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Question
Solve the following system of equations by the method of inversion.
x – y + z = 4, 2x + y – 3z = 0, x + y + z = 2
Solution
Matrix form of the given system of equations is:
`[(1, -1, 1),(2, 1, -3),(1, 1, 1)][(x),(y),(z)] = [(4),(0),(2)]`
This is of the form AX= 8, where
A = `[(1, -1, 1),(2, 1, -3),(1, 1, 1)]`, X = `[(x),(y),(z)]` and B = `[(4),(0),(2)]`
To determine X, we have to find A–1.
|A| = `|(1, -1, 1),(2, 1, -3),(1, 1, 1)|` = 1(1 + 3) + 1(2 + 3) + 1(2 – 1)
= 1(4) + 1(5) + 1(1)
= 10 ≠ 0
∴ A–1 exists
Consider AA–1 = I
`[(1, -1, 1),(2, 1, -3),(1, 1, 1)] A^-1 = [(1, 0, 0),(0, 1, 0),(0, 0, 1)]`
Applying R2 → R2 – 2R1 and R3 → R3 – R1, we get
`[(1, -1, 1),(0, 3, -5),(0, 2, 0)] A^-1 = [(1, 0, 0),(-2, 1, 0),(-1, 0, 1)]`
Applying R2 → `1/3`(R2), we get
`[(1, -1, 1),(0, 1, (-5)/3),(0, 2, 0)] A^-1 = [(1, 0, 0),((-2)/3, 1/3, 0),(-1, 0, 1)]`
Applying R1 → R1 + R2 and R3 → R3 – 2R2, we get
`[(1, 0, (-2)/3),(0, 1, (-5)/3),(0, 0, 10/3)] A^-1 = [(1/3, 1/3, 0),((-2)/3, 1/3, 0),(1/3, (-2)/3, 1)]`
Applying R1 → `3/10`R3, we get
`[(1, 0, (-2)/3),(0, 1, (-5)/3),(0, 0, 1)] A^-1 = [(1/3, 1/3, 0),((-2)/3, 1/3, 0),(1/10, (-1)/5, 3/10)]`
Applying R1 → R1 + `(2/3)`R3 and R2 → R2 + `5/3`R3, we get
`[(1, 0, 0),(0, 1, 0),(0, 0, 1)] A^-1 = [(2/5, 1/5, 1/5),(-1/2, 0, 1/2),(1/10, (-1)/5, 3/10)]`
∴ A–1 = `1/10[(4, 2, 2),(-5, 0, 5),(1, -2, 3)]`
Now, AX = B
∴ A–1(AX) = A–1(B)
∴ IX = A–1B
∴ X = A–1B
∴ X = `1/10 [(4, 2, 2),(-5, 0, 5),(1, -2, 3)][(4),(0),(2)]`
∴ `[(x),(y),(z)] = 1/10 [(16 + 0 + 4),(-20 + 0 + 10),(4 - 0 + 6)]`
= `1/10 [(20),(-10),(10)]`
= `[(2),(-1),(1)]`
Hence. by equality of matrices x = 2, y = – 1, z = 1.
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