Question

The prices of three commodities P, Q and R are rupees $x,y$ and $z$ unit respectively. A purchases 4 units of R and sells 3 units of P and 5 units of Q. B purchases 3 units of Q and sells 2 units of P and 1 unit of R, C purchases 1 unit of P and sells 4 units of Q and 6 units of R. In the process A, B, C earns Rs.5000, 6000 and 13,000 respectively. If selling the units is positive earnings and buying the units is negative earnings, find the price per unit of three commodities by using the matrix method.

Answer 1

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

Complete step-by-step answer:

Now we can see in question,

Let us take price per unit commodity P be $x$, price per unit commodity Q be$y$and price per unit commodity R be$y$.

A purchases 4 units of R and sells 3 units of P and 5 units of Q.

So, for A we get the equation as,

$3x+5y-4z=6000$……….. (1)

Now B purchases 3 units of Q and sells 2 units of P and 1 unit of R, so for B we get the equation as,

$2x-3y+z=5000$………… (2)

So, for C we get the equation as,

$-x+4y+6z=13000$………… (3)

Now equations (1), (2) and (3), will be written in matrix form as,

$DX=B$

Where $D=\left[ \begin{matrix}

3 & 5 & -4 \\

2 & -3 & 1 \\

-1 & 4 & 6 \\

\end{matrix} \right]$,$X=\left[ \begin{matrix}

x \\

y \\

z \\

\end{matrix} \right]$and$$B=\left[ \begin{matrix}

6000 \\

5000 \\

13000 \\

\end{matrix} \right]$$,

Now to find the price per unit of three commodities, we have to find $x,y$ and $z$, so we have to find the matrix $X$.

We have, $DX=B$, we know$D{{D}^{-1}}=I$

Here $I$ is an identity matrix.

$X={{D}^{-1}}B$

Let us check if ${{D}^{-1}}$ exists or not.

$\left| D \right|=3(-18-4)-5(12+1)-4(8-3)=-151\ne 0$

Therefore, we can see above that D is a non-singular matrix and the inverse of D will exist.

Now we have to find 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}

-3 & 1 \\

4 & 6 \\

\end{matrix} \right|=-18-4=-22 \\

& {{D}_{12}}={{(-1)}^{1+2}}\left| \begin{matrix}

2 & 1 \\

-1 & 6 \\

\end{matrix} \right|=-(12+1)=-13 \\

& {{D}_{13}}={{(-1)}^{1+3}}\left| \begin{matrix}

2 & -3 \\

-1 & 4 \\

\end{matrix} \right|=8-3=5 \\

& {{D}_{21}}={{(-1)}^{2+1}}\left| \begin{matrix}

5 & -4 \\

4 & 6 \\

\end{matrix} \right|=-(30+16)=-46 \\

& {{D}_{22}}={{(-1)}^{2+2}}\left| \begin{matrix}

3 & -4 \\

-1 & 6 \\

\end{matrix} \right|=18-4=14 \\

& {{D}_{23}}={{(-1)}^{2+3}}\left| \begin{matrix}

3 & 5 \\

-1 & 4 \\

\end{matrix} \right|=-(12+5)=-17 \\

& {{D}_{31}}={{(-1)}^{3+1}}\left| \begin{matrix}

5 & -4 \\

-3 & 1 \\

\end{matrix} \right|=5-12=-7 \\

& {{D}_{32}}={{(-1)}^{3+2}}\left| \begin{matrix}

3 & -4 \\

2 & 1 \\

\end{matrix} \right|=-(3+8)=-11 \\

& {{D}_{33}}={{(-1)}^{3+3}}\left| \begin{matrix}

3 & 5 \\

2 & -3 \\

\end{matrix} \right|=-9-10=-19 \\

\end{aligned}$

So the matrix of cofactor,

$D$ $=\left[ \begin{matrix}

-22 & -13 & 5 \\

-46 & 14 & -17 \\

-7 & -11 & -19 \\

\end{matrix} \right]$

So, when we take transpose of matrix of cofactor D, we get,

$adj(D)={{\left[ \begin{matrix}

-22 & -13 & 5 \\

-46 & 14 & -17 \\

-7 & -11 & -19 \\

\end{matrix} \right]}^{T}}=\left[ \begin{matrix}

-22 & -46 & -7 \\

-13 & 14 & -11 \\

5 & -17 & -19 \\

\end{matrix} \right]$

Now adj(D) becomes,

$adj(D)=\left[ \begin{matrix}

-22 & -46 & -7 \\

-13 & 14 & -11 \\

5 & -17 & -19 \\

\end{matrix} \right]$

We know,

${{D}^{-1}}=\dfrac{1}{\left| D \right|}adjD$

Substituting the corresponding values, we get

${{D}^{-1}}=-\dfrac{1}{151}\left[ \begin{matrix}

-22 & -46 & -7 \\

-13 & 14 & -11 \\

5 & -17 & -19 \\

\end{matrix} \right]$

Now we have got ${{D}^{-1}}$, so now we know $X={{D}^{-1}}B$, we get,

& \left[ \begin{matrix} x \\\ y \\\ z \\\ \end{matrix} \right]=-\dfrac{1}{151}\left[ \begin{matrix} -22 & -46 & -7 \\\ -13 & 14 & -11 \\\ 5 & -17 & -19 \\\ \end{matrix} \right]\left[ \begin{matrix} 6000 \\\ 5000 \\\ 13000 \\\ \end{matrix} \right] \\\ & \left[ \begin{matrix} x \\\ y \\\ z \\\ \end{matrix} \right]=-\dfrac{1}{151}\left[ \begin{matrix} -22\times 6000-46\times 5000-7\times 13000 \\\ -13\times 6000+14\times 5000-11\times 13000 \\\ 5\times 6000-17\times 5000-19\times 13000 \\\ \end{matrix} \right] \\\ & \\\ & \left[ \begin{matrix} x \\\ y \\\ z \\\ \end{matrix} \right]=-\dfrac{1}{151}\left[ \begin{matrix} -453000 \\\ -151000 \\\ -302000 \\\ \end{matrix} \right] \\\ \end{aligned}$$ Simplifying in simple manner we get, $$\left[ \begin{matrix} x \\\ y \\\ z \\\ \end{matrix} \right]=\left[ \begin{matrix} 3000 \\\ 1000 \\\ 2000 \\\ \end{matrix} \right]$$ Now we have to get the three prices per commodities. So, price per unit commodity, P = Rs.3000, Q = Rs.1000, R = Rs.2000 Note: Read the question carefully. You should know how to convert a matrix into its inverse. You should be familiar with the inverse ${{D}^{-1}}=\dfrac{1}{\left| D \right|}adj\left( D \right)$. Most of the mistakes occur while finding adjoint. The more mistakes will occur in finding out the cofactor that is of minus sign so take care of it. Another approach to solve the three equations is elimination or substitution method.