Cayley-Hamilton Theorem (2x2)

The power of A, A³, is given.
So use the Cayley-Hamilton theorem.

Write A²

minus,
a + d, 2 + 0
A.

plus,
ad - bc, 2⋅0 - 3⋅1
I

is equal to O.

So
A² - (2 + 0)A + (2⋅0 - 3⋅1)I = O.

-(2 + 0)A = -2A
+(2⋅0 - 3⋅1)I = +(0 - 3)I

+(0 - 3)I = -3I

Move -2A - 3I to the right side.

Then A² = 2A + 3I.

See the given equation
A³ = 7A + 6I.

Start from the left side A³.

A³ = AA²

A² = 2A + 3I

Then
AA² = A(2A + 3I).

Substitution Method

A(2A + 3I)
= 2A² + 3AI
= 2A² + 3A

Multiply a Monomial and a Polynomial

A² = 2A + 3I

Then
2A² + 3A = 2(2A + 3I) + 3A.

2(2A + 3I) = 4A + 6I

4A + 3A = 7A

A³ = 7A + 6I

So
A³ = 7A + 6I
is true.

This is the solution of this example.

The power of A, A¹⁰, is given.
So use the Cayley-Hamilton theorem.

Write A²

minus,
a + d, -1 + 1
A.

plus,
ad - bc, (-1)⋅1 - 1⋅1
I

is equal to O.

So
A² - (-1 + 1)A + ((-1)⋅1 - 1⋅1)I = O.

-(-1 + 1)A = -0A
+((-1)⋅1 - 1⋅1)I = +(-1 - 1)I

+(-1 - 1)I = -2I

Move -2I to the right side.

Then A² = 2I.

It says to find A¹⁰.

To use A² = 2I,
change A¹⁰ to (A²)⁵.

Power of a Power

A² = 2I

Then
(A²)⁵ = (2I)⁵.

(2I)⁵ = 2⁵I⁵

Power of a Product

2⁵ = 32
I⁵ = I

So A¹⁰ = 32I.