Question

A shopkeeper has 3 varieties of pens 'A', 'B' and 'C'. Meenu purchased 1 pen of each variety for a total of Rs 21. Jeevan purchased 4 pens of 'A' variety 3 pens of 'B' variety and 2 pens of 'C' variety for Rs 60. While Shikha purchased 6 pens of 'A' variety, 2 pens of 'B' variety and 3 pens of 'C' variety for Rs 70. Using matrix method, find cost of each variety of pen.

 

Answer 1

As there are 3 varieties of pen A, B and C
Meenu purchased 1 pen of each variety which costs her Rs 21
Therefore, 
\[A + B + C = 21\]
Similarly, 
For Jeevan
\[4A + 3B + 2C = 60\]
For Shikha 
\[6A + 2B + 3C = 70\]
\[\begin{bmatrix}1 & 1 & 1 \\ 4 & 3 & 2 \\ 6 & 2 & 3\end{bmatrix}\begin{bmatrix}A \\ B \\ C\end{bmatrix} = \begin{bmatrix}21 \\ 60 \\ 70\end{bmatrix}\]
\[\text{ where }P = \begin{bmatrix}1 & 1 & 1 \\ 4 & 3 & 2 \\ 6 & 2 & 3\end{bmatrix}, Q = \begin{bmatrix}21 \\ 60 \\ 70\end{bmatrix}\]
\[\left| P \right| = 1\left( 9 - 4 \right) - 1\left( 12 - 12 \right) + 1\left( 8 - 18 \right)\]
\[ = - 5 \neq 0\]
\[ \therefore P^{- 1}\text{ exists }\]
\[X = P^{- 1} Q\]
\[ C_{11} = 5 C_{12} = 0 C_{13} = - 10\]
\[ C_{21} = - 1 C_{22} = - 3 C_{23} = 4\]
\[ C_{31} = - 1 C_{32} = 2 C_{33} = - 1\]
\[adj P = \begin{bmatrix}5 & 0 & - 10 \\ - 1 & - 3 & 4 \\ - 1 & 2 & - 1\end{bmatrix}^T = \begin{bmatrix}5 & - 1 & - 1 \\ 0 & - 3 & 2 \\ - 10 & 4 & - 1\end{bmatrix}\]
\[ P^{- 1} = \frac{1}{- 5}\begin{bmatrix}5 & - 1 & - 1 \\ 0 & - 3 & 2 \\ - 10 & 4 & - 1\end{bmatrix}\]
\[X = P^{- 1} Q\]

\[\frac{1}{- 5}\begin{bmatrix}5 & - 1 & - 1 \\ 0 & - 3 & 2 \\ - 10 & 4 & - 1\end{bmatrix}\begin{bmatrix}21 \\ 60 \\ 70\end{bmatrix}\]
\[ = \frac{1}{- 5}\begin{bmatrix}105 - 60 - 70 \\ 0 - 180 + 140 \\ - 210 + 240 - 70\end{bmatrix}\]
\[ = \frac{1}{- 5}\begin{bmatrix}- 25 \\ - 40 \\ - 40\end{bmatrix}\]
\[ \therefore X = \begin{bmatrix}5 \\ 8 \\ 8\end{bmatrix}\]
Therefore, cost of A variety of pens = Rs 5
Cost of B variety of pens = Rs 8
Cost of C variety of pens = Rs 8