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Determinant identity proof

Short answer

Characteristic equation λ³ − 6λ² + 11λ − 6 = 0 gives eigenvalues 1, 2, 3. Determinant = 1·2·3 = 6. Eigenvalues of A⁻¹: 1, 1/2, 1/3; eigenvalues of adj A = |A|/λ: 6, 3, 2.

Find the eigenvalues of the matrix A = [1 0 −1; 1 2 1; 2 2 3] and use them to compute (i) eigenvalues of A⁻¹ (ii) determinant of A (iii) eigenvalues of adj A.

A = [1 0 −1; 1 2 1; 2 2 3]

Eigenvalues

  • Trace = 1 + 2 + 3 = 6
  • Sum of principal minors = (6 − 2) + (3 + 2) + (2 − 0) = 4 + 5 + 2 = 11
  • |A| = 1(6 − 2) − 0 + (−1)(2 − 4) = 4 + 2 = 6
λ³ − 6λ² + 11λ − 6 = 0  ⇒  (λ − 1)(λ − 2)(λ − 3) = 0

Eigenvalues of A: 1, 2, 3.

(i) Eigenvalues of A⁻¹

If λ is an eigenvalue of a non-singular A, then 1/λ is an eigenvalue of A⁻¹.

Eigenvalues of A⁻¹: 1, 1/2, 1/3.

(ii) Determinant of A

The determinant is the product of the eigenvalues:

|A| = 1 × 2 × 3 = 6

(This agrees with direct expansion.)

(iii) Eigenvalues of adj A

Since adj A = |A| A⁻¹, its eigenvalues are |A|/λ:

Eigenvalues of adj A: 6/1, 6/2, 6/3 = 6, 3, 2.

✓ Answer checked by Claude Opus 5.5 (high effort)

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