Question

The cost of 4 dozen pencils, 3 dozen pens and 2 dozen erasers is Rs. 60. The cost of 2 dozen pencils, 4 dozen pens and 6 dozen erasers is Rs. 90 whereas the cost of 6 dozen pencils, 2 dozen pens and 3 dozen erasers is Rs. 70. Find the cost of each item per dozen by using matrices.

Answer 1

Hint:In this question take cost of 1 dozen pencil as Rs. x, cost of 1 dozen pens as Rs. y and cost of 1 dozen eraser as Rs. z then using the given information make the equations and using the method of matrix row operation find the costs of pencil, pen and eraser.

Complete step-by-step answer:
According to the given information we have to find the costs of 4 dozen pencils, cost of 3 dozen pens and cost of 2 dozen erasers
So to find the costs let suppose that the cost of 1 dozen pencils is Rs. x, cost of 1 dozen pens as Rs. y and cost of 1 dozen erasers as Rs. z
So by the given information we know that the cost of 4 dozen pencils, 3 dozen pens and 2 dozens of erasers is Rs. 60
So the equation will be $ 4x + 3y + 2z = 60 $
We know that cost of 2 dozen pencils, 4 dozen pens and cost of 5 dozen eraser is Rs. 90
Therefore the equation is $ 2x + 4y + 5z = 90 $
And at last we have the information that the cost of 6 dozen pencils, 2 dozen pens and 3 dozen erasers is Rs. 70
So the equation that represent the above statement is given by $ 6x + 2y + 3z = 70 $
As we have 3 equations $ 4x + 3y + 2z = 60 $ , $ 2x + 4y + 5z = 90 $ and $ 6x + 2y + 3z = 70 $ which can be expressed in the form of matrix as
\[\left[ \begin{gathered}
  4\,\,\,3\,\,\,2 \\
  2\,\,\,4\,\,\,6 \\
  6\,\,\,2\,\,\,3 \\
\end{gathered} \right]\left[ \begin{gathered}
  x \\
  y \\
  z \\
\end{gathered} \right] = \left[ \begin{gathered}
  60 \\
  90 \\
  70 \\
\end{gathered} \right]\]
Now using the matrix row operation to find the value of x, y and z
For matrix \[\left[ \begin{gathered}
  4\,\,\,3\,\,\,2 \\
  2\,\,\,4\,\,\,6 \\
  6\,\,\,2\,\,\,3 \\
\end{gathered} \right]\] and $ \left[ \begin{gathered}
  60 \\
  60 \\
  40 \\
\end{gathered} \right] $
Let’s changing $ {R_3} $ as \[{R_3} \to {R_3} - \dfrac{3}{2}{R_1}\] and changing $ {R_2} $ as \[{R_2} \to {R_2} - \dfrac{1}{2}{R_1}\]
Using this row operation in matrix \[\left[ \begin{gathered}
  4\,\,\,3\,\,\,2 \\
  2\,\,\,4\,\,\,6 \\
  6\,\,\,2\,\,\,3 \\
\end{gathered} \right]\] and matrix $ \left[ \begin{gathered}
  60 \\
  60 \\
  40 \\
\end{gathered} \right] $
We get \[\left[ \begin{gathered}
  \,\,\,\,\,\,\,\,\,\,\,4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2 \\
  \left( {2 - \dfrac{1}{2} \times 4} \right)\,\,\,\left( {4 - \dfrac{1}{2} \times 3} \right)\,\,\,\left( {6 - \dfrac{1}{2} \times 2} \right) \\
  \left( {6 - \dfrac{3}{2} \times 4} \right)\,\,\,\left( {2 - \dfrac{3}{2} \times 3} \right)\,\,\,\left( {3 - \dfrac{3}{2} \times 2} \right) \\
\end{gathered} \right]\left[ \begin{gathered}
  x \\
  y \\
  z \\
\end{gathered} \right] = \left[ \begin{gathered}
  \,\,\,\,\,\,\,\,\,60 \\
  \left( {90 - \dfrac{1}{2} \times 60} \right) \\
  \left( {70 - \dfrac{3}{2} \times 60} \right) \\
\end{gathered} \right]\]
 $ \Rightarrow $ \[\left[ \begin{gathered}
  4\,\,\,\,3\,\,\,\,2 \\
  0\,\,\,\,\dfrac{5}{2}\,\,\,\,5 \\
  0\,\,\,\,\dfrac{{ - 5}}{2}\,\,\,\,0 \\
\end{gathered} \right]\left[ \begin{gathered}
  x \\
  y \\
  z \\
\end{gathered} \right] = \left[ \begin{gathered}
  60 \\
  60 \\
   - 20 \\
\end{gathered} \right]\]
Now using the row operation $ {R_3} \to {R_3} + {R_2} $ , we get
\[\left[ \begin{gathered}
  \,\,\,\,\,4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2 \\
  \,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\dfrac{5}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5 \\
  \left( {0 + 0} \right)\,\,\,\,\left( {\dfrac{{ - 5}}{2} + \dfrac{5}{2}} \right)\,\,\,\,\left( {0 + 5} \right) \\
\end{gathered} \right]\left[ \begin{gathered}
  x \\
  y \\
  z \\
\end{gathered} \right] = \left[ \begin{gathered}
  \,\,\,\,\,\,60 \\
  \,\,\,\,\,\,60 \\
   - 20 + 60 \\
\end{gathered} \right]\]
 $ \Rightarrow $ \[\left[ \begin{gathered}
  4\,\,\,\,3\,\,\,\,2 \\
  0\,\,\,\,\dfrac{5}{2}\,\,\,\,5 \\
  0\,\,\,\,0\,\,\,\,5 \\
\end{gathered} \right]\left[ \begin{gathered}
  x \\
  y \\
  z \\
\end{gathered} \right] = \left[ \begin{gathered}
  60 \\
  60 \\
  40 \\
\end{gathered} \right]\]
Constructing equations using the above matrix
4x + 3y + 2z = 60 (equation 1)
0x + $ \dfrac{5}{2} $ y + 5z = 60 (equation 2)
0x + 0y + 5z =40 (equation 3)
By equation 3 we get z = \[\dfrac{{40}}{5}\]
 $ \Rightarrow $ z = \[8\]
Substituting the value of z in equation 2 we get
\[0x + \dfrac{5}{2}y + \left( {5 \times 8} \right) = 60\]
 $ \Rightarrow $ \[\dfrac{5}{2}y + 40 = 60\]
y = \[\dfrac{{2\left( {60-40} \right)}}{5}\]
 $ \Rightarrow $ y = 8
Now substituting the value of y and z in equation 1
\[4x + \left( {3 \times 8} \right) + \left( {2 \times 8} \right) = 60\]
 $ \Rightarrow $ \[4x + 24 + 16 = 60\]
 $ \Rightarrow $ \[x = \dfrac{{60 - 40}}{4}\]
x = 5
Hence the cost of 1 dozen of pencils is Rs. 5, cost of 1 dozen pens is Rs. 8 and cost of 1 dozen erasers is Rs. 8

Note: The concept of matrix we used to find the cost of each item can be explained as the arrangement of the numbers in some rows and columns in a rectangular form where the numbers of this arrangement are named as the elements of the matrix. Whereas the arrangement of elements in row and column in square form is called determinant.