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प्रश्न
Solve the following equations by method of inversion : x – y + z = 4, 2x + y – 3z = 0 , x + y + z = 2
उत्तर
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 = B, 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)
= 4 + 5 + 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 R3 → `(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)/3, 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)]`
Pre-multiplying AX = B by A–1, we get
A–1(AX) = A–1B
∴ (A–1A) X = A–1B
∴ 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)]`
∴ By equality of matrices, we get
x = 2, y = – 1 and z = 1.
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