Question

The cost of 2 books, 6 notebooks and 3 pens is Rs.40. The cost of 3 books, 4 notebooks and 2 pens is Rs.35. While the cost of 5 books, 7 notebooks and 4 pens is Rs.61. Using this information and matrix method, find the cost of 2 books, 3 notebooks and 2 pens separately.

Answer 1

Hint: This is a question about matrices. To solve the above equation, first we will derive the equations from given data containing variables, then we will convert them into matrix form. And solving that matrix we will get the values of variables. And from that we will derive the answer.

Complete step-by-step answer:
Let the price of a book be x.
The price of a notebook be y.
The price of a pen be z.
According to the question, the cost of 2 books, 6 notebooks and 3 pens is Rs.40.
Converting the above statement into an equation we get,
 $ 2x + 6y + 3z = 40 $ ………………….. (1)
Again it is given that the cost of 3 books, 4 notebooks and 2 pens is Rs.35.
Converting the above statement into an equation we get,
 $ 3x + 4y + 2z = 35 $ …………………. (2)
Again it is given that, the cost of 5 books, 7 notebooks and 4 pens is Rs.61.
Converting the above statement into an equation we get,
 $ 5x + 7y + 4z = 61 $ ………………. (3)
Representing above 3 equation in matrix form we get,
 $ \left( {\begin{array}{*{20}{c}}
  {2x}&{6y}&{3z} \\
  {3x}&{4y}&{2z} \\
  {5x}&{7y}&{4z}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  {40} \\
  {35} \\
  {61}
\end{array}} \right) $
Taking x, y and z common from column 1, 2 and 3 respectively, we can write the above equation as,
 $ \left( {\begin{array}{*{20}{c}}
  2&6&3 \\
  3&4&2 \\
  5&7&4
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  x \\
  y \\
  z
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  {40} \\
  {35} \\
  {61}
\end{array}} \right) $
Replacing $ {R_1} \to {R_1} - {R_2} $ in the first term of the left hand side and right hand side of the equation we get,
 $ \left( {\begin{array}{*{20}{c}}
  { - 1}&2&1 \\
  3&4&2 \\
  5&7&4
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  x \\
  y \\
  z
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  5 \\
  {35} \\
  {61}
\end{array}} \right) $
Again replacing $ {R_1} \to ( - {R_1}) $ in the first term of the left hand side and right hand side of the equation we get,
 $ \left( {\begin{array}{*{20}{c}}
  1&{ - 2}&{ - 1} \\
  3&4&2 \\
  5&7&4
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  x \\
  y \\
  z
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  { - 5} \\
  {35} \\
  {61}
\end{array}} \right) $
Again replacing $ {R_2} \to ({R_2} - 3{R_1}) $ in the first term of the left hand side and right hand side of the equation we get,
 $ \left( {\begin{array}{*{20}{c}}
  1&{ - 2}&{ - 1} \\
  0&{10}&5 \\
  5&7&4
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  x \\
  y \\
  z
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  { - 5} \\
  {50} \\
  {61}
\end{array}} \right) $
Again replacing $ {R_3} \to ({R_3} - 5{R_1}) $ in the first term of the left hand side and right hand side of the equation we get,
 $ \left( {\begin{array}{*{20}{c}}
  1&{ - 2}&{ - 1} \\
  0&{10}&5 \\
  0&{17}&9
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  x \\
  y \\
  z
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  { - 5} \\
  {50} \\
  {86}
\end{array}} \right) $
Replacing $ {R_2} \to \dfrac{1}{{10}}{R_2} $ in the first term of the left hand side and right hand side of the equation we get,
 $ \left( {\begin{array}{*{20}{c}}
  1&{ - 2}&{ - 1} \\
  0&1&{\dfrac{1}{2}} \\
  0&{17}&9
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  x \\
  y \\
  z
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  { - 5} \\
  5 \\
  {86}
\end{array}} \right) $
Replacing $ {R_2} \to \dfrac{1}{{10}}{R_2} $ in the first term of the left hand side and right hand side of the equation we get,
 $ \left( {\begin{array}{*{20}{c}}
  1&{ - 2}&{ - 1} \\
  0&1&{\dfrac{1}{2}} \\
  0&0&{\dfrac{1}{2}}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  x \\
  y \\
  z
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  { - 5} \\
  5 \\
  1
\end{array}} \right) $
Converting above equation from matrix form to normal form we get 3 equation such as,
 $ x - 2y - z = - 5 $ ………. (4)
 $ y + \dfrac{z}{2} = 5 $ ………… (5)
 $ \dfrac{z}{2} = 1 $ ……………… (6)
From equation 6 we get, $ z = 2 $
Putting value of z in equation 5 we get,
 $ y + \dfrac{2}{2} = 5 \Rightarrow y + 1 = 5 \Rightarrow y = 4 $
Putting values of y and z in equation 4 we get,
 $ x - (2 \times 4) - 2 = - 5 \Rightarrow x = 5 $
As we have taken, the cost of a book x, the cost of a notebook y and the cost of a pen z.
Hence the cost of 2 books, $ 2x = 2 \times 5 = 10rs. $
The cost of 3 notebooks, $ 3y = 3 \times 4 = 12rs. $
The cost of 2 pens, $ 2z = 2 \times 2 = 4rs. $
 $ \therefore $ The cost of 2 books, 3 notebooks and 2pens are Rs.10, Rs.12 and Rs.4 respectively.

Note: Matrix is a rectangular array of numbers, symbols or expressions arranged in rows and columns
Let A be a $ 3 \times 3 $ matrix, $ A = \left( {\begin{array}{*{20}{c}}
  a&b&c \\
  d&e&f \\
  g&h&i
\end{array}} \right) $
You can solve the matrix equation by exchanging columns and rows.
You should remember all the properties and rules of the matrix.
You should practice more such questions.