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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.
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Solution
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
\[D = \begin{vmatrix}2 & 3 & 4 \\ 1 & 1 & 2 \\ 3 & 2 & 1\end{vmatrix}\]
\[ = 2(1 - 4) - 3(1 - 6) + 4(2 - 3)\]
\[ = - 6 + 15 - 4\]
\[ = 5\]
\[ D_1 = \begin{vmatrix}29 & 3 & 4 \\ 13 & 1 & 2 \\ 16 & 2 & 1\end{vmatrix}\]
\[ = 29(1 - 4) - 3(13 - 32) + 4(26 - 16)\]
\[ = - 87 + 57 + 40\]
\[ = 10\]
\[ D_2 = \begin{vmatrix}2 & 29 & 4 \\ 1 & 13 & 2 \\ 3 & 16 & 1\end{vmatrix}\]
\[ = 2(13 - 32) - 29(1 - 6) + 4(16 - 39)\]
\[ = - 38 + 145 - 92\]
\[ = 15\]
\[ D_3 = \begin{vmatrix}2 & 3 & 29 \\ 1 & 1 & 13 \\ 3 & 2 & 16\end{vmatrix}\]
\[ = 2(16 - 26) - 3(16 - 39) + 29(2 - 3)\]
\[ = - 20 + 69 - 29\]
\[ = 20\]
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 C3 cars can be produced using the three types of steel.
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