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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
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Solution
A = `[(1,2,1),(2,3,1),(1,1,5)]`
∴ |A| = 1(3 × 5 − 1 × 1) − 2(2 × 5 − 1 × 1) + 1(2 × 1 − 3 × 1)
= 1(15 − 1) −2(10 − 1) + 1(2 − 3)
= 14 − 18 − 1
= −5
Now, A11 = 14, A12 = −9, A13 = −1
A21 = −9, A22 = 4, A23 = 1
A31 = −1, A32 = 1, A33 = −1
∴ adj A = `[(14,-9,-1),(-9,4,1),(-1,1,-1)]`
∴ A−1 = `1/|A|` (adj A)
= `-1/5[(14,-9,-1),(-9,4,1),(-1,1,-1)]`
(i) L.H.S. = |adj A|
= 14(−4 − 1) + 9(9 + 1) − 1(−9 + 4)
= 14(−5) + 9(10) − 1(−5)
= 70 − 90 + 5
= 25
We have,
adj(adj A) = `[(-5,-10,-5),(-10,-15,-5),(-5,-5,-25)]`
Hence its inverse exists;
[adj A]−1 = `1/|adj A|` [adj(adj A)]
= `1/25[(-5,-10,-5),(-10,-15,-5),(-5,-5,-25)]`
L.H.S. = (adj A)−1 = `[(-1/5,-2/5,-1/5),(-2/5,-3/5,-1/5),(-1/5,-1/5,-1)]`
Now finding R.H.S.:
A−1 = `[((-14)/5,9/5,1/5),(9/5,(-4)/5,(-1)/5),(1/5,(-1)/5,1/5)]`
adj(A–1) = `[(-1/5,-2/5,-1/5),(-2/5,-3/5,-1/5),(-1/5,-1/5,-1)]`
Hence, [adj A]–1 = adj(A–1)
(ii) We have shown that:
A−1 = `-1/5[(14,-9,-1),(-9,4,1),(-1,1,-1)]`
A−1 = `1/|A| = 1/(-5)`
adj(A–1) = `[(-1/5,-2/5,-1/5),(-2/5,-3/5,-1/5),(-1/5,-1/5,-1)]`
Then,
(A−1)−1 = `1/|A^-1|` adj(A−1)
= `-1/(1/5) [(-1/5,-2/5,-1/5),(-2/5,-3/5,-1/5),(-1/5,-1/5,-1)]`
= `[(1,2,1),(2,3,1),(1,1,5)]`
Hence, (A–1)–1 = A
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