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.
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
$4 A+3 B+2 C=60$
For Shikha
$6 A+2 B+3 C=70$
$\left[\begin{array}{lll}1 & 1 & 1 \\ 4 & 3 & 2 \\ 6 & 2 & 3\end{array}\right]\left[\begin{array}{l}A \\ B \\ C\end{array}\right]=\left[\begin{array}{l}21 \\ 60 \\ 70\end{array}\right]$
where $P=\left[\begin{array}{lll}1 & 1 & 1 \\ 4 & 3 & 2 \\ 6 & 2 & 3\end{array}\right], Q=\left[\begin{array}{l}21 \\ 60 \\ 70\end{array}\right]$
$|P|=1(9-4)-1(12-12)+1(8-18)$
$=-5 \neq 0$
$\therefore P^{-1}$ exists
$X=P^{-1} Q$
$C_{11}=5 \quad C_{12}=0 \quad C_{13}=-10$
$C_{21}=-1 \quad C_{22}=-3 \quad C_{23}=4$
$C_{31}=-1 \quad C_{32}=2 \quad C_{33}=-1$
adj $P=\left[\begin{array}{rrr}5 & 0 & -10 \\ -1 & -3 & 4 \\ -1 & 2 & -1\end{array}\right]^{T}=\left[\begin{array}{rrr}5 & -1 & -1 \\ 0 & -3 & 2 \\ -10 & 4 & -1\end{array}\right]$
$P^{-1}=\frac{1}{-5}\left[\begin{array}{rrr}5 & -1 & -1 \\ 0 & -3 & 2 \\ -10 & 4 & -1\end{array}\right]$
$X=P^{-1} Q$
$\frac{1}{-5}\left[\begin{array}{ccc}5 & -1 & -1 \\ 0 & -3 & 2 \\ -10 & 4 & -1\end{array}\right]\left[\begin{array}{l}21 \\ 60 \\ 70\end{array}\right]$
$=\frac{1}{-5}\left[\begin{array}{c}105-60-70 \\ 0-180+140 \\ -210+240-70\end{array}\right]$
$=\frac{1}{-5}\left[\begin{array}{l}-25 \\ -40 \\ -40\end{array}\right]$
$\therefore X=\left[\begin{array}{l}5 \\ 8 \\ 8\end{array}\right]$
Therefore, cost of A variety of pens $=\mathrm{Rs} 5$
Cost of B variety of pens $=$ Rs 8
Cost of $\mathrm{C}$ variety of pens $=\mathrm{Rs} 8$