Question

If A = `[(1, -1, 2),(3, 0, -2),(1, 0, 3)]`, verify that A(adj A) = (adj A)A

Answer 1

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

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

A₁₂ = (−1)¹⁺² M₁₂ = `1|(3, 0),(1, 0)|` = −1(9 + 2) = −11

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

A₂₁ = (−1)²⁺¹ M₂₁ = `-1|(-1, 2),(0, 3)|` = −1(−3 − 0) = 3

A₂₂ = (−1)²⁺² M₂₂ = `1|(1, 2),(1, 3)|` = 1(3 − 2) = 1

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

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

A³² = (−1)³⁺² M₃₂ = `-1|(1, 2),(3, -2)|` = −1(−2 − 6) = 8

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

Hence, matrix of the co-factors is

`[("A"_11, "A"_12, "A"_13),("A"_21, "A"_22, "A"_23),("A"_31, "A"_32, "A"_33)] = [(0, -11, 0),(3, 1, -1),(2, 8, 3)]`

= `["A"_"ij"]_(3 xx 3)`

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

= `[(0, 3, 2),(-11, 1, 8),(0, -1, 3)]`

∴ A(adj A) = `[(1, -1, 2),(3, 0, -2),(1, 0, 3)] [(0,3, 2),(-11,1,8),(0, -1, 3)]`

= `[(0 + 11 + 0, 3 - 1 - 2, 2 - 8 + 6),(0 + 0 + 0, 9 + 0 + 2, 6 + 0 - 6),(0 + 0 + 0, 3 + 0 - 3, 2 + 0 + 9)]`

= `[(11, 0, 0),(0, 11, 0),(0, 0, 11)]`    .......(i)

(adj A)A = `[(0, 3, 2),(-11, 1, 8),(0, -1, 3)] [(1, -1, 2),(3, 0, -2),(1, 0, 3)]`

= `[(0 + 9 + 2, 0 + 0 + 0, 0 - 6 + 6),(-11 + 3 + 8, 11 + 0 + 0, -22 - 2 + 24),(0 - 3 + 3, 0 - 0 + 0, 0 + 2 + 9)]`

= `[(11, 0, 0,(0, 11, 0),(0 0 11)]`   .......(ii)

From equations (i) and (ii), we get

A(adj A) = (adj A)A