix}1/11 & 5/11 & -4/11\\5/22 & -19/22 & 13/22\\-3/22 & 7/22 & 1/22\end{bmatrix}\)
- Since det(A)=−22≠0, use A⁻¹=adj(A)/det(A).
- The cofactor matrix is \(\begin{bmatrix}-2 & -5 & 3\\-10 & 19 & -7\\8 & -13 & -1\end{bmatrix}\); transpose it to get adj(A)=\(\begin{bmatrix}-2 & -10 & 8\\-5 & 19 & -13\\3 & -7 & -1\end{bmatrix}\).
- Therefore A⁻¹=\(\begin{bmatrix}1/11 & 5/11 & -4/11\\5/22 & -19/22 & 13/22\\-3/22 & 7/22 & 1/22\end{bmatrix}\).
Part (4) 2 Marks
Alternatively, find P=A²−5A.
Answer: \(\begin{bmatrix}7 & 5 & 13\\5 & 8 & 2\\8 & 3 & 3\end{bmatrix}\)
- Compute A²=\(\begin{bmatrix}32 & 20 & 18\\15 & 13 & 17\\13 & 13 & 23\end{bmatrix}\).
- Subtract 5A=\(\begin{bmatrix}25 & 15 & 5\\10 & 5 & 15\\5 & 10 & 20\end{bmatrix}\) entry by entry.
- Therefore P=\(\begin{bmatrix}7 & 5 & 13\\5 & 8 & 2\\8 & 3 & 3\end{bmatrix}\).