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Visual SolutionPYQ 2024 · Jan 27 Shift 2Easy
Question
Let AA be a 2×22 \times 2 real matrix and II be the identity matrix of order 2. If the roots of the equation ∣A−xI∣=0|A - xI| = 0 are −1-1 and 33, then the sum of the diagonal elements of the matrix A2A^2 is:
Solution Path
Eigenvalues of A2A^2 are squares of eigenvalues of AA. Since tr(A2)=\text{tr}(A^2) = sum of eigenvalues of A2A^2, we get tr(A2)=(−1)2+32=1+9=10\text{tr}(A^2) = (-1)^2 + 3^2 = 1 + 9 = 10.
01Question Setup
1/4
2×2 matrix A with characteristic equation |A - xI| = 0 having roots -1 and 3. Find tr(A²).
Eigenvalues: λ1=−1,  λ2=3\lambda_1 = -1, \; \lambda_2 = 3
02Eigenvalue Properties
2/4
From eigenvalues: tr(A) = sum of eigenvalues = 2, det(A) = product of eigenvalues = -3.
tr(A)=2,  ∣A∣=−3\text{tr}(A) = 2, \; |A| = -3
03Trace of A² via EigenvaluesKEY INSIGHT
3/4
If Av = λv, then A²v = λ²v. Eigenvalues of A² are squares of eigenvalues of A. Trace = sum of eigenvalues.
tr(A2)=(−1)2+32=1+9=10\text{tr}(A^2) = (-1)^2 + 3^2 = 1 + 9 = 10
04Final Answer
4/4
The sum of diagonal elements of A² equals the sum of squared eigenvalues.
tr(A2)=10\text{tr}(A^2) = \boxed{10}
Concepts from this question3 concepts unlocked
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Eigenvalues, Trace & Determinant

EASY

For a square matrix A: tr(A) = sum of eigenvalues, det(A) = product of eigenvalues.

tr(A)=∑λi,∣A∣=∏λi\text{tr}(A) = \sum \lambda_i, \quad |A| = \prod \lambda_i

Converts matrix questions into simple algebra on eigenvalues. Saves you from computing A² explicitly.

Trace problemsDeterminant shortcutsCharacteristic equation
Practice (10 Qs) →
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Eigenvalue Power Rule

EASY

If λ is an eigenvalue of A, then λⁿ is an eigenvalue of Aⁿ. So tr(Aⁿ) = sum of λᵢⁿ.

Av=λv  ⇒  Anv=λnvA\mathbf{v} = \lambda\mathbf{v} \;\Rightarrow\; A^n\mathbf{v} = \lambda^n\mathbf{v}

The key shortcut for 'find tr(A²)' or 'find det(A³)' style questions. No need to multiply matrices.

Matrix powersTrace of AⁿCayley-Hamilton applications
Practice (3 Qs) →

Cayley-Hamilton Theorem

STANDARD

Every square matrix satisfies its own characteristic equation: if p(x) = |A - xI|, then p(A) = 0.

A2−tr(A)⋅A+∣A∣⋅I=0(for 2×2)A^2 - \text{tr}(A) \cdot A + |A| \cdot I = 0 \quad (\text{for } 2 \times 2)

Reduces higher powers of A to lower ones. Essential for expressing A⁻¹ in terms of A and I.

Matrix inverseHigher powers of AMatrix polynomial
Practice (14 Qs) →
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