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Question
Solve the following system of linear equations by matrix inversion method:
x + y + z – 2 = 0, 6x – 4y + 5z – 31 = 0, 5x + 2y + 2z = 13
Solution
`[(1, 1, 1),(6, -4, 5),(5, 2, 2)][(x),(y),(z)] = [(2),(31),(13)]`
AX = B
X = A-1B
A = `[(1, 1, 1),(6, -4, 5),(5, 2, 2)]`
A| = 1(– 8 – 10) – 1(12 – 25) + 1(12 + 20)
= 18 + 13 +32 = 27
≠ 0
A–1 exists
adj A = `[((-8 - 10), -(12 - 25), (12 + 20)),(-(2 - 2), (2 - 5), -(2 - 5)),((5 + 4), -(5 - 6), (- 4 - 6))]^"T"`
= `[(- 18, 13, 32),(0, -3, 3),(9, 1, -10)]^"T"`
= `[(-18, 0, 9),(13, -3, 1),(32, 3, -10)]`
A–1 = `1/|"A"|` adj A
= `1/27 [(-18, 0, 9),(13, -3, 1),(32, 3, -10)]`
X = `"A"^-1"B"`
`[(x),(y),(z)] = 1/27 [(-18, 0, 9),(13, -3, 1),(32, 3, 10)][(2),(31),(13)]`
= `1/27 [(- 36 + 0 + 117),(26 - 93 + 13),(64 + 93 - 130)]`
`[(x),(y),(z)] = 1/27 [(81),(-54),(27)]`
= `[(3),(-2),(1)]`
∴ x = 3, y = – 2, z = 1
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