Cayley-Hamilton theorem verification [−2,2,−3;2,1,−6;…]

Cayley–Hamilton theorem: every square matrix satisfies its own characteristic equation. If |A − λI| = λ³ + c₁λ² + c₂λ + c₃ = 0, then A³ + c₁A² + c₂A + c₃I = O. It lets us write A⁻¹ as a polynomial in A.

A = [−2 2 −3; 2 1 −6; −1 −2 0]

Characteristic equation

• Trace = −1
• Sum of principal minors = −12 − 3 − 6 = −21
• |A| = 45

λ³ + λ² − 21λ − 45 = 0

So A³ + A² − 21A − 45I = O.

Inverse

|A| = 45 ≠ 0, so A⁻¹ exists. Multiply by A⁻¹:

A² + A − 21I − 45A⁻¹ = O   ⇒   A⁻¹ = (1/45)(A² + A − 21I)

Compute A²:

A² = [−2 2 −3][−2 2 −3]   [11   4  −6]
     [ 2 1 −6][ 2 1 −6] = [ 4  17 −12]
     [−1 −2 0][−1 −2 0]   [−2  −4  15]

Then A² + A − 21I:

[11−2−21    4+2     −6−3 ]   [−12   6  −9]
[ 4+2     17+1−21  −12−6 ] = [  6  −3 −18]
[−2−1      −4−2    15+0−21]   [ −3  −6  −6]

Answer: A⁻¹ = (1/45) [−12 6 −9; 6 −3 −18; −3 −6 −6].

Check: row 1 of A times column 1 of the matrix: (−2·(−12) + 2·6 + (−3)(−3))/45 = (24 + 12 + 9)/45 = 1 ✓; row 1 times column 2: (−12 − 6 + 18)/45 = 0 ✓.