# Cayley-Hamilton Theorem > Cayley-Hamilton theorem for JEE Mathematics: matrices satisfy their characteristic equation, computing A^2, A inverse and higher powers without determinants. - Canonical URL: https://prepelephant.com/topics/jee/mathematics/cayley-hamilton - Exam / course: JEE · Subject: Mathematics - Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team - First published: 2026-10-02 - Last updated: 2026-10-02 - How to cite: "Cayley-Hamilton Theorem", PrepElephant, https://prepelephant.com/topics/jee/mathematics/cayley-hamilton ## Direct answer Every square matrix satisfies its own characteristic equation. If the characteristic polynomial of A (from det(A - λI) = 0 or det(λI - A) = 0) is λ^n + c1 λ^(n-1) + ... + cn, then the matrix identity A^n + c1 A^(n-1) + ... + cn I = O holds — the theorem substitutes the matrix for the scalar variable, with the constant term landing on cn I. For a 2 × 2 matrix this reads A^2 - (tr A) A + (det A) I = O, and rearranging it produces the two workhorse results: A^2 = (tr A) A - (det A) I, and when det A ≠ 0, A^(-1) = [(tr A) I - A]/det A — the inverse found without adjoints or row reduction. Higher powers reduce through polynomial division: divide λ^k by the characteristic polynomial and substitute A into the remainder. ## What you must remember - **The theorem:** every square matrix obeys its own characteristic equation; the substitution replaces λ by A and the constant term by (constant) × I. - **2 × 2 identity:** A^2 - (tr A)A + (det A)I = O; from it, A^2 = (tr A)A - (det A)I and A^(-1) = ((tr A)I - A)/det A for det A ≠ 0. - **Characteristic polynomial conventions:** det(A - λI) and det(λI - A) differ by (-1)^n; pick one and stay consistent, or the middle signs flip. - **Higher powers:** to get A^k, divide x^k by the characteristic polynomial and evaluate the remainder at A — the quotient's contribution collapses by the theorem. - **Trace and determinant as fingerprints:** eigenvalues satisfy Σλi = tr A and Πλi = det A; the theorem is the polynomial side of the same coin. - **3 × 3 form:** A^3 - (tr A)A^2 + (sum of principal 2 × 2 minors)A - (det A)I = O — the middle coefficient is the sum of the three principal minors. - **Singularity read instantly:** a zero constant term in the characteristic polynomial means det A = 0, so the theorem directly reflects invertibility. ## Inverting a matrix without adjoints Let A have rows (1, 2) and (2, 1). The trace is 2 and the determinant is -3, so the Cayley-Hamilton identity reads A^2 - 2A - 3I = O. Verify by direct multiplication: A^2 has rows (5, 4) and (4, 5), and 2A + 3I has rows (5, 4) and (4, 5) — the identity holds. Now rearrange for the inverse: A^2 - 2A = 3I factors on the left as A(A - 2I) = 3I, so A^(-1) = (A - 2I)/3. Compute A - 2I: rows (-1, 2) and (2, -1), so A^(-1) has rows (-1/3, 2/3) and (2/3, -1/3). Confirm by multiplication: the first diagonal entry of A × A^(-1) is 1 × (-1/3) + 2 × (2/3) = -1/3 + 4/3 = 1, and the off-diagonal 1 × (2/3) + 2 × (-1/3) = 0, with the second row checking by symmetry. Note the equivalent closed form: A^(-1) = ((tr A)I - A)/det A = (2I - A)/(-3), the same matrix — and note where sign discipline matters, since the constant term (-3, not 3) must travel with the rearrangement. The whole inverse took three lines and no adjoint, which is precisely the exam appeal. ## Where students slip JEE Main asks the identity itself (compute A^2 - 5A + 7I for a matrix whose characteristic polynomial is known to be λ^2 - 5λ + 7 — the answer is O) or the quick inverse via ((tr A)I - A)/det A. Advanced wants A^k for larger k (say A^5) through remainder division, the 3 × 3 version with principal minors, or Cayley-Hamilton embedded in questions about eigenvalues. The recurring errors: substituting the constant term without the identity matrix — A^2 - 2A - 3 ≠ O, only A^2 - 2A - 3I = O — the single most marked slip in this chapter; mixing the two characteristic-polynomial conventions so middle signs flip; and assuming the theorem lets you divide by A when det A = 0 (the rearrangement for the inverse needs invertibility). A subtler point worth one line of defence: the theorem is a matrix identity, not a determinant statement — det(A - A·I) = 0 is vacuously true and proves nothing. Matrices form a named unit in both syllabi, and this theorem is its standard crown jewel question. ## Frequently asked questions ### What does the Cayley-Hamilton theorem state? Every square matrix satisfies its own characteristic equation: substitute A for λ (and append I to the constant term) in det(λI - A) = 0, and the resulting matrix identity holds. ### How is the inverse of a 2 × 2 matrix found using the theorem? From A^2 - (tr A)A + (det A)I = O, rearrange to A^(-1) = ((tr A)I - A)/det A — valid whenever det A ≠ 0. ### Why must the constant term be multiplied by I? Because the other terms are matrices; only cI can be added to them, and the identity matrix plays the role of the scalar 1 in polynomial evaluation. ### How do you compute high powers like A^5 for a 2 × 2 matrix? Divide x^5 by the characteristic polynomial and substitute A into the remainder (of degree at most 1): the theorem collapses the quotient's contribution to zero. ### Does Cayley-Hamilton work for singular matrices? Yes — the identity holds for every square matrix; only the further step of dividing by A to extract an inverse requires det A ≠ 0.