Question

If A = `[(1, 0, 0),(3, 3, 0),(5, 2, -1)]`, find A⁻¹ by the adjoint method

Answer 1

Let A = `[(1,0,0),(3,3,0),(5,2,-1)]`

∴ |A| = `|(1,0,0),(3,3,0),(5,2,-1)|`

= 1(− 3 − 0) − 0 + 0

= − 3 ≠ 0

∴ A⁻¹ exist

A₁₁ = (−1)¹⁺¹ M₁₁ = `|(3,0),(2,-1)|` = −3 − 0 = −3

A₁₂ = (−1)¹⁺² M₁₂ = −`|(3,0),(5,-1)|` = −(−3 − 0) = 3

A₁₃ = (−1)¹⁺³ M₁₃ = `|(3,3),(5,2)|` = 6 − 15 = −9

A₂₁ = (−1)²⁺¹ M₂₁ = −`|(0,0),(2,-1)|` = −(0 − 0) = 0

A₂₂ = (−1)²⁺² M₂₂ = `|(1, 0),(5, -1)|` = − 1 − 0 = −1

A₂₃ = (−1)²⁺³ M₂₃ = −`|(1,0),(5,2)|` = −(2 − 0) = −2

A₃₁ =(−1)³⁺¹ M₃₁ = `|(0,0),(3,0)|` = 0 − 0 = 0

A₃₂ = (−1)³⁺² M₃₂ = −`|(1,0),(3,0)|` = −(0 − 0) = 0

A₃₃ = (−1)³⁺³ M₃₃ = `|(1,0),(3,3)|` = 3 − 0 = 3

Hence, matrix of the co-factors is

`["A"_"ij"]_(3 xx 3) = [("A"_11,"A"_12,"A"_13),("A"_21,"A"_22,"A"_23),("A"_31,"A"_32,"A"_33)]`

= `[(-3,3,-9),(0,-1,-2),(0,0,3)]`

Now, adj A = `["A"_"ij"]_(3 xx 3)^"T"`

= `[(-3,0,0),(3,-1,0),(-9,-2,3)]`

∴ A⁻¹ = `1/|"A"|` (adj A)

= `1/(-3)[(-3,0,0),(3,-1,0),(-9,-2,3)]` 

= `1/3[(3,0,0),(-3,1,0),(9,2,-3)]`