An automobile company uses three types of steel S1, S2 and S3

Solution:

Let $x, y$ and $z$ denote the number of cars that can be produced of each type. Then,

$2 x+3 y+4 z=29$

$x+y+2 z=13$

$3 x+2 y+z=16$

Using Cramer's rule, we get

$D=\left|\begin{array}{lll}2 & 3 & 4 \\ 1 & 1 & 2 \\ 3 & 2 & 1\end{array}\right|$

$=2(1-4)-3(1-6)+4(2-3)$

$=-6+15-4$

$=5$

$D_{1}=\left|\begin{array}{lll}29 & 3 & 4 \\ 13 & 1 & 2 \\ 16 & 2 & 1\end{array}\right|$

$=29(1-4)-3(13-32)+4(26-16)$

$=-87+57+40$

$=10$

$D_{2}=\left|\begin{array}{lll}2 & 29 & 4 \\ 1 & 13 & 2 \\ 3 & 16 & 1\end{array}\right|$

$=2(13-32)-29(1-6)+4(16-39)$

$=-38+145-92$

$=15$

$D_{3}=\left|\begin{array}{lll}2 & 3 & 29 \\ 1 & 1 & 13 \\ 3 & 2 & 16\end{array}\right|$

$=2(16-26)-3(16-39)+29(2-3)$

$=-20+69-29$

$=20$

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, $2 C_{1}$ cars, $3 C_{2}$ cars and $4 C_{3}$ cars can be produced using the three types of steel.