Trace Of A Matrix
If A is a square matrix of order 2 with trace equal to 5 and determinant equal to 6, then the sum of the eigenvalues and product of the eigenvalues of A are respectively which pair?
Select the correct option:
Solution
5 and 6
For any square matrix the trace equals the sum of its eigenvalues and the determinant equals their product, an invariant relationship JEE Advanced exploits. For an order-2 matrix A with trace 5 and determinant 6, the two eigenvalues lambda_1 and lambda_2 satisfy lambda_1 + lambda_2 = trace = 5 and lambda_1 · lambda_2 = determinant = 6. Hence the sum of eigenvalues is 5 and the product is 6. These eigenvalues are roots of the characteristic equation lambda^2 - 5 lambda + 6 = 0, namely 2 and 3. Option 6 and 5 swaps the roles of trace and determinant. Option 7 and 6 miscomputes the sum. Option 5 and 11 confuses the product with a sum of squares. Hence the pair is 5 and 6. Plausibility check: the eigenvalues 2 and 3 indeed sum to 5 and multiply to 6, and they satisfy lambda^2 - 5 lambda + 6 = 0, confirming the trace-determinant correspondence with eigenvalues. Idempotent matrices are precisely projection operators, mapping space onto a subspace while fixing that subspace pointwise, which geometrically explains why their only possible determinants are zero and one. The eigenvalues of any idempotent matrix are confined to zero and one as well, so its trace counts the dimension of the image, a fact often paired with the determinant constraint in problems.
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About This Question
- Subject
- mathematics
- Chapter
- matrices and determinants
- Topic
- trace of a matrix
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
5 and 6
For any square matrix the trace equals the sum of its eigenvalues and the determinant equals their product, an invariant relationship JEE Advanced exploits. For an order-2 matrix A with trace 5 and determinant 6, the two eigenvalues lambda_1 and lambda_2 satisfy lambda_1 + lambda_2 = trace = 5 and lambda_1 · lambda_2 = determinant = 6. Hence the sum of eigenvalues is 5 and the product is 6. These eigenvalues are roots of the characteristic equation lambda^2 - 5 lambda + 6 = 0, namely 2 and 3. Option 6 and 5 swaps the roles of trace and determinant. Option 7 and 6 miscomputes the sum. Option 5 and 11 confuses the product with a sum of squares. Hence the pair is 5 and 6. Plausibility check: the eigenvalues 2 and 3 indeed sum to 5 and multiply to 6, and they satisfy lambda^2 - 5 lambda + 6 = 0, confirming the trace-determinant correspondence with eigenvalues. Idempotent matrices are precisely projection operators, mapping space onto a subspace while fixing that subspace pointwise, which geometrically explains why their only possible determinants are zero and one. The eigenvalues of any idempotent matrix are confined to zero and one as well, so its trace counts the dimension of the image, a fact often paired with the determinant constraint in problems.
This medium difficulty mathematics question is from the chapter matrices and determinants, covering the topic of trace of a matrix. It appeared in the 2025 exam.
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