# Represent the following problem by a system of linear equations. Rs.188000 is the cost of 3 cycles, 1 motorbike and 1 car, the cost of 2 cycles, 2 motorbikes and 1 car is Rs.228000. The cost of 4 cycles 2 cars and 1 motorbike is Rs. 330000. Use matrix to find the cost of each.

Toolbox:
• If the value of the determinant of a $3\times 3$ matrix is not equal to zero,then it is a non-singular matrix.
• If it is a non-singular matrix, then inverse exists.
• $A^{-1}=\frac{1}{|A|}(adj A)$
• $A^{-1}X=B$
Step 1:
Let the cost of one cycle =x
Let the cost of one bike =y
Let the cost of one car =z
According to the given information,the equation can be formed as follows:
3x+y+z=188000
2x+2y+z=228000
4x+2y+z=330000
This is of the form AX=B.
Where $A=\begin{bmatrix}3 & 1&1\\2 & 2 & 1\\4 & 2 &1\end{bmatrix},X=\begin{bmatrix}x\\y\\z\end{bmatrix}$ and $B=\begin{bmatrix}188000\\228000\\330000\end{bmatrix}$
Now first let us find the determinant of A,by expanding along $R_1$
$|A|=3(2\times 1-2\times 1)-1(2\times 1-4\times 1)+1(2\times 2-4\times 2)$
$\;\;=0+2-4=-2\neq 0.$
Hence $A^{-1}$ exists.
Step 2:
Next let us find the cofactors of A,
$A_{11}=(-1)_{1+1}\begin{vmatrix}2 & 1\\2 & 1\end{vmatrix}$=2-2=0.
$A_{12}=(-1)_{1+2}\begin{vmatrix}2 & 1\\4 & 1\end{vmatrix}$=-(2-4)=2
$A_{13}=(-1)_{1+3}\begin{vmatrix}2 & 2\\4 & 2\end{vmatrix}$=4-8=-4.
$A_{21}=(-1)_{2+1}\begin{vmatrix}1 & 1\\2 & 1\end{vmatrix}$=-(1-2)=1.
$A_{22}=(-1)_{2+2}\begin{vmatrix}3 & 1\\4 & 1\end{vmatrix}$=3-4=-1.
$A_{23}=(-1)_{2+3}\begin{vmatrix}3 & 1\\4 & 2\end{vmatrix}$=-(6-4)=-2.
$A_{31}=(-1)_{3+1}\begin{vmatrix}1 & 1\\2 & 1\end{vmatrix}$=1-2=-1
$A_{32}=(-1)_{3+2}\begin{vmatrix}3 & 1\\2 & 1\end{vmatrix}$=-(3-2)=-1.
$A_{33}=(-1)_{1+1}\begin{vmatrix}3 & 1\\2 & 2\end{vmatrix}$=6-2=4.
Hence the adjoint of A is $\begin{bmatrix}A_{11} & A_{21} & A_{31}\\A_{12} & A_{22} & A_{32}\\A_{13} & A_{23} & A_{33}\end{bmatrix}$
$\qquad\qquad\qquad\qquad=\begin{bmatrix}0 & 1 & -1\\2 & -1 & -1\\-4& -2& 4\end{bmatrix}$
$A^{-1}=\frac{1}{|A|}adj(A)$,we know |A|=-2.
$A^{-1}=\frac{1}{-2}\begin{bmatrix}0 & 1 & -1\\2 & -1 & -1\\-4 & -2 & 4\end{bmatrix}$
Step 3:
$A^{-1}B=X$,substituting for $A^{-1}$,B and X we get
$\begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{1}{-2}\begin{bmatrix}0 & 1& -1\\2 &-1 & -1\\-4 & -2 & 4\end{bmatrix}\begin{bmatrix}188000\\228000\\330000\end{bmatrix}$
$\begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{-1}{2}\begin{bmatrix}0+228000-330000\\376000-228000-330000\\-752000-456000+1320000\end{bmatrix}$
$\begin{bmatrix}x\\y\\z\end{bmatrix}=\frac{-1}{2}\begin{bmatrix}-102000\\-182000\\-112000\end{bmatrix}$
x=51000
y=91000
z=56000
Cost of 1 cycle is Rs. 51000
Cost of 1 bike is Rs. 91000
Cost of 1 car is Rs. 56000