A shopkeeper has 3 varieties of pens 'A', 'B' and 'C'.

Solution:

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 $=\mathrm{Rs} 8$