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Question
Find inverse of the following matrices (if they exist) by elementary transformations :
`[(2, 0, -1),(5, 1, 0),(0, 1, 3)]`
Solution
Let A = `[(2, 0, -1),(5, 1, 0),(0, 1, 3)]`
∴ |A| = `|(2, 0, -1),(5, 1, 0),(0, 1, 3)|`
= 2(3 – 0) –0 –1(5 – 0)
= 6 – 0 – 5
= 1 ≠ 0
∴ A–1 existts.
Consider AA–1 = I
∴ `[(2, 0, 1),(5, 1, 0),(0, 1, 3)] "A"^-1 = [(1, 0, 0),(0, 1, 0),(0, 0, 1)]`
Applying R1 ↔ R2, we get
`[(5, 1, 0),(2, 0, -1),(0, 1, 3)] "A"^-1 = [(-2, 1, 0),(1, 0, 0),(0, 0, 1)]`
Applying R1 → R1 – 2R2, we get
`[(1, 1, 2),(2, 0, -1),(0, 1, 3)] "A"^-1 = [(-2, 1, 0),(1, 0, 0),(0, 0, 1)]`
Applying R2 → R2 – 3R3, we get
`[(1, 1, 2),(0, 1, 4),(0, 1, 3)] "A"^-1 = [(-2, 1, 0),(5, -2, 3),(0, 0, 1)]`
Applying R1 → R1 – R2 and R3 → R3 – R2 , we get
`[(1, 0, -2),(0, 1, 4),(0, 0, -1)] "A"^-1 = [(-7, 3, -3),(5, -2, 3),(-5, 2, -2)]`
Applying R3 → (– 1) R3, we get
`[(1, 0, -2),(0, 1, 4),(0, 0, 1)] "A"^-1 = [(-7, 3, -3),(5, -2, 3),(5, -2, 2)]`
Applying R1 → R1 + 2R3 and R2 → R2 – 4R3 , we get
`[(1, 0, 0),(0, 1, 0),(0, 0, 1)] "A"^-1 = [(3, -1, 1),(-15, 6, -5),(5, -2, 2)]`
∴ A–1 = `[(3, -1, 1),(-15, 6, -5),(5, -2, 2)]`.
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