Question

Find the adjoint of matrix A = `[(2, 0, -1),(3, 1, 2),(-1, 1, 2)]`

Answer 1

A₁₁ = (–1)¹⁺¹ M₁₁ = `1|(1, 2),(1, 2)|` = 1(2 – 2) = 0

A₁₂ = (–1)¹⁺² M₁₂ = `(-1)|(3, 2),(-1, 2)|` = (–1)(6 + 2) = –8

A₁₃ = (–1)¹⁺³ M₁₃ = `1|(3, 1),(-1, 1)|` = 1(3 + 1) = 4

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

A22 = (–1)²⁺² M₂₂ = `1|(2, -1),(-1, 2)|` = 1(4 – 1) = 3

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

A₃₁ = (–1)³⁺¹ M₃₁ = `1|(0, -1),(1, 2)|` = 1(0 + 1) = 1

A₃₂ = (–1)³⁺² M₃₂ =  `(-1)|(2, -1),(3, 2)|` = (–1)(4 + 3) = –7

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

∴ adj (A) = `[("A"_11, "A"_12, "A"_13),("A"_21, "A"_22, "A"_23),("A"_31, "A"_32, "A"_33)]^"T"`

= `[(0, -8, 4),(-1, 3, -2),(1, -7, 2)]^"T"`

= `[(0, -1, 1),(-8, 3, -7),(4, -2, 2)]`