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Question
If A = `[(2,3),(1,2)]`, B = `[(1,0),(3,1)]`, find AB and (AB)-1 . Verify that (AB)-1 = B-1.A-1.
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Solution
AB = `[(2,3),(1,2)] [(1,0),(3,1)]`
`= [(2+9,0+3),(1+6,0+2)] = [(11,3),(7,2)]`
∴ |AB| = `|(11,3),(7,2)| = 22 - 21 = 1 ne 0`
∴ (AB)-1 exists.
Now, (AB)(AB)-1 = I
∴ `[(11,3),(7,2)]("AB")^-1 = [(1,0),(0,1)]`
By 2R1, we get,
`[(22,6),(7,2)] ("AB")^-1 = [(2,0),(0,1)]`
By R1 - 3R2, we get,
`[(1,0),(7,2)] ("AB")^-1 = [(2,-3),(0,1)]`
By R2 - 7R1, we get,
`[(1,0),(0,2)] ("AB")^-1 = [(2,-3),(-14,22)]`
By `(1/2)"R"_2,` we get,
`[(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)]`
By R1 ↔ R2, we get,
`[(1,2),(2,3)] "A"^-1 = [(0,1),(1,0)]`
By `"R"_2 - 2"R"_1,`we get,
`[(1,2),(0,-1)] "A"^-1 = [(0,1),(1,-2)]`
By `(-1)"R"_2`, we get,
`[(1,2),(0,1)] "A"^-1 = [(0,1),(-1,2)]`
By `"R"_1 - 2"R"_2`, we get,
`[(1,0),(0,1)] "A"^-1 = [(2,-3),(-1,2)]`
∴ A-1 = `[(2,-3),(-1,2)]`
|B| = `|(1,0),(3,1)| = 1 - 0 = 1 ne 0`
∴ B-1 exists.
Consider, BB-1 = I
∴ `[(1,0),(3,1)] "B"^-1 = [(1,0),(0,1)]`
By `"R"_2 - 3"R"_1`, we get,
`[(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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