Question

The cost of 4 kg potato, 3 kg wheat and 2 kg of rice is Rs. 60. The cost of 1 kg potato, 2 kg wheat, and 3 kg rice is Rs. 45. The cost of 6 kg potato, 2 kg wheat, and 3 kg rice is Rs. 70. Find the cost of each item per kg by matrix method.

Answer 1

Hint: Assume that the cost of potato, wheat, and rice per kg is Rs. x, Rs. y, and Rs. z respectively. It is given that the cost of 4 kg potato, 3 kg wheat, and 2 kg of rice is Rs. 60. The cost of 1 kg potato, 2 kg wheat, and 3 kg rice is Rs. 45. The cost of 6 kg potato, 2 kg wheat, and 3 kg rice is Rs. 70. Now, form mathematical equations. Our system of linear equations is \[4x+3y+2z=60\] , \[x+2y+3z=45\] , and \[6x+2y+3z=70\] . Express it in the form of \[AX=B\] , where \[A=\left[ \begin{align}
  & \begin{matrix}
   4 & 3 & 2 \\
\end{matrix} \\
 & \begin{matrix}
   1 & 2 & 3 \\
\end{matrix} \\
 & \begin{matrix}
   6 & 2 & 3 \\
\end{matrix} \\
\end{align} \right]\] , \[X=\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]\] , and \[B=\left[ \begin{matrix}
   60 \\
   45 \\
   70 \\
\end{matrix} \right]\] . Now check for consistency of these equations by getting the determinant value of matrix A. Solving further the equation \[AX=B\] , we get \[X={{A}^{-1}}B\] . Now, we need the value of \[{{A}^{-1}}\] . We know the formula, \[{{A}^{-1}}=\dfrac{adj\left[ A \right]}{\left| A \right|}\] . The adjoint matrix is obtained after interchanging the rows with the columns of the cofactor matrix. Now, put the value of \[{{A}^{-1}}\] , X and B in the equation \[X={{A}^{-1}}B\] and solve it further.

Complete step by step solution:
First of all, let us assume the cost of potato, wheat, and rice per kg be Rs. x, Rs. y, and Rs. z respectively.
We have the cost of 4 kg potato, 3 kg wheat, and 2 kg of rice are Rs. 60. Expressing it in a mathematical equation, we get
\[4x+3y+2z=60\] ……………………………….(1)
We have the cost of 1 kg potato, 2 kg wheat, and 3 kg of rice is Rs. 45. Expressing it in a mathematical equation, we get
\[x+2y+3z=45\] ……………………………….(2)
We have the cost of 6 kg potato, 2 kg wheat, and 3 kg of rice is Rs. 70. Expressing it in a mathematical equation, we get
\[6x+2y+3z=70\] ……………………………….(3)
Now, from equation (1), equation (2), and equation (3), we have the system of equations,
\[4x+3y+2z=60\]
\[x+2y+3z=45\]
\[6x+2y+3z=60\]
We can express these systems of equations in the form of matrices.
\[\left[ \begin{align}
  & \begin{matrix}
   4 & 3 & 2 \\
\end{matrix} \\
 & \begin{matrix}
   1 & 2 & 3 \\
\end{matrix} \\
 & \begin{matrix}
   6 & 2 & 3 \\
\end{matrix} \\
\end{align} \right]\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\left[ \begin{matrix}
   60 \\
   45 \\
   70 \\
\end{matrix} \right]\] ………………………………(4)
The system of the linear equation is of the form, \[AX=B\] ……………….(5)
Now, comparing equation (4) with \[AX=B\] , we get
\[A=\left[ \begin{align}
  & \begin{matrix}
   4 & 3 & 2 \\
\end{matrix} \\
 & \begin{matrix}
   1 & 2 & 3 \\
\end{matrix} \\
 & \begin{matrix}
   6 & 2 & 3 \\
\end{matrix} \\
\end{align} \right]\] …………………………..(6)
\[X=\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]\] ………………………………(7)
\[B=\left[ \begin{matrix}
   60 \\
   45 \\
   70 \\
\end{matrix} \right]\] …………………………….(8)
Now, we have to decide whether the system of linear equations is consistent or not. For that, we need to get the determinant value of matrix A.
Now, getting the determinant value of matrix A.
\[\left| A \right|=\left| \begin{align}
  & \begin{matrix}
   4 & 3 & 2 \\
\end{matrix} \\
 & \begin{matrix}
   1 & 2 & 3 \\
\end{matrix} \\
 & \begin{matrix}
   6 & 2 & 3 \\
\end{matrix} \\
\end{align} \right|\]
\[\begin{align}
  & \Rightarrow \left| A \right|=4\left( 6-6 \right)-3\left( 3-18 \right)+2\left( 2-12 \right) \\
 & \Rightarrow \left| A \right|=4.0-3\left( -15 \right)+2\left( -10 \right) \\
 & \Rightarrow \left| A \right|=45-20 \\
 & \Rightarrow \left| A \right|=25 \\
\end{align}\]
The determinant value of matrix A is 25 ………….(9)
Here, we got \[\left| A \right|\ne 0\] .
It means that our system of linear equations is consistent and has a unique solution.
From equation (5), we have
\[AX=B\]
\[\Rightarrow X={{A}^{-1}}B\] ………………………(10)
In equation (10), we can see that we need the value of the matrix \[{{A}^{-1}}\] .
We know the formula, \[{{A}^{-1}}=\dfrac{adj\left[ A \right]}{\left| A \right|}\] …………………(11)
Here, we need the value of \[adj\left[ A \right]\] and the determinant value of matrix A.
Our matrix A is, \[\left| A \right|=\left| \begin{align}
  & \begin{matrix}
   4 & 3 & 2 \\
\end{matrix} \\
 & \begin{matrix}
   1 & 2 & 3 \\
\end{matrix} \\
 & \begin{matrix}
   6 & 2 & 3 \\
\end{matrix} \\
\end{align} \right|\] .
To obtain the adjoint of matrix A we need the cofactor of each element.
For the first row, we have
Cofactor of 4 = \[2\times 3-2\times 3=6-6=0\] .
Cofactor of 3 = \[-\left( 3\times 1-6\times 3 \right)=-\left( 3-18 \right)=15\] .
Cofactor of 2 = \[1\times 2-6\times 2=2-12=-10\] .
For the second row, we have
Cofactor of 1 = \[-\left( 3\times 3-2\times 2 \right)=-\left( 9-4 \right)=-5\] .
Cofactor of 2 = \[4\times 3-6\times 2=12-12=0\] .
Cofactor of 3 = \[-\left( 4\times 2-6\times 3 \right)=-\left( 8-18 \right)=10\] .
For the third row, we have
Cofactor of 6 = \[\left( 3\times 3-2\times 2 \right)=\left( 9-4 \right)=5\] .
Cofactor of 2 = -\[-\left( 4\times 3-2\times 1 \right)=-\left( 12-2 \right)=-10\] .
Cofactor of 3 = \[\left( 4\times 2-3\times 1 \right)=\left( 8-3 \right)=5\] .
Now, our matrix with the cofactors is \[\left[ \begin{align}
  & \begin{matrix}
   0 & 15 & -10 \\
\end{matrix} \\
 & \begin{matrix}
   -5 & 0 & 10 \\
\end{matrix} \\
 & \begin{matrix}
   5 & -10 & 5 \\
\end{matrix} \\
\end{align} \right]\] .
We know that for the adjoint of a matrix we interchange the rows with the columns of a cofactor matrix.
On interchanging we get,
\[adj\left[ A \right]=\left[ \begin{align}
  & \begin{matrix}
   0 & -5 & 5 \\
\end{matrix} \\
 & \begin{matrix}
   15 & 0 & -10 \\
\end{matrix} \\
 & \begin{matrix}
   -10 & 10 & 5 \\
\end{matrix} \\
\end{align} \right]\] ……………………………….(12)
Now, from equation (9), equation (11), and equation (12), we get
\[{{A}^{-1}}=\dfrac{\left[ \begin{align}
  & \begin{matrix}
   0 & -5 & 5 \\
\end{matrix} \\
 & \begin{matrix}
   15 & 0 & -10 \\
\end{matrix} \\
 & \begin{matrix}
   -10 & 10 & 5 \\
\end{matrix} \\
\end{align} \right]}{25}=\dfrac{1}{25}\times 5\times \left[ \begin{align}
  & \begin{matrix}
   0 & -1 & 1 \\
\end{matrix} \\
 & \begin{matrix}
   3 & 0 & -2 \\
\end{matrix} \\
 & \begin{matrix}
   -2 & 2 & 1 \\
\end{matrix} \\
\end{align} \right]\]
\[{{A}^{-1}}=\dfrac{1}{5}\times \left[ \begin{align}
  & \begin{matrix}
   0 & -1 & 1 \\
\end{matrix} \\
 & \begin{matrix}
   3 & 0 & -2 \\
\end{matrix} \\
 & \begin{matrix}
   -2 & 2 & 1 \\
\end{matrix} \\
\end{align} \right]\] ……………………………….(13)
Now, from equation (7), equation (8), equation (10), and equation (13), we get
\[X={{A}^{-1}}B\]
\[\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\dfrac{1}{5}\times \left[ \begin{align}
  & \begin{matrix}
   0 & -1 & 1 \\
\end{matrix} \\
 & \begin{matrix}
   3 & 0 & -2 \\
\end{matrix} \\
 & \begin{matrix}
   -2 & 2 & 1 \\
\end{matrix} \\
\end{align} \right]\left[ \begin{matrix}
   60 \\
   45 \\
   70 \\
\end{matrix} \right]\]
On solving, we get
\[\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\dfrac{1}{5}\times \left[ \begin{matrix}
   0-45+70 \\
   180+0-140 \\
   -120+90+70 \\
\end{matrix} \right]\]
\[\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\dfrac{1}{5}\times \left[ \begin{matrix}
   25 \\
   40 \\
   40 \\
\end{matrix} \right]\]
\[\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\left[ \begin{matrix}
   \dfrac{1}{5}\times 25 \\
   \dfrac{1}{5}\times 40 \\
   \dfrac{1}{5}\times 40 \\
\end{matrix} \right]\]
\[\left[ \begin{matrix}
   x \\
   y \\
   z \\
\end{matrix} \right]=\left[ \begin{matrix}
   5 \\
   8 \\
   8 \\
\end{matrix} \right]\]
Therefore, the cost of potato, wheat, and rice per kg is Rs. 5, Rs. 8, and Rs.8 respectively.

Note: We can solve this question by the substitution method.
Let us assume the cost of potato, wheat, and rice per kg be x, y, and z respectively.
It is given that the cost of 4 kg potato, 3 kg wheat, and 2 kg of rice is Rs. 60. The cost of 1 kg potato, 2 kg wheat, and 3 kg rice is Rs. 45. The cost of 6 kg potato, 2 kg wheat, and 3 kg rice is Rs. 70. Expressing it in mathematical form, we get
\[4x+3y+2z=60\] ……………………………….(1)
\[x+2y+3z=45\] ……………………………….(2)
\[6x+2y+3z=70\] ……………………………….(3)
Solving equation (2), we get
\[x+2y+3z=45\]
\[\Rightarrow x=45-2y-3z\] …………………………..(4)
Now, from equation (1) and (4), we get
\[\begin{align}
  & \Rightarrow 4\left( 45-2y-3z \right)+3y+2z=60 \\
 & \Rightarrow 180-8y-12z+3y+2z=60 \\
 & \Rightarrow 120=5y+10z \\
 & \Rightarrow 24=y+2z \\
\end{align}\]
\[\Rightarrow 24-2z=y\] ………………………….(5)
Now, from equation (3), equation (4), and equation (5), we get
\[\begin{align}
  & 6\left\{ 45-2\left( 24-2z \right)-3z \right\}+2\left( 24-2z \right)+3z=70 \\
 & \Rightarrow 6\left( 45-48+4z-3z \right)+48-4z+3z=70 \\
 & \Rightarrow 6\left( -3+z \right)-z=70-48 \\
 & \Rightarrow -18+6z-z=22 \\
 & \Rightarrow 5z=40 \\
 & \Rightarrow z=8 \\
\end{align}\]
Putting the value of z in equation (5), we get
\[\begin{align}
  & \Rightarrow 24-2\left( 8 \right)=y \\
 & \Rightarrow 24-16=y \\
 & \Rightarrow 8=y \\
\end{align}\]
Now, putting the value of y and z in equation (4), we get
\[\begin{align}
  & \Rightarrow x=45-2\left( 8 \right)-3\left( 8 \right) \\
 & \Rightarrow x=45-16-24 \\
 & \Rightarrow x=45-40 \\
 & \Rightarrow x=5 \\
\end{align}\]
Therefore, the cost of potato, wheat, and rice per kg is Rs. 5, Rs. 8, and Rs.8 respectively.