Question
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...
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.
Solution
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 3×3 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.
∴ we will get the following equation:-
⇒2x+6y+3z=40 ……….(1)
Now, since we are given that the cost of 3 books, 4 notebooks and 2 pens is Rs. 35.
∴ we will get the following equation:-
⇒3x+4y+2z=35 ……….(2)
And, since we are given that the cost of 5 books, 7 notebooks & 4 pens is Rs. 61.
∴ we will get the following equation:-
⇒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 R1→R1−R2. 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 R1→−R1. 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 R2→R2−3R1 and R3→R3−5R1. 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 R2→10R2 . Then, we will get:-