Skip to main content

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

Short answer

Cayley–Hamilton: every square matrix satisfies its characteristic equation. Here λ³ − 32λ + 56 = 0, so A³ − 32A + 56I = O; multiplying by A⁻¹ gives A⁻¹ = (32I − A²)/56. The inverse is (1/56)[24 8 12; 2 10 −6; 10 −6 −2].

State Cayley-Hamilton theorem. Use it to find the inverse of the matrix [1 1 3; 1 3 −3; 2 −4 −4].

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 can be used to find A⁻¹ by multiplying the relation by A⁻¹.

A = [1 1 3; 1 3 −3; 2 −4 −4]

Characteristic equation

  • Trace = 1 + 3 − 4 = 0
  • Sum of principal minors = (3·(−4) − (−3)(−4)) + (1·(−4) − 3·2) + (1·3 − 1·1) = −24 − 10 + 2 = −32
  • |A| = 1(−12 − 12) − 1(−4 + 6) + 3(−4 − 6) = −24 − 2 − 30 = −56
λ³ − 0·λ² − 32λ + 56 = 0   i.e.   λ³ − 32λ + 56 = 0

Apply the theorem

A³ − 32A + 56I = O

Since |A| = −56 ≠ 0, A⁻¹ exists. Multiply by A⁻¹:

A² − 32I + 56A⁻¹ = O   ⇒   A⁻¹ = (1/56)(32I − A²)

Compute A²

A² = [ 8  −8 −12]
     [−2  22   6]
     [−10  6  34]

Inverse

32I − A² = [ 24   8  12]
           [  2  10  −6]
           [ 10  −6  −2]

A⁻¹ = (1/56) [24   8  12]   [3/7   1/7   3/14 ]
             [ 2  10  −6] = [1/28  5/28  −3/28]
             [10  −6  −2]   [5/28 −3/28  −1/28]

Check: first row of A times first column of A⁻¹: (24 + 2 + 30)/56 = 1 ✓; first row times second column: (8 + 10 − 18)/56 = 0 ✓.

Answer: A⁻¹ = (1/56)[24 8 12; 2 10 −6; 10 −6 −2].

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Mathematics III chapter 1 questions →