Question

If A = `[(1,-1),(2,3)]` show that A² - 4A + 5I₂ = 0 and also find A^-1.

Answer 1

Given A = `[(1,-1),(2,3)] |"A"| = |(1,-1),(2,3)|` = 3 + 2 = 5 ≠ 0

∴ A^-1 exists.

LHS = A² - 4A + 5I

Premultiply by A^-1 we get

LHS = `"A"^-1. "A"^2 - 4"A"^-1."A" + 5"A"^-1."I"`

`= ("A"^-1 "A")."A" - 4"I" + 5"A"^-1`   ...[∵ A^-1A = I]

= IA - 4I + 5A^-1

= A - 4I + 5A^-1

Now `"A"^-1 = 1/|"A"| "adj  A"`

`= 1/5 [(3,1),(-2,1)]`

∴ LHS = `[(1,-1),(2,3)] - 4[(+1,0),(0,1)] + 5 . 1/5 [(3,1),(-2,1)]`

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

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

= RHS