Question

Find the inverse of `[(3, 1, 5),(2, 7, 8),(1, 2, 5)]` by adjoint method.

Answer 1

Let A = `[(3, 1, 5),(2, 7, 8),(1, 2, 5)]`

|A| = `|(3, 1, 5),(2, 7, 8),(1, 2, 5)|`

= 3(35 – 16) – 1(10 – 8) + 5(4 – 7)

= 3(19) – 1(2) + 5(– 3)

= 57 – 2 – 15

= 40 ≠ 0

∴ A^–1 exists

A₁₁ = (–1)¹⁺¹ × M₁₁

= `1|(7, 8),(2, 5)|`

= 1(35 – 16)

= 19

A₁₂ = (–1)¹⁺² M₁₂

= `-1|(2, 8),(1, 5)|`

= 1(10 – 8) 

= –2

A₁₃ = (–1)¹⁺³ × M₁₃ 

`1|(2, 7),(1, 2)|`

= 1(4 – 7)

= –3

A₂₁ = (–1)²⁺¹ × M₂₁

= `-1|(1, 5),(2, 5)|`

= (–1)(5 – 10)

= 5

A₂₂ = (– 1)²⁺² × M₂₂

= `1|(3, 5),(1, 5)|`

= (1)(15 – 5)

= 10

A₂₃ = (–1)²⁺³ × M₂₃

= `(-1)|(3, 1),(1, 2)|`

= (–1)(6 – 1)

= –5

A₃₁ = (–1)³⁺¹ × M₃₁

= `(1)|(1, 5),(7, 8)|`

= (1)(8 – 35)

= –27

A₃₂ = (–1)³⁺² M₃₂

= `(-1)|(3, 5),(2, 8)|`

= (–1)(24 – 10)

= –14

A₃₃ = (–1)³⁺³ M₃₃

= `(+1)|(3, 1),(2, 7)|`

= 1(21 – 2)

= 19

∴ The matrix of the co-factors is

A = `[("A"_11, "A"_12, "A"_13),("A"_21, "A"_22, "A"_23),("A"_31, "A"_32, "A"_33)] = [(19, -2, -3),(5, 10, -5),(-27, -14, 19)]`

Now, adj A = `[(19, 5, -27),(-2, 10, -14),(-3, -5, 19)]`

∴ A^–1 = `(1)/|"A"| ("adj A")`

∴ A^–1 = `(1)/(40)[(19, 5, -27),(-2, 10, -14),(-3, -5, 19)]`