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A JEE question says "A is a square matrix with A² = A" and asks for (I + A)⁵. Multiplying out a general matrix five times is not the plan. The condition A² = A is the whole question. This post explains three such conditions, what each one tells you, and works through examples including two straight from NCERT.
Where is this in the syllabus?
The NCERT Class 12 chapter Matrices (2026-27 reprint) does not use the words nilpotent, idempotent or involutory, but its miscellaneous exercise asks about a matrix with A² = I (question 9) and one with A² = A (question 11). The JEE (Main) 2026 syllabus lists "algebra of matrices, type of matrices" in Unit 3, and the JEE Advanced 2026 brochure lists "properties of these matrix operations". Check the syllabus for your exam year.
What do the three names mean?
Idempotent: A² = A. Squaring changes nothing, so every higher power equals A too.
Involutory: A² = I. The matrix is its own inverse.
Nilpotent: Aᵏ = O for some positive integer k. The smallest such k is the index.
What do they tell you about the determinant?
Take determinants of each condition, using det(AB) = det(A) det(B).
Idempotent: (det A)² = det A, so det A = 0 or 1.
Involutory: (det A)² = 1, so det A = ±1, and A⁻¹ = A.
Nilpotent: (det A)ᵏ = 0, so det A = 0. A nilpotent matrix is never invertible.
Worked example 1: checking an idempotent matrix
Let A have rows (2, −2, −4), (−1, 3, 4) and (1, −2, −3). Multiply A by itself. The first row of A² is (4 + 2 − 4, −4 − 6 + 8, −8 − 8 + 12) = (2, −2, −4), the same as A's first row. The other two rows also come back unchanged, so A² = A. Its determinant works out to 0, which fits the rule above.
Worked example 2: NCERT question 11
If A² = A, find (I + A)³ − 7A.
I and A commute, so the binomial expansion is allowed: (I + A)³ = I + 3A + 3A² + A³. Every power of A equals A, so this is I + 7A. Subtract 7A and the answer is I.
The same steps give a general rule: (I + A)ⁿ = I + (2ⁿ − 1)A, because the binomial coefficients after the first add up to 2ⁿ − 1. For n = 5, (I + A)⁵ = I + 31A.
Worked example 3: NCERT question 9
A has rows (α, β) and (γ, −α), and A² = I. Which condition holds?
Multiply: the top-left entry of A² is α² + βγ, the bottom-right is γβ + α², and both off-diagonal entries are αβ − βα = 0. So A² = (α² + βγ)I. For A² = I you need α² + βγ = 1, which is option (C) 1 − α² − βγ = 0.
Worked example 4: a nilpotent matrix and its inverse trick
Let N have rows (1, 1, 3), (5, 2, 6) and (−2, −1, −3). Then N² has rows (0, 0, 0), (3, 3, 9) and (−1, −1, −3), which is not zero, and N³ = O. So N is nilpotent with index 3.
Now I − N is invertible even though N is not. Just as (1 − x)(1 + x + x²) = 1 − x³, you get (I − N)(I + N + N²) = I − N³ = I. So (I − N)⁻¹ = I + N + N², found without any cofactors.
How are involutory and idempotent linked?
If A² = I, then ((I + A)/2)² = (I + 2A + A²)/4 = (2I + 2A)/4 = (I + A)/2. So (I + A)/2 is idempotent. This is a common "show that" question.
Which mistakes cost marks?
Cancelling. A² = A does not mean A = I or A = O, because you cannot divide by a matrix that may have no inverse. Example 1 is neither.
Expanding non-commuting sums as if they were numbers.
Thinking nilpotent means "has zero entries". Example 4 has no zero entries at all.
Giving the wrong index. It is the smallest k with Nᵏ = O, so check N² before N³.
Writing det A = 1 for every idempotent matrix. It can be 0, as in example 1.
For related properties, see our posts on symmetric and skew-symmetric matrices and the adjoint and inverse of a matrix.
If your matrix products keep going wrong, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer, so you can find the exact row where a sign slipped.
A practice routine
Learn the three definitions and the three determinant facts as one card.
Verify the matrices in examples 1 and 4 by hand once. It builds speed for row-by-column products.
Do NCERT miscellaneous questions 9 and 11 without looking back here.
Practise two "show that" questions a week, such as the involutory link above.
Before expanding any matrix expression, write "commute?" in the margin and answer it.
Try this now: if A² = A, simplify (I − A)² and show that I − A is idempotent too.
