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Question
Verify A (adj A) = (adj A) A = |A|I.
`[(2,3),(-4,-6)]`
Solution
Let, `A = [(2,3),(-4,-6)]`
`A_11 = (- 1)^(1 + 1) M_11 = - 6`
`A_12 = (- 1)^(1 + 2) M_12 = - (- 4) = 4`
`A_21 = (- 1)^(2 + 1) M_21 = - 3`
`A_22 = (- 1)^(2 + 2) M_22 = 2`
adjA = `[(-6,-3),(4,2)]`
|A| = `|(2,3),(-4,-6)|`
|A| = - 12 + 12
|A| = 0
L.H.S. = A(adj A) = `[(2,3),(-4,-6)] [(-6,-3),(4,2)]`
`= [(2 xx (- 6) + 3 xx 4, 2 xx (-3) + 3 xx 2),(- 4 xx (- 6) + (- 6) xx 4, - 4 xx (- 3) + (- 6) xx 2)]`
`= [(-12 + 12, -6 + 6),(24 - 24, 12 - 12)]`
`= [(0,0),(0,0)]`
= 0 = |A| · I
R.H.S. = (adj A) A `= [(-6,-3),(4,2)][(2,3),(-4,-6)]`
`= [(-12 + 12,-18 + 18),(8 - 8, 12 - 12)]`
`= [(0,0),(0,0)]`
`abs A . I = 0. [(1,0),(0,1)]`
`= [(0,0),(0,0)]`
= 0 = |A| I = 0
Hence, A(adj A) = (adj A) A = `abs A. I`
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