Question

Let A = `[(1,2,1),(2,3,1),(1,1,5)]` verify that

1. [adj A]^–1 = adj(A^–1)
2. (A^–1)^–1 = A

Answer 1

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