Question

The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs.80. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is Rs.70.Find the cost of each item per kg by matrix method.

Answer 1

Now we can frame according to the given information as

$4x+3y+2z=60-----\left(1\right)$

$2x+4y+3z=10-----\left(2\right)$

$6x+2y+3z=70-----\left(3\right)$

This is of the form $AX=B$ then $X=A^{-1}B$

$\left[\begin{array}{ccc}4& 3& 2\\ 2& 4& 6\\ 6& 2& 3\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}60\\ 10\\ 70\end{array}\right]$

where $A=\left[\begin{array}{ccc}4& 3& 2\\ 2& 4& 6\\ 6& 2& 3\end{array}\right]X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]B=\left[\begin{array}{c}60\\ 10\\ 70\end{array}\right]$

Now lets find (adj $A$)

$m_{11}=\left|\begin{array}{cc}4& 6\\ 2& 3\end{array}\right|=12-12=0,m_{11}=\left|\begin{array}{cc}2& 6\\ 6& 3\end{array}\right|=6-36=-30m_{21}=\left|\begin{array}{cc}2& 4\\ 6& 2\end{array}\right|=4-24=-20,m_{21}=\left|\begin{array}{cc}3& 2\\ 2& 3\end{array}\right|=9-4=5m_{31}=\left|\begin{array}{cc}4& 2\\ 6& 3\end{array}\right|=12-12=0,m_{21}=\left|\begin{array}{cc}4& 3\\ 6& 2\end{array}\right|=8-18=-10m_{33}=\left|\begin{array}{cc}3& 2\\ 4& 6\end{array}\right|=18-8=10,m_{32}=\left|\begin{array}{cc}4& 2\\ 2& 6\end{array}\right|=24-4=20$

$m_{33}=\left|\begin{array}{cc}4& 3\\ 2& 4\end{array}\right|=16-6=10$

$A_{11}=(-1{)}^{1+1}(0)=0$

$A_{12}=(-1{)}^{1+2}(-30)=30$

$A_{13}=(-1{)}^{1+3}(-20)=-20$

$A_{21}=(-1{)}^{2+1}(5)=-5$

$A_{22}=(-1{)}^{2+2}(0)=0$

$A_{23}=(-1{)}^{2+3}(-10)=10$

$A_{31}=(-1{)}^{3+1}(10)=10$

$A_{32}=(-1{)}^{3+2}(20)=-20$

$A_{33}=(-1{)}^{3+3}(10)=10$

$\therefore adjA=\left[\begin{array}{ccc}{A}_{11}& A_{21}& A_{31}\\ A_{12}& A_{22}& A_{32}\\ A_{13}& A_{23}& A_{33}\end{array}\right]=\left[\begin{array}{ccc}0& -5& 10\\ 30& 0& -20\\ -20& 10& 10\end{array}\right]$

where $A^{-1}=\frac{1}{|A|}$ ( adj $A$) we know $|A|=50$

$A^{-1}=\frac{1}{50}\left[\begin{array}{ccc}0& -5& 10\\ 30& 0& -20\\ -20& 10& 10\end{array}\right]$

$X=A^{-1}B$ to solve the equation

$\therefore \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{50}\left[\begin{array}{ccc}0& -5& 10\\ 30& 0& -20\\ -20& 10& 10\end{array}\right]\left[\begin{array}{c}60\\ 90\\ 70\end{array}\right]$

$\therefore \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{50}\left[\begin{array}{ccc}0& -450& 10\\ 6& 0& -20\\ -20& 10& 10\end{array}\right]\left[\begin{array}{c}60\\ 90\\ 70\end{array}\right]$

$\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{50}\left[\begin{array}{c}250\\ 400\\ 400\end{array}\right]$

$\therefore \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}5\\ 8\\ 8\end{array}\right]$

$x=5,y=8,z=8$