Question

Find the inverse of matrix B = `[(3,1, 5),(2, 7, 8),(1, 2, 5)]` by using adjoint method

Answer 1

B = `[(3,1, 5),(2, 7, 8),(1, 2, 5)]` 

∴ |B| = `|(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

∴ B^–1 exists.

Here,

b₁₁ = 3

∴ M₁₁ = `|(7, 8),(2, 5)|`

= 35 – 16

= 19

and B₁₁ = (–1)¹⁺¹ M₁₁ = 19

b₁₂ = 1

∴ M₁₂ = `|(2, 8),(1, 5)|`

= 10 – 8

= 2

and B₁₂ = (–1)¹⁺² M₁₂ = – 2

b₁₃ = 5

∴ M₁₃ = `|(2, 7),(1, 2)|`

= 4 – 7

= – 3

and B₁₃ = (–1)¹⁺³ M₁₃ = – 3

b₂₁ = 2

∴ M₂₁ = `|(1, 5),(2, 5)|`

= 5 – 10

= – 5

and B₂₁ = (–1)²⁺¹ M₂₁ = 5

b₂₂ = 7

∴ M₂₂ = `|(3, 5),(1, 5)|`

= 15 – 5

= 10

and B₂₂ = ( –1)²⁺² M₂₂ = 10

b₂₃ = 8

∴ M₂₃ = `|(3, 1),(1, 2)|`

= 6 – 1

= 5

and B₂₃ = ( –1)²⁺³ M₂₃ = – 5

b₃₁ = 1

∴ M₃₁ = `|(1, 5),(7, 8)|`

= 8 – 35

= – 27

and B₃₁ = ( –1)³⁺¹ M₃₁ = – 27

b₃₂ = 2

∴ M₃₂ = `|(3, 5),(2, 8)|`

= 24 – 10

= 14

and B₃₂ = ( –1)³⁺² M₃₂ = – 14

b₃₃ = 5

∴ M₃₃ = `|(3, 1),(2, 7)|`

= 21 – 2

= 19

and B₃₃ = ( –1)³⁺³ M₃₃ = 19

∴ The matrix of the co-factors is

[Bij]_3×3 = `[("B"_11, "B"_12, "B"_13),("B"_21, "B"_22, "B"_23),("B"_31, "B"_32, "B"_33)]`

= `[(19, -2, -3),(5, 10, -5),(-27, -4, 19)]`

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

= `[(19, 5, -27),(-2, 10, -14),(-3, -5, 9)]`

∴ B^–1 = `1/|"B"|` (adj B)

= `1/40 [(19, 5, -27),(-2, 10, -14),(-3, -5, 9)]`