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# 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 & 4 pens is Rs.61. Using this information and matrix method, find the cost of two books, three notebooks and two separately.

Last updated date: 17th Jun 2024
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Hint: We will first assume the cost of book, notebook and pen to be Rs. x, y and z respectively. After this, we will firm 3 equations as per the given data and thus, we will form a $3 \times 3$ matrix. Thus, on solving it we will get the answer.

Complete step-by-step solution:
Let us assume that the cost of a book is given by Rs. x, the cost of a notebook be Rs. y and the cost of a pen is Rs. z.
First, since we are given that the cost of 2 books, 6 notebooks and 3 pens is Rs. 40.
$\therefore$ we will get the following equation:-
$\Rightarrow 2x + 6y + 3z = 40$ ……….(1)
Now, since we are given that the cost of 3 books, 4 notebooks and 2 pens is Rs. 35.
$\therefore$ we will get the following equation:-
$\Rightarrow 3x + 4y + 2z = 35$ ……….(2)
And, since we are given that the cost of 5 books, 7 notebooks & 4 pens is Rs. 61.
$\therefore$ we will get the following equation:-
$\Rightarrow 5x + 7y + 4z = 61$ ……….(3)
Now, let us represent (1), (2) and (3) in matrix form to get the following expression:-
$\Rightarrow \left[ {\begin{array}{*{20}{c}} 2&6&3 \\ 3&4&2 \\ 5&7&4 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {40} \\ {35} \\ {61} \end{array}} \right]$
Now, let us apply some row transformations so as to get the answer:-
Step 1: First of all we will apply ${R_1} \to {R_1} - {R_2}$. Then, we will get:-
$\Rightarrow \left[ {\begin{array}{*{20}{c}} {2 - 3}&{6 - 4}&{3 - 2} \\ 3&4&2 \\ 5&7&4 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {40 - 35} \\ {35} \\ {61} \end{array}} \right]$
On simplifying it, we will get:-
$\Rightarrow \left[ {\begin{array}{*{20}{c}} { - 1}&2&1 \\ 3&4&2 \\ 5&7&4 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 5 \\ {35} \\ {61} \end{array}} \right]$
Step 2: Now, we will apply ${R_1} \to - {R_1}$. Then, we will get:-
$\Rightarrow \left[ {\begin{array}{*{20}{c}} { - ( - 1)}&{ - 2}&{ - 1} \\ 3&4&2 \\ 5&7&4 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 5} \\ {35} \\ {61} \end{array}} \right]$
On simplifying it, we will get:-
$\Rightarrow \left[ {\begin{array}{*{20}{c}} 1&{ - 2}&{ - 1} \\ 3&4&2 \\ 5&7&4 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 5} \\ {35} \\ {61} \end{array}} \right]$
Step 3: Now, we will apply ${R_2} \to {R_2} - 3{R_1}$ and ${R_3} \to {R_3} - 5{R_1}$. Then, we will get:-
$\Rightarrow \left[ {\begin{array}{*{20}{c}} 1&{ - 2}&{ - 1} \\ {3 - 3}&{4 + 6}&{2 + 3} \\ {5 - 5}&{7 + 10}&{4 + 5} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 5} \\ {35 + 15} \\ {61 + 25} \end{array}} \right]$
On simplifying it, we will get:-
$\Rightarrow \left[ {\begin{array}{*{20}{c}} 1&{ - 2}&{ - 1} \\ 0&{10}&5 \\ 0&{17}&9 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 5} \\ {50} \\ {86} \end{array}} \right]$
Step 4: Now, we will apply ${R_2} \to \dfrac{{{R_2}}}{{10}}$ . Then, we will get:-
$\Rightarrow \left[ {\begin{array}{*{20}{c}} 1&{ - 2}&{ - 1} \\ {\dfrac{0}{{10}}}&{\dfrac{{10}}{{10}}}&{\dfrac{5}{{10}}} \\ 0&{17}&9 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 5} \\ {\dfrac{{50}}{{10}}} \\ {86} \end{array}} \right]$
On simplifying it, we will get:-
$\Rightarrow \left[ {\begin{array}{*{20}{c}} 1&{ - 2}&{ - 1} \\ 0&1&{\dfrac{1}{2}} \\ 0&{17}&9 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 5} \\ 5 \\ {86} \end{array}} \right]$
Step 5: Now, we will apply ${R_3} \to {R_3} - 17{R_2}$ . Then, we will get:-
$\Rightarrow \left[ {\begin{array}{*{20}{c}} 1&{ - 2}&{ - 1} \\ 0&1&{\dfrac{1}{2}} \\ 0&{17 - 17}&{9 - \dfrac{{17}}{2}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 5} \\ 5 \\ {86 - 85} \end{array}} \right]$
On simplifying it, we will get:-
$\Rightarrow \left[ {\begin{array}{*{20}{c}} 1&{ - 2}&{ - 1} \\ 0&1&{\dfrac{1}{2}} \\ 0&0&{\dfrac{1}{2}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 5} \\ 5 \\ 1 \end{array}} \right]$
Now, we see that we have the following equation from the above matrix:-
$\Rightarrow x - 2y - z = - 5$ …………….(4)
$\Rightarrow y + \dfrac{z}{2} = 5$ …………….(5)
$\Rightarrow \dfrac{z}{2} = 1$ …………….(6)
Now, we can see that from the equation (6), we get:
$\Rightarrow z = 2$ …………..(7)
Substituting this in (5), we will get:-
$\Rightarrow y + \dfrac{2}{2} = 5$
$\Rightarrow y = 4$ ………….(8)
Substituting (7) and (8) in (4):-
$\Rightarrow x - 2(4) - 2 = - 5$
$\Rightarrow x - 8 - 2 = - 5$
$\Rightarrow x = 5$
$\therefore$ We get:- $x = 5,y = 4,z = 2$

Note: The students must notice that we need as many equations as many number of unknown variables we have. Like if we take the example of this question, we have 3 unknown variables and 3 equations.
We can use the matrix method only for solving linear equations (equations with unitary power of any variable). This uses back substitution. We basically try to make an identity matrix or an upper triangular matrix to make easy equations for us.