Question

Find the inverse of the following matrices by the adjoint method `[(1, 2, 3),(0, 2, 4),(0, 0, 5)]`.

Answer 1

Let A = `[(1, 2, 3),(0, 2, 4),(0, 0, 5)]`

∴ |A| = `|(1, 2, 3),(0, 2, 4),(0, 0, 5)|`

= 1(10 – 0) –2(0 –0) + 3(0 –0) = 10 ≠ 0
∴ A^–1 exists.
A₁₁ = (– 1)¹⁺¹ M₁₁ = `1|(2, 4),(0, 5)|` = 1(10 – 0) = 10

A₁₂ = (– 1)¹⁺² M₁₂ = `-1|(0, 4),(0, 5)|` = 1(0 – 0) = 0

A₁₃ = (– 1)¹⁺³ M₁₃ = `1|(0, 2),(0, 0)|` = 1(0 – 0) = 0

A₂₁ = (– 1)²⁺¹ M₂₁ = `-1|(2, 3),(0, 5)|` = – 1(10 – 0) = – 10

A₂₂ = (– 1)²⁺² M₂₂ = `1|(1, 3),(0, 5)|` = 1(5 – 0) = 5

A₂₃ = (– 1)²⁺³ M₂₃ = `-1|(1, 2),(0, 0)|` = –1(0 – 0) = 0

A₃₁ = (– 1)³⁺¹ M₃₁ = `1|(2, 3),(2, 4)|` = 1(8 – 6) = 2

A₃₂ = (– 1)³⁺² M₃₂ = `-1|(1, 3),(0, 4)|` = –1(4 – 0) = – 4

A₃₃ = (– 1)³⁺³ M₃₃ = `1|(1, 2),(0, 2)|` = 1(2 – 0) = 2

∴ The matrix of the co-factor is

[A_ij]_3x3 = `[("A"_11, "A"_12, "A"_13),("A"_21, "A"_22, "A"_23),("A"_31, "A"_32, "A"_33)] = [(10, 0, 0),(-10, 5, 0),(2, -4, 2)]`

Now,adj A = `["A"_"ij"]_(3xx3)^"T" = [(10, -10, 2),(0, 5, -4),(0, 0, 2)]`

∴ A^–1 = `(1)/|"A"| ("adj A") = (1)/(10)[(10, -10, 2),(0, 5, -4),(0, 0, 2)]`.