A shop sells pens, notebooks and files. The cost of 3 pens, 2 notebooks and 1 file is Rs. 105; the cost of 1 pen, 1 notebook and 1 file is Rs. 75; and the cost of 2 pens, 1 notebook and 3 files is Rs. 180. Using the matrix method, the prices (in Rs.) of one pen, one notebook and one file are respectively
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 5, 20, 50
Let the prices of one pen, one notebook and one file be Rs. x, Rs. y and Rs. z. Then
\( 3x + 2y + z = 105 \)
\( x + y + z = 75 \)
\( 2x + y + 3z = 180 \)
In the form AX = B:
\( \displaystyle A = \begin{bmatrix}3 & 2 & 1\\ 1 & 1 & 1\\ 2 & 1 & 3\end{bmatrix},\quad B = \begin{bmatrix}105\\ 75\\ 180\end{bmatrix} \)
\( |A| = 3(3 - 1) - 2(3 - 2) + 1(1 - 2) = 6 - 2 - 1 = 3 \)
Since \(|A| \ne 0\), the solution is unique and \(X = A^{-1}B\). The cofactors are
\( C_{11} = 2,\ C_{12} = -1,\ C_{13} = -1 \)
\( C_{21} = -5,\ C_{22} = 7,\ C_{23} = 1 \)
\( C_{31} = 1,\ C_{32} = -2,\ C_{33} = 1 \)
The adjoint is the transpose of the cofactor matrix, so
\( \displaystyle X = \frac{1}{3}\begin{bmatrix}2 & -5 & 1\\ -1 & 7 & -2\\ -1 & 1 & 1\end{bmatrix}\begin{bmatrix}105\\ 75\\ 180\end{bmatrix} \)
\( \displaystyle = \frac{1}{3}\begin{bmatrix}210 - 375 + 180\\ -105 + 525 - 360\\ -105 + 75 + 180\end{bmatrix} = \frac{1}{3}\begin{bmatrix}15\\ 60\\ 150\end{bmatrix} = \begin{bmatrix}5\\ 20\\ 50\end{bmatrix} \)
So a pen costs Rs. 5, a notebook Rs. 20 and a file Rs. 50.
Check: \(15 + 40 + 50 = 105\), \(5 + 20 + 50 = 75\) and \(10 + 20 + 150 = 180\).