--- title: "Characteristic equation and verification [1,0,2;0,2,1;…]" subject: "Mathematics III" chapter: 1 key: "semester_3/mathematics_iii:characteristic-equation-and-verification-1-0" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/characteristic-equation-and-verification-1-0 site: BCT Engineering --- # Characteristic equation and verification [1,0,2;0,2,1;…] Mathematics III · Chapter 1: Determinants and Matrices - Asked in: 2069 Chaitra, 5 marks - Video: [Cayley Hamilton : Inverse of a matrix | Step wise explanation | Example solved-1| Mathspedia |](https://www.youtube.com/watch?v=qJ-zw_meMQ0) — Mathspedia - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/characteristic-equation-and-verification-1-0 - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-1 ## Question Obtain the characteristic equation of the matrix A = [1 0 2; 0 2 1; 2 0 3] and verify that it is satisfied by A. ## Short answer The characteristic equation is λ^3 − 6λ^2 + 7λ + 2 = 0. Substituting A gives A^3 − 6A^2 + 7A + 2I = O, verifying it is satisfied. ## Answer The characteristic equation of A is |A − λI| = 0. By the Cayley–Hamilton theorem, A satisfies it. A = [1 0 2; 0 2 1; 2 0 3] #### Characteristic equation ```text |A − λI| = | 1−λ 0 2 | | 0 2−λ 1 | | 2 0 3−λ | ``` Expand along the second column (only (2−λ) is non-zero): ```text = (2 − λ)[(1 − λ)(3 − λ) − 4] = (2 − λ)(λ² − 4λ − 1) = −(λ − 2)(λ² − 4λ − 1) = −(λ³ − 6λ² + 7λ + 2) ``` So the characteristic equation is ```text λ³ − 6λ² + 7λ + 2 = 0 ``` (Check with the formula λ³ − (trace)λ² + (sum of principal minors)λ − |A| = 0: trace 6, minors 6 − 1 + 2 = 7, |A| = −2.) #### Verification: A³ − 6A² + 7A + 2I = O ```text A² = [5 0 8] A³ = [21 0 34] [2 4 5] [12 8 23] [8 0 13] [34 0 55] ``` ```text Row 1: 21 − 30 + 7 + 2 = 0 ; 0 ; 34 − 48 + 14 = 0 Row 2: 12 − 12 + 0 = 0 ; 8 − 24 + 14 + 2 = 0 ; 23 − 30 + 7 = 0 Row 3: 34 − 48 + 14 = 0 ; 0 ; 55 − 78 + 21 + 2 = 0 ``` Every entry is zero, so A³ − 6A² + 7A + 2I = O, i.e. A satisfies its characteristic equation. ∎ ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The characteristic equation of a 3x3 matrix is: A. λ^3 + (trace)λ^2 + (minors)λ + |A| = 0 B. λ^3 - (trace)λ^2 + (sum of principal minors)λ - |A| = 0 C. λ^3 - |A|λ^2 + trace = 0 D. λ^3 - (trace)λ^2 - (minors)λ + |A| = 0 Answer: B. Signs alternate: trace, minors, determinant. (Key point: Formula) 2. For this A, the characteristic equation is: A. λ^3 + 6λ^2 + 7λ + 2 = 0 B. λ^3 - 6λ^2 - 7λ - 2 = 0 C. λ^3 - 6λ^2 + 7λ - 2 = 0 D. λ^3 - 6λ^2 + 7λ + 2 = 0 Answer: D. Trace 6, minors 7, |A| = -2 so -|A| = +2. (Key point: Equation for A) 3. Why is expansion along the second column easy? A. Only (2-λ) is non-zero there B. All entries are equal C. It contains the trace D. It has no λ Answer: A. The other two entries of column 2 are zero. (Key point: Expansion) 4. Verification means showing: A. A^3 - 6A^2 + 7A + 2I = O B. A^3 = 6A^2 C. |A| = 0 D. A^2 = 7A Answer: A. A satisfies its characteristic equation, with all entries zero. 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