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# Solve the following equations by method of inversion : x + y – z = 2, x – 2y + z = 3 and 2x – y – 3z = – 1 - Mathematics and Statistics

Sum

Solve the following equations by method of inversion : x + y – z = 2, x – 2y + z = 3 and 2x – y – 3z = – 1

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#### Solution

Matrix form of the given system of equations is

[(1, 1, -1),(1, -2, 1),(2, -1, -3)] [(x),(y)] = [(2),(3),(-1)]

This is of the form AX = B, where

A = [(1, 1, -1),(1, -2, 1),(2, -1, -3)],"X" = [(x),(y)] "and B" = [(2),(3),(-1)]

To determine X, we have to find A–1.

|A|= |(1, 1, -1),(1, -2, 1),(2, -1, -3)|

= 1(6 + 1) – 1(–3 – 2) –1(–1 + 4)
= 1(7) –1(–5)–1(3)
= 7 + 5 – 3
= 9 ≠ 0
∴ A–1 exists.
Consider AA–1

∴ [(1, 1, -1),(1, -2, 1),(2, -1, -3)]"A"^-1 = [(1, 0, 0),(0, 1, 0),(0, 0, 1)]

Applying R2 → R2 – R1 and R3 → R3 – 2R1, we get

[(1, 1, -1),(0, -3, 2),(0, -3, -1)] "A"^-1 = [(1, 0, 0),(-1, 1, 0),(-2, 0, 1)]

Applying R2 → ((-1)/3) R2, we get

[(1, 1, -1),(0, 1, -2/3),(0, -3, -1)] "A"^-1 = [(1, 0, 0),(1/3, (-1)/3, 0),(-2, 0, 1)]

Applying R1 → R1 – R2 and R3 → R3 + 3R2, we get

[(1, 0, -1/3),(0, 1, (-2)/3),(0, 0, -3)] "A"^-1 = [(2/3, 1/3, 0),(1/3, -1/3, 0),(-1, -1, 1)]

Applying R3 → (-1/3) R3, we get

[(1, 0, -1/3),(0, 1, -2/3),(0, 0, 1)] "A"^-1 = [(2/3, 1/3, 0),(1/3, -1/3, 0),(1/3, 1/3, -1/3)]

Applying R1 → R1 + (-1/3) R3 and R2 → R2 + (2/3) R3, we get

[(1, 0, 0),(0, 1, 0),(0, 0, 1)] "A"^-1 = [(7/9, 4/9, -1/9),(5/9, -1/9, -2/9),(1/3, 1/3, -1/3)]

∴ A–1 = (1)/(9)[(7, 4, -1),(5, -1, -2),(3, 3, -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)/(9)[(7, 4, -1),(5, -1, -2),(3, 3, -3)][(2),(3),(-1)]

∴ [(x),(y),(z)] = (1)/(9)[(14 + 12 + 1),(10 - 3 + 2),(6 + 9 + 3)]

= (1)/(9)[(27),(9),(18)]

= [(3),(1),(2)]

∴ By equality of martices, we get
x = 3, y = 1 and z = 2.

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#### APPEARS IN

Balbharati Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board
Chapter 2 Matrices
Miscellaneous Exercise 2 | Q 4.18 | Page 85
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