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Question
Find the inverse of A = `[(1, 0, 1),(0, 2, 3),(1, 2, 1)]` by elementary column transformations.
Solution
|A| = `|(1, 0, 1),(0, 2, 3),(1, 2, 1)|`
= 1(2 – 6) – 0 + 1)0 – 2)
= – 4 – 2
= – 6 ≠ 0
∴ A–1 exists.
Consider A–1A = I
∴ A–1 = `[(1, 0, 1),(0, 2, 3),(1, 2, 1)] = [(1, 0, 0),(0, 1, 0),(0, 0, 1)]`
Applying C3 → C3 – C1, we get
A–1 = `[(1, 0, 0),(0, 2, 3),(1, 2, 0)] = [(1, 0, -1),(0, 1, 0),(0, 0, 1)]`
Applying C2 ↔ C3, we get
A–1 = `[(1, 0, 0),(0, 3, 2),(1, 0, 2)] = [(1, -1, 0),(0, 0, 1),(0, 1, 0)]`
Applying C2 → C2 – C3, we get
A–1 = `[(1, 0, 0),(0, 1, 2),(1, -2, 2)] = [(1, -1, 0),(0, -1, 1),(0, 1, 0)]`
Applying C3 → C3 – 2C2, we get
A–1 = `[(1, 0, 0),(0, 1, 0),(1, -2, 6)] = [(1, -1, 2),(0, -1, 3),(0, 1, -2)]`
Applying C3 → `(1/6)` C3, we get
A–1 = `[(1, 0, 0),(0, 1, 0),(1, -2, 1)] = [(1, -1, 1/3),(0, -1, 1/2),(0, 1, -1/3)]`
Applying C1 → C1 – C3 and C2 → C2 + 2C3, we get
A–1 = `[(1, 0, 0),(0, 1, 0),(0, 0, 1)] = [(2/3, -1/3, 1/3),(-1/2, 0, 1/2),(1/3, 1/3, -1/3)]`
∴ A–1 = `(1)/(6)[(4, -2, 2),(-3, 0, 3),(2, 2, -2)]`.
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