ISRO EC 2023 Q.2: solution and answer

ISRO 2023 · ECQ.2 · 1 mark · MCQ

Given that the Eigen values of matrix $A_{3 \times 3}$ are 1, 3 and 5. The Eigen values of $A^{3}$ are:

(A)
1, 3 and 5
(B)
3, 9 and 15
(C)
1, 27, 125
(D)
Given information is inadequate

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Answer and solution

ANSWER

The eigenvalues of $A^{3}$ are $1$ , $27$ and $125$ , option (c).

Official answer: CSource: ISRO's revised and final answer key of 16 February 2024

If $A x = λ x$ with $x \neq = 0$ , then $A^{2} x = A (λ x) = λ A x = λ^{2} x$ .
Multiplying by $A$ once more, $A^{3} x = λ^{2} A x = λ^{3} x$ . So every eigenvalue $λ$ of $A$ gives the eigenvalue $λ^{3}$ of $A^{3}$ , with the same eigenvector.
The matrix is $3 \times 3$ , so it has exactly three eigenvalues, namely $1$ , $3$ and $5$ . These are all of them, so the cubes of these three are all the eigenvalues of $A^{3}$ .
Cubes: $1^{3} = 1$ , $3^{3} = 27$ , $5^{3} = 125$ .
The eigenvalues of $A^{3}$ are $1, 27, 125$ , which is option (c). Option (a) would need $A^{3}$ to have the same eigenvalues as $A$ . Option (b) multiplies each eigenvalue by $3$ instead of cubing it. Option (d) is wrong because the eigenvalues of $A$ fix those of $A^{3}$ completely.

KEY IDEA

For any positive integer $n$ , if $λ$ is an eigenvalue of $A$ then $λ^{n}$ is an eigenvalue of $A^{n}$ , with the same eigenvector. Raise each eigenvalue to the power.

WATCH OUT FOR

Multiplying each eigenvalue by the power (giving $3, 9, 15$ ) instead of raising it to the power, or thinking more information about $A$ is needed.

From the ISRO EC 2023 paper. Every question of this paper · The full paper (PDF, ISRO) · All 3 ISRO Linear algebra questions, with notes

ISRO EC 2023 Q.2: solution and answer

Answer and solution

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