Question

If A = `[(2,-1,1),(-1,2,-1),(1,-1,2)]` verify that A³ − 6A² + 9A − 4I = 0 and hence find A⁻¹.

Answer 1

We have A = `[(2,-1,1),(-1,2,-1),(1,-1,2)]`

A² = AA

= `[(2,-1,1),(-1,2,-1),(1,-1,2)] [(2,-1,1),(-1,2,-1),(1,-1,2)]`

= `[(4+1+1,-2-2-1,2+1+2),(-2-2-1,1+4+1,-1-2-2),(2+1+2,-1-2-2,1+1+4)]`

= `[(6,-5,5),(-5,6,-5),(5,-5,6)]`

A³ = A²A

= `[(6,-5,5),(-5,6,-5),(5,-5,6)] [(2,-1,1),(-1,2,-1),(1,-1,2)]`

= `[(12+5+5,-6-10-5,6+5+10),(-10-6-5,5+12+5,-5-6-10),(10+5+6,-5-10-6,5+5+12)]`

= `[(22,-21,21),(-21,22,-21),(21,-21,22)]`

Now, A³ − 6A² + 9A − 4I

= `[(22,-21,21),(-21,22,-21),(21,-21,22)] - 6 [(6,-5,5),(-5,6,-5),(5,-5,6)] + 9 [(2,-1,1),(-1,2,-1),(1,-1,2)] - 4 [(1,0,0),(0,1,0),(0,0,1)]`

= `[(22,-21,21),(-21,22,-21),(21,-21,22)] - [(36,-30,30),(-30,36,-30),(30,-30,36)] + [(18,-9,9),(-9,18,-9),(9,-9,18)] - [(4,0,0),(0,4,0),(0,0,4)]`

= `[(22 - 36 + 18 - 4, -21 + 30 - 9 - 0, -21 - 30 + 9 - 0),(-21 + 30 - 9 - 0, 22 - 36 + 18 - 4, -21 - 30 + 9 - 0),(21 - 30 + 9 - 0, -21 + 30 - 9 - 0,22 - 36 + 18 - 4)]`

= `[(0,0,0),(0,0,0),(0,0,0)]`

= 0

Hence A³ − 6A² + 9A − 4I = 0

Now, A³ − 6A² + 9A − 4I = 0

⇒ 4I = A³ − 6A² + 9A

Multiplying both sides by A⁻¹, we get

⇒ `A^(−1) = 1/4 A^2 - 6/4 A + 9/4I`

= `1/4[(6,-5,5),(-5,6,-5),(5,-5,6)] - 6/4 [(2,-1,1),(-1,2,-1),(1,-1,2)] + 9/4 [(1,0,0),(0,1,0),(0,0,1)]`

= `1/4 [(6-12+9,-5+6+0,5-6+0),(-5+6+0,6-12+9,-5+6+0),(5-6+0,-5+6+0,6-12+9)]`

= `1/4 [(3,1,-1),(1,3,1),(-1,1,3)]`