Question

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

Answer 1

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

∴ |A| = `|(1,-1,2),(3,0,-2),(1,0,3)|`

= 1(0 + 0) + 1(9 + 2) + 2(0 − 0)
= 0 + 11 + 0
= 11

First we have to find the co-factor matrix = [A_ij]_3×3′
where A_ij = (− 1)^i+j M_ij

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

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

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

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

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

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

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

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

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

Hence the co-factor matrix

= `[("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)]`

∴ adj A = `[(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)]` .................(1)

(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)]` ...............(2)

|A| I = 11`[(1,0,0),(0,1,0),(0,0,1)]=[(11,0,0),(0,11,0),(0,0,11)]` ............(3)

From (1), (2) and (3), we get,
A (adj A) = (adj A)A = |A| I.

[Note: This relation is valid for any non-singular matrix A]