Question

A shopkeeper has three varieties of pens A, B and C. Meenu purchased 1 pen of each variety for a total of Rs. 21. Jeen 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 the matrix method, find the cost of each pen.

Answer 1

Hint: Here, first we will convert the given three statements into linear equation form. Then, we will convert this equation into matrix form i.e. given as $\left| \begin{matrix}
   {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
   {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
   {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right|\left| \begin{matrix}
   x \\
   y \\
   x \\
\end{matrix} \right|=\left| \begin{matrix}
   C1 \\
   C2 \\
   C3 \\
\end{matrix} \right|$ where x, y, z are variables and C1, C2, C3 are constant terms. Then we will apply the row transformation method in such a way that we will get three equations at the end in which all the 3 variables are there, only 2 variables are there and only one variable is there. Thus, solving those equations we will get an answer.

Complete step-by-step answer:
Here, we will first convert the given statement in linear equation form. It is given that Meenu purchased 1 pen of all the three varieties for Rs. 21. So, we can write it as
$1A+1B+1C=21$ …………………………..(1)
Similarly, Jeen purchased 4 pens of A variety, 3 pens of B variety and 2 pens of C variety for Rs. 60. So, in equation form we can write it as
$4A+3B+2C=60$ ……………………………..(2)
Also, Shikha purchased 6 pens of A variety, 2 pens of B variety and 3 pens of C variety for Rs. 70. So, we will write it as
$6A+2B+3C=70$ ……………………………….(3)
Now, we will write this coefficient of all three equations in matrix form i.e. $\left| \begin{matrix}
   {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\
   {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\
   {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\
\end{matrix} \right|\left| \begin{matrix}
   x \\
   y \\
   x \\
\end{matrix} \right|=\left| \begin{matrix}
   C1 \\
   C2 \\
   C3 \\
\end{matrix} \right|$ where x, y, z are variables and C1, C2, C3 are constant terms.
We will fill the first matrix with coefficients of the variables A, B and C. Second matrix we will fill as A, B and C. Then in the third matrix, we will fill in the constant terms from the equations formed.
From equations 1, 2 and 3, we can write matrix as
$\left| \begin{matrix}
   1 & 1 & 1 \\
   4 & 3 & 2 \\
   6 & 2 & 3 \\
\end{matrix} \right|\left| \begin{matrix}
   A \\
   B \\
   C \\
\end{matrix} \right|=\left| \begin{matrix}
   21 \\
   60 \\
   70 \\
\end{matrix} \right|$ ………………………….(4)
Here, we have to find values for A, B and C. So, for this we will use a row transformation method. We have to perform operations like addition, subtraction, multiplication, or division to make any of the coefficients equals to zero so it can be easier to solve.
So, here we are performing operation on Row2 i.e. $Row2=Row2-4Row1$ . So, here we will multiply row1 with 4 and subtract it from row 2. So, be doing this we will get matrix as
$\left| \begin{matrix}
   1 & 1 & 1 \\
   4-4 & 3-4 & 2-4 \\
   6 & 2 & 3 \\
\end{matrix} \right|\left| \begin{matrix}
   A \\
   B \\
   C \\
\end{matrix} \right|=\left| \begin{matrix}
   21 \\
   60-84 \\
   70 \\
\end{matrix} \right|$
$\left| \begin{matrix}
   1 & 1 & 1 \\
   0 & -1 & -2 \\
   6 & 2 & 3 \\
\end{matrix} \right|\left| \begin{matrix}
   A \\
   B \\
   C \\
\end{matrix} \right|=\left| \begin{matrix}
   21 \\
   -24 \\
   70 \\
\end{matrix} \right|$
Same way, we are performing operation on Row3 i.e. $Row3=Row3-6Row1$ . So, here we will multiply row1 with 6 and subtract it from row 3. So, be doing this we will get matrix as
$\left| \begin{matrix}
   1 & 1 & 1 \\
   0 & -1 & -2 \\
   6-6 & 2-6 & 3-6 \\
\end{matrix} \right|\left| \begin{matrix}
   A \\
   B \\
   C \\
\end{matrix} \right|=\left| \begin{matrix}
   21 \\
   -24 \\
   70-126 \\
\end{matrix} \right|$
\[\left| \begin{matrix}
   1 & 1 & 1 \\
   0 & -1 & -2 \\
   0 & -4 & -3 \\
\end{matrix} \right|\left| \begin{matrix}
   A \\
   B \\
   C \\
\end{matrix} \right|=\left| \begin{matrix}
   21 \\
   -24 \\
   -56 \\
\end{matrix} \right|\]
Now, we will apply again on Row3 i.e. $Row3=Row3+\left( -4 \right)Row2$ . So, on doing this we will get matrix as
$\left| \begin{matrix}
   1 & 1 & 1 \\
   0 & -1 & -2 \\
   0 & -4+4 & -3+8 \\
\end{matrix} \right|\left| \begin{matrix}
   A \\
   B \\
   C \\
\end{matrix} \right|=\left| \begin{matrix}
   21 \\
   -24 \\
   -56+96 \\
\end{matrix} \right|$
$\left| \begin{matrix}
   1 & 1 & 1 \\
   0 & -1 & -2 \\
   0 & 0 & 5 \\
\end{matrix} \right|\left| \begin{matrix}
   A \\
   B \\
   C \\
\end{matrix} \right|=\left| \begin{matrix}
   21 \\
   -24 \\
   40 \\
\end{matrix} \right|$
Thus, from the above equation, we have linear equation as
$A+B+C=21$
$-B-2C=-24$
$5C=40$
From this we get the value of C as $C=\dfrac{40}{5}=8$.
Substituting the value of C in $-B-2C=-24$ , we get value of B as
$-B-2\left( 8 \right)=-24$
$-B=-24+16=-8$
$\therefore B=8$
Similarly putting values of B and C and finding the value of A. We will get as
$A+8+8=21$
$A=21-16=5$
Thus, price of pen A is Rs. 5, price of variety of pen B is Rs. 8, price of variety of pen C is Rs. 8.

Note: Remember that we can make any row transformation applying proper operation. Here, we have made changes in row2 and row3, but we can also make changes in row1 and can find the answer. Just we have to make sure that we have 3 equations in which all 3 variables are there, in another equation there should be only 2 variables and in the next equation, there should be only 1 variable present. Thus, it will be very easy to solve.