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A JEE question gives you a 2×2 matrix and asks for A⁵⁰. Multiplying fifty times is not the plan, and a wrong guess from two powers is the usual way marks go. This post gives five patterns that cover most of these questions, each worked with checked numbers, plus a one-line test that catches a bad guess.
Where is this in NCERT and the JEE Main syllabus?
The JEE (Main) 2026 syllabus lists "Matrices, algebra of matrices, type of matrices" in Unit 3. Powers of a matrix sit under the algebra of matrices; the syllabus does not name them separately. The NCERT Class 12 chapter Matrices (2026-27 reprint) proves in Miscellaneous Example 23 that the matrix with rows (cos θ, sin θ) and (−sin θ, cos θ) has Aⁿ with rows (cos nθ, sin nθ) and (−sin nθ, cos nθ), using mathematical induction.
Note the mismatch. The rationalised NCERT Class 11 book no longer has a chapter on induction, and the JEE Main 2026 syllabus does not list it. So in JEE Main you mostly spot a pattern and use it. If a board question says "prove", follow the induction step in Example 23.
Pattern 1: compute three powers, then guess
Take A with rows (1, 2) and (0, 1). Then A² has rows (1, 4) and (0, 1), and A³ has rows (1, 6) and (0, 1). The guess is Aⁿ with rows (1, 2n) and (0, 1).
Check before you trust it. Multiply A by your formula: the top-right entry is 1 × 2n + 2 × 1 = 2n + 2, which is the formula for n + 1. So A⁵⁰ has rows (1, 100) and (0, 1).
Why three powers and not two? Take M with rows (1, 1) and (0, 2). M² has rows (1, 3) and (0, 4). From 1 and 3 you might guess the top-right entry is 2n − 1. But M³ has rows (1, 7) and (0, 8), so that guess fails at n = 3. The true rule is Mⁿ with rows (1, 2ⁿ − 1) and (0, 2ⁿ), and the check works: 1 × (2ⁿ − 1) + 1 × 2ⁿ = 2ⁿ⁺¹ − 1.
Pattern 2: λI plus a matrix whose square is zero
B with rows (2, 1) and (0, 2) equals 2I + N, where N has rows (0, 1) and (0, 0), and N² = O. Because I commutes with every matrix, the binomial theorem applies, and every term with N² or higher vanishes:
Bⁿ = 2ⁿI + n × 2ⁿ⁻¹N.
For n = 10: 2¹⁰ = 1024 and 10 × 2⁹ = 5120, so B¹⁰ has rows (1024, 5120) and (0, 1024). Our post on nilpotent and idempotent matrices covers more of these conditions.
Pattern 3: A² is a multiple of A
Let J be the 3×3 matrix with every entry 1. Each entry of J² is 1 + 1 + 1 = 3, so J² = 3J. Then J³ = 3J² = 9J, and in general Jⁿ = 3ⁿ⁻¹J. So J⁵ = 81J, a matrix of 81s.
Pattern 4: powers that repeat
R with rows (0, −1) and (1, 0) gives R² = −I and R⁴ = I. The powers repeat every four steps. Since 2026 = 4 × 506 + 2, R²⁰²⁶ = R² = −I. Divide the exponent by the cycle length and keep only the remainder; a remainder of 0 means the answer is I, not R.
Pattern 5: use the matrix's own equation
NCERT's Miscellaneous Exercise asks you to show that A with rows (3, 1) and (−1, 2) satisfies A² − 5A + 7I = O. Use it to lower powers: A² = 5A − 7I, so
A³ = 5A² − 7A = 5(5A − 7I) − 7A = 18A − 35I, which has rows (19, 18) and (−18, 1).
A⁴ = 18A² − 35A = 18(5A − 7I) − 35A = 55A − 126I, which has rows (39, 55) and (−55, −16).
Check one entry by direct multiplication: the top-left of A⁴ is 55 × 3 − 126 = 39.
Which mistakes cost the most marks?
Guessing from two powers. The M example shows how two terms can fit a wrong rule.
Using the binomial theorem on A + B when AB ≠ BA. It works for λI + N only because I commutes with N.
Dropping the I. Bⁿ = 2ⁿI + n2ⁿ⁻¹N, not 2ⁿ + n2ⁿ⁻¹N; a number cannot be added to a matrix.
Off-by-one cycles. For R, the remainder 0 gives I, remainder 1 gives R.
Why explain each step to yourself?
In a classic study of physics students, Chi and colleagues (1989) found that the "Good" students generated many explanations while studying worked examples and related each step to principles, while "Poor" students explained little and leaned on the examples later. For matrix powers, the explanation is the check line: say why A times Aⁿ gives Aⁿ⁺¹.
A practice routine for this week
Take five matrices, one per pattern above, and compute up to the fourth power by hand.
Write the general formula and the one-line check for each.
Mix in two questions from the chapter's time-saving properties and traps, so you also practise choosing the pattern.
Redo any you got wrong two days later without notes.
If the pattern is right but the algebra keeps slipping, the Fermi Problem Library has a step-by-step solver that explains the reasoning rather than printing an answer, so you can see which multiplication went wrong.
Start now: find Mⁿ for M with rows (1, 1) and (0, 2) yourself, and run the check.
