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Question
Find the inverse of A = `[(2, -3, 3),(2, 2, 3),(3, -2, 2)]` by using elementary row transformations.
Solution
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 R2 → R2 – R1 and R3 → 2R3 – 3R1, we get
`[(2, -3, 3),(0, 5, 0),(0, 5, -5)]` A−1 = `[(1, 0, 0),(-1, 1, 0),(-3, 0, 2)]`
Applying R3 → R3 – R2, we get
`[(2, -3, 3),(0, 5, 0),(0, 0, -5)]` A−1 = `[(1, 0, 0),(-1, 1, 0),(-2, -1, 2)]`
Applying `R_2 → (1/5) R_2` and `R_3 → (1/5) R_3`, we get
`[(2, -3, 3),(0, 1, 0),(0, 0, 1)]` A−1 = `[(1, 0, 0),((-1)/5, 1/5, 0),(2/5, 1/5, (-2)/5)]`
Applying R1 → R1 + 3R2, we get
`[(2, 0, 3),(0, 1, 0),(0, 0, 1)]` A−1 = `[(2/5, 3/5, 0),((-1)/5, 1/5, 0),(2/5, 1/5, (-2)/5)]`
Applying R1 → R1 – 3R3, we get
`[(2, 0, 0),(0, 1, 0),(0, 0, 1)]` A−1 = `1/5[(-4, 0, 6),(-1, 1, 0),(2, 1, -2)]`
Applying `R_1 → (1/2) R_1`, we get
`[(1, 0, 0),(0, 1, 0),(0, 0, 1)]` A−1 = `1/5[(-2, 0, 3),(-1, 1, 0),(2, 1, -2)]`
∴ A−1 = `1/5[(-2, 0, 3),(-1, 1, 0),(2, 1, -2)]`
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