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Question
Using the matrix method, solve the following system of linear equations:
`2/x + 3/y + 10/z` = 4, `4/x - 6/y + 5/z` = 1, `6/x + 9/y - 20/z` = 2.
Solution
The given system of equations can be written in the form AX = B,
Where, A = `[(2, 3, 10),(4, -6, 5),(6, 9, -20)]`, X = `[(1//x),(1//y),(1//z)]` and B = `[(4),(1),(2)]`
Now, |A| = `|(2, 3, 10),(4, -6, 5),(6, 9, -20)|`
= 2(120 – 45) – 3(–80 – 30) + 10(36 + 36)
= 2(75) – 3(–100) + 10(72)
= 150 + 330 + 720
= 1200 ≠ 0
∴ A–1 exists.
∴ adj A = `[(75, 110, 72),(150, -100, 0),(75, 30, -24)]^T = [(75, 150, 75),(110, -100, 30),(72, 0, -24)]`
Hence, A–1 = `1/|A| (adjA) = 1/1200[(75, 150, 75),(110, -100, 30),(72, 0, -24)]`
As, AX = B
`\implies` X = A–1B
`\implies [(1/x),(1/y),(1/z)] = 1/1200[(75, 150, 75),(110, -100, 30),(72, 0, -24)][(4),(1),(2)]`
= `1/1200[(300 + 150 + 150),(440 - 100 + 60),(288 + 0 - 48)]`
`\implies [(1/x),(1/y),(1/z)] = 1/1200[(600),(400),(240)] = [(1/2),(1/3),(1/5)]`
Thus, `1/x = 1/2, 1/y = 1/3, 1/z = 1/5`
Hence, x = 2, y = 3, z = 5.
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