Question

If A = `[(5, 3),(-1, -2)]`, show that A² – 3A – 7I₂ = O₂. Hence find A^–1 

Answer 1

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

A² = A × A

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

= `[(25 - 3, 15 - 6),(-5 + 2, -3 + 4)]`

= `[(22, 9),(-3, 1)]`

 – 3A = `-3[(5, 3),(-1, -2)]`

= `[(-15, -9),(3, 6)]`

– 7I₂ = `-7[(1, 0),(0, 1)]`

= `[(-7, 0),(0, -7)]`

A² – 3A – 7I₂ 

= `[(22, 9),(-3, 1)] + [(15, -9),(3, 6)] + [(-7, 0),(0, -7)]`

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

∴ A² – 3A – 7I₂ = O₂

Post multiply this equation by A^–1

A²A^–1 – 3A A^–1 – 7I₂ A^–1 = 0

A – 3I – 7A^–1 = 0

A – 3I = 7 A^–1 

A^–1 = `1/7 ("A" - 3"I")`

= `1/7 [((5, 3),(-1, -2)), -3((1, 0),(0, 1))]`

= `1/7 [((5, 3),(-1, -2)) + ((-3, 0),(0, -3))]`

A^–1 = `1/7[(2, 3),(-1, -5)]`