Question

The cost of 4 kg potato, 3 kg wheat and 2 kg rice is Rs.150. The cost of 1 kg potato, 2 kg wheat and 3 kg Rice is Rs.125. The cost of 6 kg potato, 2 kg wheat and 3 kg Rice is Rs.175. Find the cost of each item per kg, by using Cramer's Rule.

Answer 1

Hint: We will first assume the cost of potato, wheat and rice be some variables. After that, we will form equations so that we can form a matrix to use Cramer’s rule. After that, we will find determinants which we require for all the variables.

Complete step-by-step solution:
Let us assume the cost of 1 kg potato, 1 kg wheat and1 kg rice be x, y and z respectively.
Now, since in the initial data, we are given that:
The cost of 4 kg potato, 3 kg wheat and 2 kg rice is Rs.150.
According to the assumptions of the cost of them individually, we will get the equation given as follows:-
\[ \Rightarrow 4x + 3y + 2z = 150\] ……………(1)
Now, we also have the data: the cost of 1 kg potato, 2 kg potato, 2 kg wheat and 3 kg Rice is Rs. 125.
So, we get the following equation:-
\[ \Rightarrow x + 2y + 3z = 125\] ……………(2)
Now, we also have the data: the cost of 6 kg potato, 2 kg wheat and 3 kg Rice is Rs.175.
So, we get the following equation:-
\[ \Rightarrow 6x + 2y + 3z = 175\] ……………(3)
Now, we have three equations as we required as (1), (2) and (3).
Now, writing them in matrix form, we will get the following expression:-
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  4&3&2 \\
  1&2&3 \\
  6&2&3
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  x \\
  y \\
  z
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {150} \\
  {125} \\
  {175}
\end{array}} \right]$
If we compare this to $AX = Y$, we have $A = \left[ {\begin{array}{*{20}{c}}
  4&3&2 \\
  1&2&3 \\
  6&2&3
\end{array}} \right],X = \left[ {\begin{array}{*{20}{c}}
  x \\
  y \\
  z
\end{array}} \right]$ and $Y = \left[ {\begin{array}{*{20}{c}}
  {150} \\
  {125} \\
  {175}
\end{array}} \right]$.
Now, according to Cramer’s method, we have:-
$ \Rightarrow x = \dfrac{{{D_1}}}{D},y = \dfrac{{{D_2}}}{D}$ and $z = \dfrac{{{D_3}}}{D}$, where ${D_i}$ is as similar as A matrix with just ${i^{th}}$ column replaces with Y and D = |A|.
Now, let us find all the ${D_i}'s$ now and D as well to get the required answer.
Since $D = \left| {\begin{array}{*{20}{c}}
  4&3&2 \\
  1&2&3 \\
  6&2&3
\end{array}} \right|$ .
Now, let us find its value:-
$ \Rightarrow D = 4\left( {2 \times 3 - 2 \times 3} \right) - 3\left( {1 \times 3 - 6 \times 3} \right) + 2\left( {1 \times 2 - 6 \times 2} \right)$
Simplifying the values to get the following:-
$ \Rightarrow D = 4\left( {6 - 6} \right) - 3\left( {3 - 18} \right) + 2\left( {2 - 12} \right)$
Simplifying the values further to get the following:-
$ \Rightarrow D = 45 - 20$
$ \Rightarrow D = 25$ ……………(4)
Since $A = \left[ {\begin{array}{*{20}{c}}
  4&3&2 \\
  1&2&3 \\
  6&2&3
\end{array}} \right]$ and $Y = \left[ {\begin{array}{*{20}{c}}
  {150} \\
  {125} \\
  {175}
\end{array}} \right]$.
$ \Rightarrow {D_1} = \left| {\begin{array}{*{20}{c}}
  {150}&3&2 \\
  {125}&2&3 \\
  {175}&2&3
\end{array}} \right|$
Now, let us find its value:-
$ \Rightarrow {D_1} = 150\left( {2 \times 3 - 2 \times 3} \right) - 125\left( {3 \times 3 - 2 \times 2} \right) + 175\left( {3 \times 3 - 2 \times 2} \right)$
Simplifying the values to get the following:-
$ \Rightarrow {D_1} = 150\left( {6 - 6} \right) - 125\left( {9 - 4} \right) + 175\left( {9 - 4} \right)$
Simplifying the values further to get the following:-
$ \Rightarrow {D_1} = 250$ ……………(5)
Since $A = \left[ {\begin{array}{*{20}{c}}
  4&3&2 \\
  1&2&3 \\
  6&2&3
\end{array}} \right]$ and $Y = \left[ {\begin{array}{*{20}{c}}
  {150} \\
  {125} \\
  {175}
\end{array}} \right]$.
\[ \Rightarrow {D_2} = \left| {\begin{array}{*{20}{c}}
  4&{150}&2 \\
  1&{125}&3 \\
  6&{175}&3
\end{array}} \right|\]
Now, let us find its value:-
$ \Rightarrow {D_2} = 4\left( {125 \times 3 - 175 \times 3} \right) - 1\left( {150 \times 3 - 175 \times 2} \right) + 6\left( {150 \times 3 - 125 \times 2} \right)$
Simplifying the values to get the following:-
$ \Rightarrow {D_2} = 4\left( {375 - 525} \right) - 1\left( {450 - 350} \right) + 6\left( {450 - 250} \right)$
Simplifying the values further to get the following:-
$ \Rightarrow {D_2} = 500$ ……………(6)
Since $A = \left[ {\begin{array}{*{20}{c}}
  4&3&2 \\
  1&2&3 \\
  6&2&3
\end{array}} \right]$ and $Y = \left[ {\begin{array}{*{20}{c}}
  {150} \\
  {125} \\
  {175}
\end{array}} \right]$.
\[ \Rightarrow {D_3} = \left| {\begin{array}{*{20}{c}}
  4&3&{150} \\
  1&2&{125} \\
  6&2&{175}
\end{array}} \right|\]
Now, let us find its value:-
$ \Rightarrow {D_3} = 4\left( {175 \times 2 - 125 \times 2} \right) - 1\left( {175 \times 3 - 150 \times 2} \right) + 6\left( {125 \times 3 - 150 \times 2} \right)$
Simplifying the values to get the following:-
$ \Rightarrow {D_3} = 4\left( {350 - 250} \right) - \left( {525 - 300} \right) + 6\left( {375 - 300} \right)$
Simplifying the values further to get the following:-
$ \Rightarrow {D_3} = 625$ ……………(7)
Now, putting (4), (5), (6) and (7) in $x = \dfrac{{{D_1}}}{D},y = \dfrac{{{D_2}}}{D}$ and $z = \dfrac{{{D_3}}}{D}$, we will get:-
$ \Rightarrow x = \dfrac{{{D_1}}}{D} = \dfrac{{250}}{{25}} = 10$
$ \Rightarrow y = \dfrac{{{D_2}}}{D} = \dfrac{{500}}{{25}} = 20$
$ \Rightarrow z = \dfrac{{{D_3}}}{D} = \dfrac{{625}}{{25}} = 25$

The cost of potato per kg is 10rs, cost of wheat per kg is 20rs and the cost of rice per kg is 25rs.

Note: The students must note that though the Cramer’s rule is very famous but until not mentioned, do not perform it because it requires a lot of hassle and a lot of calculations.
The students must also note that this method cannot be used on the matrices which are not invertible. Just give it a bit of thinking. Let us understand why not!
This is because, we know that a matrix is not invertible if and only if |A| = 0. If we have |A| = 0 and we are trying to apply Cramer’s rule, then we will divide by 0 in order to get the value of the required variables.