Question

The cost of 4kg onion, 3kg wheat, and 2kg rice is Rs 60. The cost of 2kg onion, 4kg wheat, and 6kg rice is Rs 90. The cost of 6kg onion, 2kg wheat, and 3kg rice is Rs 70. Find the cost of each item per kg by matrix method.

Answer 1

Hint: Write the equations. Then convert it in the form of a matrix $X={{D}^{-1}}B$ and find the inverse. And so place the inverse matrix and solve it you will get the price per unit of three commodities.

Now we can see in question.
Let the cost of each item per kg be $x$,$y$ and $z$.
So for first, we get the equation as,
$4x+3y+2z=60$……….. (1)
So for a second, we get the equation as,
$2x+4y+6z=90$………… (2)
So for the third, we get the equation as,
$6x+2y+3z=70$ ………… (3)
So now equation (1), (2) and (3) can be written in matrix form that is,
$DX=B$
Where $D=\left[ \begin{matrix} 4 & 3 & 2 \\\ 2 & 4 & 6 \\\ 6 & 2 & 3 \\\ \end{matrix} \right]$,
$X=\left[ \begin{matrix} x \\\ y \\\ z \\\ \end{matrix} \right]$and$B=\left[ \begin{matrix} 60 \\\ 90 \\\ 70 \\\ \end{matrix} \right]$,
So now to find the cost of each item per kg, we have to find $x$, $y$ and $z$.
So we have to find a matrix $X$.
So we have, $DX=B$
We know $D{{D}^{-1}}=I$
where $I$ is the identity matrix.
So $X={{D}^{-1}}B$
So let us check if${{D}^{-1}}$it exists or not.
So$\left| D \right|=4(12-12)-3(6-36)+2(4-24)=50\ne 0$
Therefore we can see above that $D$ is a nonsingular matrix and the inverse of $D$ exists.
So now we have to find an adjoint $D$.
Let $A=\left[ {{a}_{ij}} \right]$ be a square matrix of order $n$ . The adjoint of a matrix $A$ is the transpose of the cofactor matrix of $A$ . It is denoted by adj $A$ .
Given a square matrix A, the transpose of the matrix of the cofactor of A is called adjoint of A and is denoted by adj A. An adjoint matrix is also called an adjugate matrix.
In other words, we can say that matrix A is another matrix formed by replacing each element of the current matrix by its corresponding cofactor and then taking the transpose of the new matrix formed.

So now we have to find the inverse of $D$.
$\begin{aligned} & {{D}_{11}}={{(-1)}^{1+1}}\left| \begin{matrix} 4 & 6 \\\ 2 & 3 \\\ \end{matrix} \right|=12-12=0 \\\ & {{D}_{12}}={{(-1)}^{1+2}}\left| \begin{matrix} 2 & 6 \\\ 6 & 3 \\\ \end{matrix} \right|=-(6-36)=30 \\\ & {{D}_{13}}={{(-1)}^{1+3}}\left| \begin{matrix} 2 & 4 \\\ 6 & 2 \\\ \end{matrix} \right|=4-24=-20 \\\ & {{D}_{21}}={{(-1)}^{2+1}}\left| \begin{matrix} 3 & 2 \\\ 2 & 3 \\\ \end{matrix} \right|=-(9-4)=-5 \\\ & {{D}_{22}}={{(-1)}^{2+2}}\left| \begin{matrix} 4 & 2 \\\ 6 & 3 \\\ \end{matrix} \right|=12-12=0 \\\ & {{D}_{23}}={{(-1)}^{2+3}}\left| \begin{matrix} 4 & 3 \\\ 6 & 2 \\\ \end{matrix} \right|=-(8-18)=10 \\\ & {{D}_{31}}={{(-1)}^{3+1}}\left| \begin{matrix} 3 & 2 \\\ 4 & 6 \\\ \end{matrix} \right|=18-8=10 \\\ & {{D}_{32}}={{(-1)}^{3+2}}\left| \begin{matrix} 4 & 2 \\\ 2 & 6 \\\ \end{matrix} \right|=-(24-4)=-20 \\\ & {{D}_{33}}={{(-1)}^{3+3}}\left| \begin{matrix} 4 & 3 \\\ 2 & 4 \\\ \end{matrix} \right|=16-6=10 \\\ & \\\ \end{aligned}$
So the matrix of cofactors of $D$ $=\left[ \begin{matrix} 0 & 30 & -20 \\\ -5 & 0 & 10 \\\ 10 & -20 & 10 \\\ \end{matrix} \right]$.
So now adj $D$ becomes,
adj $D$ $=\left[ \begin{matrix} 0 & -5 & 10 \\\ 30 & 0 & -20 \\\ -20 & 10 & 10 \\\ \end{matrix} \right]$
So we know ${{D}^{-1}}=\dfrac{1}{\left| D \right|}adjD$.
So now, ${{D}^{-1}}=\dfrac{1}{50}\left[ \begin{matrix} 0 & -5 & 10 \\\ 30 & 0 & -20 \\\ -20 & 10 & 10 \\\ \end{matrix} \right]$.
So now we have got ${{D}^{-1}}$, so now we know $X={{D}^{-1}}B$.
So now we get,

& \left[ \begin{matrix} x \\\ y \\\ z \\\ \end{matrix} \right]=\dfrac{1}{50}\left[ \begin{matrix} 0 & -5 & 10 \\\ 30 & 0 & -20 \\\ -20 & 10 & 10 \\\ \end{matrix} \right]\left[ \begin{matrix} 60 \\\ 90 \\\ 70 \\\ \end{matrix} \right] \\\ & \left[ \begin{matrix} x \\\ y \\\ z \\\ \end{matrix} \right]=\dfrac{1}{50}\left[ \begin{matrix} 0-5\times 90+10\times 70 \\\ 30\times 60+0-20\times 70 \\\ -20\times 60+10\times 90+10\times 70 \\\ \end{matrix} \right] \\\ & \\\ & \left[ \begin{matrix} x \\\ y \\\ z \\\ \end{matrix} \right]=\dfrac{1}{50}\left[ \begin{matrix} 250 \\\ 400 \\\ 400 \\\ \end{matrix} \right] \\\ \end{aligned}$$ So simplifying the above equation by dividing each element inside the matrix by $50$, we get, $$\left[ \begin{matrix} x \\\ y \\\ z \\\ \end{matrix} \right]=\left[ \begin{matrix} 5 \\\ 8 \\\ 8 \\\ \end{matrix} \right]$$ So we have got the cost of each item per kg of $x$, $y$ and $z$ as $5$ rupees,$8$ rupees, and$8$ rupees respectively. Note: Read the question carefully. You should know how to convert a matrix into its inverse. You should be familiar with the inverse ${{D}^{-1}}=\dfrac{1}{\left| D \right|} adjD$. Most of the mistakes occur while finding adjoint. The more mistakes are seen in finding out the cofactor that is of the minus sign so take care of it.