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Question
Find inverse of the following matrices (if they exist) by elementary transformations :
`[(2, -3, 3),(2, 2, 3),(3, -2, 2)]`
Solution
Let A = `[(2, -3, 3),(2, 2, 3),(3, -2, 2)]`
∴ |A| = `|(2, -3, 3),(2, 2, 3),(3, -2, 2)|`
= 2(4 + 6) + 3(4 – 9) + 3(–4 – 6)
= 20 – 15 – 30
= – 25 ≠ 0
∴ A–1 exists.
Consider AA–1 = I
∴ `[(2, -3, 3),(2, 2, 3),(3, -2, 2)] "A"^-1 = [(1, 0, 0),(0, 1, 0),(0, 0, 1)]`
Applying R1 → 2R1 – R3, we get
`[(1, -4, 4),(2, 2, 3),(3, -2, 2)] "A"^1 = [(2, 0, -1),(0, 1, 0),(0, 0, 1)]`
Applying R2 → R1 – 2R1 and R3 → R1 – 2R1, we get
`[(1, -4, 4),(0, 10, -5),(0, 10, -10)] "A"^-1 [(2, 0, -1),(-4, 1, 2),(-6, 0, 4)]`
Applying R2 → `(1/10)` R2 and R3 → `(-1/10)`R3, we get
`[(1, -4, 4),(0, 1, -1/2),(0, -1, 1)] "A"^-1 = [(2, 0, -1),(-4/10, 1/10, 2/10),(6/10, 0, -4/10)]`
Applying R1 → R1 – 4R2 and R3 → R3 + R2, we get
`[(1, 0, 2),(0, 1, -1/2),(0, 0, 1/2)] "A"^-1 = [(4/10, 4/10, -2/10),(-4/10, 1/10, 2/10),(2/10, 1/10, -2/10)]`
Applying R3 → 2R3 , we get
`[(1, 0, 2),(0, 1, -1/2),(0, 0, 1)] "A"^-1 = [(4/10, 4/10, -2/10),(-4/10, 1/10, 2/10),(4/10, 2/10, -4/10)]`
Applying R1 → R1 – 2R3 and R2 → R2 + `(1/2)`R3, we get
`[(1, 0, 0),(0, 1, 0),(0, 0,1)] "A"^-1 [(-4/10, 0, 6/10),(-2/10, 2/10, 0),(4/10, 2/10, -4/10)]`
∴ A–1 = `[(-2/5, 0, 3/5),(-1/5, 1/5, 0),(2/5, 1/5, -2/5)]`.
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