Question

An automobile company uses three types of steel S₁, S₂ and S₃ for producing three types of cars C₁, C₂and C₃. Steel requirements (in tons) for each type of cars are given below : 

	Cars C1	C2	C3
Steel S1	2	3	4
S2	1	1	2
S3	3	2	1

Using Cramer's rule, find the number of cars of each type which can be produced using 29, 13 and 16 tons of steel of three types respectively.

Answer 1

Let x, y and z denote the number of cars that can be produced of each type . Then, 
$$2x+3y+4z=29$$ 
$$x+y+2z=13$$ 
$$3x+2y+z=16$$ 
Using Cramer's rule, we get
$$D=\left|\begin{array}{ccc}2& 3& 4\\ 1& 1& 2\\ 3& 2& 1\end{array}\right|$$ 
$$=2(1-4)-3(1-6)+4(2-3)$$ 
$$=-6+15-4$$ 
$$=5$$ 
$$D_1=\left|\begin{array}{ccc}29& 3& 4\\ 13& 1& 2\\ 16& 2& 1\end{array}\right|$$ 
$$=29(1-4)-3(13-32)+4(26-16)$$ 
$$=-87+57+40$$ 
$$=10$$ 
$$D_2=\left|\begin{array}{ccc}2& 29& 4\\ 1& 13& 2\\ 3& 16& 1\end{array}\right|$$ 
$$=2(13-32)-29(1-6)+4(16-39)$$ 
$$=-38+145-92$$ 
$$=15$$ 
$$D_3=\left|\begin{array}{ccc}2& 3& 29\\ 1& 1& 13\\ 3& 2& 16\end{array}\right|$$ 
$$=2(16-26)-3(16-39)+29(2-3)$$ 
$$=-20+69-29$$ 
$$=20$$ 
Thus,
$$x=\frac{D_1}{D}=\frac{10}{5}=2$$ 
$$y=\frac{D_2}{D}=\frac{15}{5}=3$$ 
$$z=\frac{D_3}{D}=\frac{20}{5}=4$$
Therefore, 2 C₁ cars, 3 C₂ cars and 4 C_​3 cars can be produced using the three types of steel.