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प्रश्न
If A = `[(2, 3),(1, 2)]`, B = `[(1, 0),(3, 1)]`, find AB and (AB)−1
उत्तर
AB = `[(2, 3),(1, 2)][(1, 0),(3, 1)]``
= `[(2 + 9, 0 + 3),(1 + 6, 0 + 2)]`
= `[(11, 3),(7, 2)]`
Now, |AB| = `|(11, 3),(7, 2)|`
= 22 − 21
= 1 ≠ 0
∴ (AB)−1 exists.
Consider, (AB)(AB)−1 = I
∴ `[(11, 3),(7, 2)]` (AB)−1 = `[(1, 0),(0, 1)]`
Applying R1 → 2R1,
`[(22, 6),(7, 2)]` (AB)−1 = `[(2, 0),(0, 1)]`
Applying R1 → R1 − 3R2,
`[(1, 0),(7, 2)]` (AB)−1 = `[(2, -3),(0, 1)]`
Applying R2 → R2 − 7R1,
`[(1, 0),(0, 2)]` (AB)−1 = `[(2, -3),(-14, 22)]`
Applying R2 → `(1/2)` R2
`[(1, 0),(0, 1)]` (AB)−1 = `[(2, -3),(-7, 11)]`
∴ (AB)−1 = `[(2, -3),(-7, 11)]` .......(1)
|A| = `|(2,3),(1,2)| = 4 - 3 = 1 ne 0`
∴ A−1 exists.
Consider, AA−1 = I
∴ `[(2, 3),(1, 2)]` A−1 = `[(1, 0),(0, 1)]`
Applying R1 ↔ R2,
`[(1, 2),(2, 3)]` A−1 = `[(0, 1),(1, 0)]`
Aplying 2R2 → R1,
`[(1, 2),(0, -1)]` A−1 = `[(0, 1),(1, -2)]`
Applying (−1)R2 ↔ R1,
`[(1, 2),(0, 1)]` A−1 = `[(0, 1),(-1, 2)]`
Applying R1 ↔ R2,
`[(1, 0),(0, 1)]` A−1 = `[(2, -3),(-1, 2)]`
∴ A−1 = `[(2, -3),(-1, 2)]`
|B| = `|(1, 0),(3, 1)|`
= 1 − 0
= 1 ≠ 0
∴ B−1 exists.
Consider, BB−1 = I
∴ `[(1, 0),(3, 1)]` B−1 = `[(1, 0),(0, 1)]`
Applying R2 ↔ 3R1,
`[(1, 0),(0, 1)]` A−1 = `[(1, 0),(-3, 1)]`
∴ B−1 = `[(1, 0),(-3, 1)]`
∴ B−1. A−1 = `[(1, 0),(-3, 1)],[(2, -3),(-1, 2)]`
= `[(2 - 0, -3+0),(-6-1, 9+2)]`
= `[(2, -3),(-7,11)]` .......(2)
From (1) and (2), (AB)−1 = B−1. A−1
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