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Question
If A = `[(1, 2),(3, 5)]` B = `[(0, 4),(2, -1)]`, show that AB ≠ BA, but |AB| = IAl·IBI
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Solution
AB = `[(1, 2),(3, 5)] [(0, 4),(2, -1)]`
= `[(0 + 4, 4 - 2),(0 + 10, 12 - 5)]`
= `[(4, 2),(10, 7)]` ...(1)
BA = `[(0, 4),(2, -1)] [(1, 2),(3, 5)]`
= `[(0 + 12, 0 + 20),(2 - 3, 4 - 5)]`
= `[(12, 20),(-1, -1)]` ...(2)
From (1) and (2), AB ≠ BA.
Now, |AB| = `|(4, 2),(10, 7)|`
= 28 – 20 = 8 ...(3)
|A| = `|(1, 2),(3, 5)|` = 5 – 6 = – 1
|B| = `|(0, 4),(2, -1)|` = 0 – 8 = – 8
∴ |A|·|B| = (– 1)(– 8) = 8 ...(4)
From (3) and (4),
IABI = IAl·IBI.
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