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Question
Let A = `[(1,-2,1),(-2,3,1),(1,1,5)]` verify that
- [adj A]–1 = adj (A–1)
- (A–1)–1 = A
Solution
A = `[(1,-2,1),(-2,3,1),(1,1,5)]`
∴ |A| = 1(15 - 1) + 2(-10 - 1) + 1(-2 - 3)
= 14 - 22 - 5
= -13
Now, `A_11 = 14, A_12 = 11, A_13 = -5`
`A_21 = 11, A_22 = 4, A_23 = -3`
`A_31 = -5, A_32 = -3, A_33 = -1`
∴ adj A = `[(14,11,-5),(11,4,-3),(-5,-3,-1)]`
∴ `A^-1 = 1/|A|`(adj A)
= `-1/13[(14,11,-5),(11,4,-3),(-5,-3,-1)] = 1/13[(-14,-11,5),(-11,-4,3),(5,3,1)]`
(i) |adj A| = 14(-4 - 9) - 11(-11 - 15)-5(-33 + 20)
= 14(-13) - 11(-26) - 5(-13)
= -183 + 286 + 65 = 169
we have,
adj(adj A) = `[(-13,26,-13),(26,-39,-13),(-13,-13,-65)]`
∴ `[adj A]^-1 = 1/|adj A|(adj(adjA))`
= `1/169[(-13,26,-13),(26,-39,-13),(-13,-13,-65)]`
= `1/13[(-1,2,-1),(2,-3,-1),(-1,-1,-5)]`
Now, `A^-1 = 1/13[(-14,-11,5),(-11,-4,3),(5,3,1)] = [(-14/13, -11/13, 5/13),(-11/13,-4/13,3/13),(5/13,3/13,1/13)]`
∴ `adj(A^-1) = [(-4/169 - 9/169, -(-11/169 - 15/169),-33/169 + 20/169),(-(-11/169 - 15/169),(-14/169 - 25/169), -(42/169 + 55/169)),(-33/169 + 20/169, -(42/169 + 55/169), 56/169 - 121/169)]`
= `1/169[(-13,26,-13),(26,-39,-13),(-13,-13,-65)] = 1/13[(-1,2,-1),(2,-3,-1),(-1,-1,-5)]`
So, `[adjA]^-1 = adj(A^-1)`.
(ii) We have shown that:
`A^-1 = 1/13[(-14,-11,5),(-11,-4,3),(5,3,1)]`
or, `adjA^-1 = 1/13[(-1,2,-1),(2,-3,-1),(-1,-1,-5)]`
Now,
`|A^-1| = (1/13)^3[-14 xx (-13) + 11 xx (-26) + 5 xx (-13)]`
`= (1/13)^3 xx (-169) `
`= -1/13`
∴ `(A^-1)^-1 = (adjA^-1)/|A^-1| `
`= 1/((-1/13)) xx 1/13[(-1,2,-1),(2,-3,-1),(-1,-1,-5)] `
= `[(1,-2,1),(-2,3,1),(1,1,5)]` = A
Hence, (A-1)-1 = A
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