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. Find the cost of each variety of pen using the Matrix method?

Answer 1

Hint: We start solving the problem by assigning the variables for cost of the pen of each variety given. We then write the given cost in terms of equations using these variables. We then convert the obtained linear set of equations to the form $AX=B$. We then apply the row transformations and make necessary calculations to get the values of cost of pens of each variety.

Complete step-by-step answer:
According to the problem, we are given that the shopkeeper sells 3 varieties of pens ‘A’, ‘B’ and ‘C’. Meenu purchased 1 pen of each variety for a total of Rs.21 while 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. We need to find the cost of each variety of pen using the Matrix method.
Let us assume the cost of pens of varieties of pens ‘A’, ‘B’ and ‘C’ be ‘x’, ‘y’ and ‘z’.
According to the problem, Meenu purchased 1 pen of each variety for Rs.21.
So, we get $x+y+z=21$ ---(1).
According to the problem, Jeevan purchased 4 pens of ‘A’ variety, 3 pens of ‘B’ variety and 2 pens of ‘C’ variety for Rs.60.
So, we get $4x+3y+2z=60$ ---(2).
According to the problem, Shikha purchased 6 pens of ‘A’ variety, 2 pens of ‘B’ variety and 3 pens of ‘C’ variety for Rs.70.
So, we get $6x+2y+3z=70$ ---(3).
So, we need to solve the following linear set of equations obtained from equations (1), (2) and (3) to find the costs ‘x’, ‘y’, ‘z’.
$x+y+z=21$.
$4x+3y+2z=60$.
$6x+2y+3z=70$.
Let us write these linear set of equation in form of $AX=B$, where
$\Rightarrow A=coefficient\ matrix=\left[ \begin{matrix}
   1 & 1 & 1 \\
   4 & 3 & 2 \\
   6 & 2 & 3 \\
\end{matrix} \right]$.
$\Rightarrow X=\operatorname{variable} matrix=\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]$.
$\Rightarrow B=\operatorname{constant} matrix=\left[ \begin{matrix}
   21 \\
   60 \\
   70 \\
\end{matrix} \right]$.
So, we get \[\left[ \begin{matrix}
   1 & 1 & 1 \\
   4 & 3 & 2 \\
   6 & 2 & 3 \\
\end{matrix} \right]\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\left[ \begin{matrix}
   21 \\
   60 \\
   70 \\
\end{matrix} \right]\] ---(4).
Let us apply row transformations in the matrix of equation (4) to get the values of ‘x’, ‘y’ and ‘z’.
Let us apply ${{R}_{2}}\to {{R}_{2}}-4{{R}_{1}}$ and ${{R}_{3}}\to {{R}_{3}}-6{{R}_{1}}$.
$\Rightarrow \begin{matrix}
   {{R}_{1}}\to {{R}_{1}} \\
   {{R}_{2}}\to {{R}_{2}}-4{{R}_{1}} \\
   {{R}_{3}}\to {{R}_{3}}-6{{R}_{1}} \\
\end{matrix}\left[ \begin{matrix}
   1 & 1 & 1 \\
   4-4 & 3-4 & 2-4 \\
   6-6 & 2-6 & 3-6 \\
\end{matrix} \right]\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\left[ \begin{matrix}
   21 \\
   60-84 \\
   70-126 \\
\end{matrix} \right]$.
$\Rightarrow \left[ \begin{matrix}
   1 & 1 & 1 \\
   0 & -1 & -2 \\
   0 & -4 & -3 \\
\end{matrix} \right]\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\left[ \begin{matrix}
   21 \\
   -24 \\
   -56 \\
\end{matrix} \right]$.
Let us apply ${{R}_{3}}\to {{R}_{3}}-4{{R}_{2}}$.
$\Rightarrow \begin{matrix}
   {{R}_{1}}\to {{R}_{1}} \\
   {{R}_{2}}\to {{R}_{2}} \\
   {{R}_{3}}\to {{R}_{3}}-4{{R}_{2}} \\
\end{matrix}\left[ \begin{matrix}
   1 & 1 & 1 \\
   0 & -1 & -2 \\
   0-0 & -4+4 & -3+8 \\
\end{matrix} \right]\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\left[ \begin{matrix}
   21 \\
   -24 \\
   -56+96 \\
\end{matrix} \right]$.
$\Rightarrow \left[ \begin{matrix}
   1 & 1 & 1 \\
   0 & -1 & -2 \\
   0 & 0 & 5 \\
\end{matrix} \right]\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\left[ \begin{matrix}
   21 \\
   -24 \\
   40 \\
\end{matrix} \right]$.
So, we get the linear equations converted as
$x+y+z=21$ ---(5).
$-y-2z=-24$ ---(6).
$5z=40$ ---(7).
From equation (7), we get $z=\dfrac{40}{5}=8$.
Let us substitute the value of z in equation (6). We get $-y-2\left( 8 \right)=-24$.
$\Rightarrow y=24-16=8$.
Let us substitute the values of y and z in equation (5). We get $x+8+8=21$.
$\Rightarrow x=21-16=5$.
So, we have found the cost of pens of variety ‘A’, ‘B’ and ‘C’ as Rs.5, Rs.8, Rs.8.
∴ The cost of pens of variety ‘A’, ‘B’ and ‘C’ is Rs.5, Rs.8, Rs.8.

Note: We can also solve this problem by using Cramer’s method, Matrix inversion method. We can also take the row echelon form of matrix to reduce the confusion while applying the row transformations. We should not apply row transformations to the variable matrix as it will give the same linear set of equations again. As there are good amounts of calculations required, there is a high chance of making mistakes. To avoid such mistakes, we need to calculate each step properly.