Let A be the 2 × 2 matrix with elements a11 = a12 = a21 = +1 and a22 = -1. Then the eigenvalues of the matrix A19 are A) 1024 and -1024 B) 1024√(2) and -1024√(2) C) 4√(2) and -4√(2) D) 512√(2) and -512√(2)

Official answer: D) 512√(2) and -512√(2). Explanation: The matrix A is given by: A = 1 & 1 1 & -1 To find the eigenvalues of A, solve the characteristic equation det(A - I) = 0: 1 - & 1 1 & -1 - = 0 -(1 - )(1 + ) - 1 = 0 - (1 - 2) - 1 = 0 2 - 2 = 0 = ±√(2) If is an eigenvalue of A, then k is an eigenvalue of Ak. For A19, the eigenvalues are: 119 = (√(2))19 = (√(2))18 · √(2) = 29 √(2) = 512√(2) 219 = (-√(2))19 = -512√(2) Therefore, the eigenvalues of A19 are 512√(2) and -512√(2), which matches option (D).

Linear Algebra

GATE CSE 2012 · Q11

MCQ1 markMedium

Time00:00

Let A be the $2 \times 2$ matrix with elements $a_{11} = a_{12} = a_{21} = +1$ and $a_{22} = -1$ . Then the eigenvalues of the matrix $A^{19}$ are

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Official answer: D) 512√(2) and -512√(2)

Linear Algebra

The matrix $A$ is given by:

A = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}

To find the eigenvalues of $A$ , solve the characteristic equation $\det(A - \lambda I) = 0$ :

\begin{vmatrix} 1 - \lambda & 1 \\ 1 & -1 - \lambda \end{vmatrix} = 0

-(1 - \lambda)(1 + \lambda) - 1 = 0

- (1 - \lambda^2) - 1 = 0 \implies \lambda^2 - 2 = 0 \implies \lambda = \pm\sqrt{2}

If $\lambda$ is an eigenvalue of $A$ , then $\lambda^k$ is an eigenvalue of $A^k$ .

For $A^{19}$ , the eigenvalues are:

\lambda_1^{19} = (\sqrt{2})^{19} = (\sqrt{2})^{18} \cdot \sqrt{2} = 2^9 \sqrt{2} = 512\sqrt{2}

\lambda_2^{19} = (-\sqrt{2})^{19} = -512\sqrt{2}

Therefore, the eigenvalues of $A^{19}$ are $512\sqrt{2}$ and $-512\sqrt{2}$ , which matches option (D).

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This Linear Algebra previous year question is Q11 of the GATE CSE 2012 paper. Find more GATE CSE PYQs on Linear Algebra in Engineering Mathematics, or browse all GATE CSE PYQs.

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Independent practice explanation verified by the GateAI team. GATE is conducted by the IITs; question text follows the official paper.