Answer to Question #216780 in Linear Algebra for sahal

Question

A company that manufactures cars used three different types of steels S1, S2, and S3 for producing three cars C1, C2, and C3. The tons of steel requirement in each type of car is given in below table Cars Steel C1 C2 C3 S1 1 1 1 S2 1 2 4 S3 4 2 3 Determine the number of cars of each type which can be produced using 45, 120 and 130 tones of steel of the three types respectively. (Hint: use gaussian method)

Answer 1

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c} & C_1 & C_2 & C_3 \\ \hline S_1 & 1 & 1 & 1 \\ \hdashline S_2 & 1 & 2 & 4 \\ \hdashline S_3 & 4 & 2 & 3 \end{array}$$

$$\begin{matrix} x+y+z=45 \\ x+2y+4z=120 \\ 4x+2y+3z=130 \end{matrix}$$

$$\begin{pmatrix} 1 & 1 & 1 & & 45 \\ 1 & 2 & 4 & & 120 \\ 4 & 2 & 3 & & 130\\ \end{pmatrix}$$

$R_2=R_2-R_1$

$$\begin{pmatrix} 1 & 1 & 1 & & 45 \\ 0 & 1 & 3 & & 75 \\ 4 & 2 & 3 & & 130\\ \end{pmatrix}$$

$R_3=R_3-4R_1$

$$\begin{pmatrix} 1 & 1 & 1 & & 45 \\ 0 & 1 & 3 & & 75 \\ 0 & -2 & -1 & & -50\\ \end{pmatrix}$$

$R_1=R_1-R_2$

$$\begin{pmatrix} 1 & 0 & -2 & & -30 \\ 0 & 1 & 3 & & 75 \\ 0 & -2 & -1 & & -50\\ \end{pmatrix}$$

$R_3=R_3+2R_2$

$$\begin{pmatrix} 1 & 0 & -2 & & -30 \\ 0 & 1 & 3 & & 75 \\ 0 & 0 & 5 & & 100\\ \end{pmatrix}$$

$R_3=R_3/5$

$$\begin{pmatrix} 1 & 0 & -2 & & -30 \\ 0 & 1 & 3 & & 75 \\ 0 & 0 & 1 & & 20\\ \end{pmatrix}$$

$R_1=R_1+2R_3$

$$\begin{pmatrix} 1 & 0 &0 & & 10 \\ 0 & 1 & 3 & & 75 \\ 0 & 0 & 1 & & 20\\ \end{pmatrix}$$

$R_2=R_2-3R_3$

$$\begin{pmatrix} 1 & 0 &0 & & 10 \\ 0 & 1 & 0 & & 15 \\ 0 & 0 & 1 & & 20\\ \end{pmatrix}$$

$$x=10, y=15, z=20$$

Number of cars produced by type $C_1,C_2$ and $C_3$ are $10,15$ and $20$ respectively.