Idempotent Matrix
A square matrix A is called idempotent when it satisfies A squared equal to A; for such a matrix the determinant of A can only take which set of values?
Select the correct option:
Solution
0 or 1
An idempotent matrix satisfies A^2 = A, a condition that strongly constrains its determinant, a subtle JEE Advanced result. Taking determinants of both sides of A^2 = A gives det(A^2) = det(A), and since the determinant is multiplicative, det(A)^2 = det(A). Letting d = det(A), this means d^2 = d, so d^2 - d = 0, that is d(d - 1) = 0. Therefore d = 0 or d = 1 are the only possibilities. Option only 1 misses the singular idempotent case like a projection onto a proper subspace, which has determinant 0. Option any real number ignores the strong constraint. Option 0 or -1 introduces an impossible value -1. Hence det(A) is 0 or 1. Plausibility check: the identity matrix is idempotent with determinant 1, and a non-trivial projection matrix is idempotent with determinant 0, so both admissible values are realized, confirming the constraint. The multiplicative property det(AB) equal to det(A) det(B) reflects that composing linear maps multiplies their volume-scaling factors, and the sign of the determinant tracks whether orientation is preserved or reversed. This property also yields det of a matrix power as the determinant raised to that power, and combined with the inverse rule it makes determinant computations for products entirely routine.
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About This Question
- Subject
- mathematics
- Chapter
- matrices and determinants
- Topic
- idempotent matrix
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
0 or 1
An idempotent matrix satisfies A^2 = A, a condition that strongly constrains its determinant, a subtle JEE Advanced result. Taking determinants of both sides of A^2 = A gives det(A^2) = det(A), and since the determinant is multiplicative, det(A)^2 = det(A). Letting d = det(A), this means d^2 = d, so d^2 - d = 0, that is d(d - 1) = 0. Therefore d = 0 or d = 1 are the only possibilities. Option only 1 misses the singular idempotent case like a projection onto a proper subspace, which has determinant 0. Option any real number ignores the strong constraint. Option 0 or -1 introduces an impossible value -1. Hence det(A) is 0 or 1. Plausibility check: the identity matrix is idempotent with determinant 1, and a non-trivial projection matrix is idempotent with determinant 0, so both admissible values are realized, confirming the constraint. The multiplicative property det(AB) equal to det(A) det(B) reflects that composing linear maps multiplies their volume-scaling factors, and the sign of the determinant tracks whether orientation is preserved or reversed. This property also yields det of a matrix power as the determinant raised to that power, and combined with the inverse rule it makes determinant computations for products entirely routine.
This medium difficulty mathematics question is from the chapter matrices and determinants, covering the topic of idempotent matrix. It appeared in the 2025 exam.
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