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

Short answer

Cayley-Hamilton theorem holds: A³ − 6A² + 9A − 4I = O. Thus A⁻¹ = (1/4)(A² − 6A + 9I) = (1/4)[3 1 −1; 1 3 1; −1 1 3].

Verify Cayley-Hamilton theorem for the matrix A = [2 −1 1; −1 2 −1; 1 −1 2] and find the inverse of A.

A = [2 −1 1; −1 2 −1; 1 −1 2]

Characteristic equation

  • Trace = 2 + 2 + 2 = 6
  • Sum of principal minors = (4 − 1) + (4 − 1) + (4 − 1) = 9
  • |A| = 2(4 − 1) + 1(−2 + 1) + 1(1 − 2) = 6 − 1 − 1 = 4
λ³ − 6λ² + 9λ − 4 = 0      [ = (λ − 1)²(λ − 4) ]

Verification: A³ − 6A² + 9A − 4I = O

A² = [ 6 −5  5]         A³ = [ 22 −21  21]
     [−5  6 −5]              [−21  22 −21]
     [ 5 −5  6]              [ 21 −21  22]

Row 1: 22 − 36 + 18 − 4 = 0 ;  −21 + 30 − 9 = 0 ;  21 − 30 + 9 = 0
Row 2: −21 + 30 − 9 = 0 ;  22 − 36 + 18 − 4 = 0 ;  −21 + 30 − 9 = 0
Row 3: 21 − 30 + 9 = 0 ;  −21 + 30 − 9 = 0 ;  22 − 36 + 18 − 4 = 0

All entries are zero, so Cayley–Hamilton is verified. ∎

Inverse

Since |A| = 4 ≠ 0, multiply A³ − 6A² + 9A − 4I = O by A⁻¹:

A² − 6A + 9I − 4A⁻¹ = O   ⇒   A⁻¹ = (1/4)(A² − 6A + 9I)

A² − 6A + 9I = [ 6−12+9   −5+6    5−6  ]   [ 3  1 −1]
               [ −5+6     6−12+9  −5+6 ] = [ 1  3  1]
               [ 5−6      −5+6    6−12+9]   [−1  1  3]

Answer: A⁻¹ = (1/4) [3 1 −1; 1 3 1; −1 1 3].

Check: row 1 of A times column 1: (2·3 − 1·1 + 1·(−1))/4 = 4/4 = 1 ✓.

✓ Answer checked by Claude Opus 5.5 (high effort)

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