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Question
Solve the following equations by method of inversion.
2x + y = 5, 3x + 5y = – 3
Solution
Matrix form of the given system of equations is
`[(2, 1),(3, 5)] [(x),(y)] = [(5),(-3)]`
This is of the form AX = B,
where A = `[(2, 1),(3, 5)], "X" = [(x),(y)] "and B" = [(5),(-3)]`
To deterrmine X, we have to find A–1.
|A| = `|(2, 1),(3, 5)|`
= 10 – 3
= 7 ≠ 0
∴ A–1 exissts.
Consider AA–1 = I
∴ `[(2, 1),(3, 5)] "A"^-1 = [(1, 0),(0, 1)]`
Applying R1 ↔ R2, we get
`[(3, 5),(2, 1)] "A"^-1 = [(0, 1),(1, 0)]`
Applying R1 → R1 – R2, we get
`[(1, 4),(2, 1)] "A"^-1 = [(-1, 1),(1, 0)]`
Applying R2 → R2 – 2R1, we get
`[(1, 4),(0, -7)] "A"^-1 = [(-1, 1),(3, -2)]`
Applying R2 → `(-1/7)` R2, we get
`[(1, 4),(0, 1)] "A"^-1 = [(-1, 1),(-3/7, 2/7)]`
Applying R1 → R1 – 4R2, we get
`[(1, 0),(0, 1)] "A"^-1 = [(5/7, -1/7),(-3/7, 2/7)]`
∴ A–1 =`[(5/7, -1/7),(-3/7, 2/7)]`
Pre-multiplying AX = B by A–1, we get
A–1(AX) = A–1B
∴ (A–1 A)X = A–1 B
∴ Ix = A–1 B
∴ X = A–1 B
∴ `[(x),(y)] = [(5/7, -1/7),(-3/7, 2/7)][(5),(-3)]`
∴ `[(x),(y)] = [(25/7, (+3)/7),(-15/7, -6/7)] = [(28/7),(-21/7)] = [(4),(-3)]`
∴ By equality of matrices, we get
x = 4 and y = – 3.
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