Cayley-Hamilton theorem verification [4,3,1;2,1,−2;…]

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.

A = [4 3 1; 2 1 −2; 1 2 1]

Characteristic equation

• Trace = 4 + 1 + 1 = 6
• Sum of principal minors = (1·1 − (−2)·2) + (4·1 − 1·1) + (4·1 − 3·2) = 5 + 3 − 2 = 6
• |A| = 4(1 + 4) − 3(2 + 2) + 1(4 − 1) = 20 − 12 + 3 = 11

λ³ − 6λ² + 6λ − 11 = 0

By the theorem: A³ − 6A² + 6A − 11I = O.

Inverse

|A| = 11 ≠ 0, so A⁻¹ exists. Multiply the relation by A⁻¹:

A² − 6A + 6I − 11A⁻¹ = O   ⇒   A⁻¹ = (1/11)(A² − 6A + 6I)

Compute A²:

A² = [4 3 1][4 3 1]   [23 17 −1]
     [2 1 −2][2 1 −2] = [ 8  3 −2]
     [1 2 1][1 2 1]   [ 9  7 −2]

Then

A² − 6A + 6I = [23−24+6   17−18    −1−6 ]   [ 5 −1 −7]
               [ 8−12     3−6+6    −2+12]  = [−4  3 10]
               [ 9−6      7−12     −2−6+6]   [ 3 −5 −2]

Answer: A⁻¹ = (1/11) [5 −1 −7; −4 3 10; 3 −5 −2].

Check: row 1 of A times column 1 of the matrix: (4·5 + 3·(−4) + 1·3)/11 = (20 − 12 + 3)/11 = 1 ✓; row 1 times column 2: (−4 + 9 − 5)/11 = 0 ✓.