A shop sells pens, notebooks and files, each at a fixed price. Customer P buys 2 pens, 3 notebooks and 1 file for Rs. 170. Customer Q buys 1 pen, 2 notebooks and 2 files for Rs. 160. Customer R buys 3 pens, 1 notebook and 2 files for Rs. 170. Using the matrix method, the total price of one pen, one notebook and one file is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Rs. 90
Let the prices of a pen, a notebook and a file be Rs. x, Rs. y and Rs. z.
\(2x + 3y + z = 170\)
\(x + 2y + 2z = 160\)
\(3x + y + 2z = 170\)
In matrix form AX = B, with:
\(\displaystyle A = \begin{bmatrix}2 & 3 & 1\\ 1 & 2 & 2\\ 3 & 1 & 2\end{bmatrix},\quad B = \begin{bmatrix}170\\ 160\\ 170\end{bmatrix}\)
\(|A| = 2(4 - 2) - 3(2 - 6) + 1(1 - 6)\)
\(|A| = 4 + 12 - 5 = 11\)
The cofactors are:
\(A_{11} = 2,\ A_{12} = 4,\ A_{13} = -5\)
\(A_{21} = -5,\ A_{22} = 1,\ A_{23} = 7\)
\(A_{31} = 4,\ A_{32} = -3,\ A_{33} = 1\)
The adjoint is the transpose of the cofactor matrix, and \(\displaystyle X = A^{-1}B = \frac{1}{|A|}(\mathrm{adj}\,A)B\):
\(\displaystyle X = \frac{1}{11}\begin{bmatrix}2 & -5 & 4\\ 4 & 1 & -3\\ -5 & 7 & 1\end{bmatrix}\begin{bmatrix}170\\ 160\\ 170\end{bmatrix}\)
\(\displaystyle X = \frac{1}{11}\begin{bmatrix}340 - 800 + 680\\ 680 + 160 - 510\\ -850 + 1120 + 170\end{bmatrix} = \frac{1}{11}\begin{bmatrix}220\\ 330\\ 440\end{bmatrix}\)
\(x = 20,\ y = 30,\ z = 40\)
\(x + y + z = 20 + 30 + 40 = 90\)
Check with Q: 20 + 60 + 80 = 160.