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Question
Find the inverse of matrix A = `[(1, 0, 1),(0, 2, 3),(1, 2, 1)]` by using elementary row transformations
Solution
A = `[(1, 0, 1),(0, 2, 3),(1, 2, 1)]`
∴ |A| = `|(1, 0, 1),(0, 2, 3),(1, 2, 1)|`
= 1(2 – 6) – 0 + 1(0 – 2)
= 1(– 4) + 1(– 2)
= – 4 – 2
= – 6 ≠ 0
∴ A–1 exists.
Consider AA–1 = I
∴ `[(1, 0, 1),(0, 2, 3),(1, 2, 1)]` A–1 = `[(1, 0, 0),(0, 1, 0),(0, 0, 1)]`
Applying R3 → R3 – R1, we get
`[(1, 0, 1),(0, 2, 3),(0, 2, 0)]` A–1 = `[(1, 0, 0),(0, 1, 0),(-1, 0, 1)]`
Applying R2 → `(1/2)` R2, we get
`[(1, 0, 1),(0, 1, 3/2),(0, 2, 0)]` A–1 = `[(1, 0, 0),(0, 1/2, 0),(-1, 0, 1)]`
Applying R3 → R3 – 2R2, we get
`[(1, 0, 1),(0, 1, 3/2),(0, 0, -3)]` A–1 = `[(1, 0, 0),(0, 1/2, 0),(-1, -1, 1)]`
Applying R3 → `(-1/3)` R3, we get
`[(1, 0, 1),(0, 1, 3/2),(0, 0, 1)]` A–1 = `[(1, 0, 0),(0, 1/2, 0),(1/3, 1/3, -1/3)]`
Applying R1 → R1 – R3, R2 → R2 – `(3/2)` R3, we get
`[(1, 0, 0),(0, 1, 0),(0, 0, 1)]` A–1 = `[(2/3, -1/3, 1/3),(-1/2, 0, 1/2),(1/3, 1/3, -1/3)]`
∴ A–1 = `[(2/3, -1/3, 1/3),(-1/2, 0, 1/2),(1/3, 1/3, -1/3)]`
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