Determinant Properties
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Direct answer
Row swaps merely flip a determinant's sign; adding a multiple of one row to another leaves the value untouched; and multiplying one row by k scales the value by exactly k. From these three operations almost every JEE determinant question is answered without expansion. The companion algebra: |A^T| = |A|, |AB| = |A||B|, |kA| = k^n |A| for an n × n matrix, |adj A| = |A|^(n−1), and a determinant vanishes whenever two rows are identical, a row is zero, or one row is a linear combination of the others. Triangular determinants collapse to the product of diagonal entries — the destination every row-reduction is driving toward when the clock is running.
What you must remember
- Row and column operations: interchange changes the sign; Ri → Ri + kRj leaves the value unchanged; Ri → kRi multiplies the determinant by k — never more, never less.
- Scaling the whole matrix: |kA| = k^n |A|, so for a 3 × 3 matrix |2A| = 8|A| — the single most-marked slip in the chapter.
- Zero tests without expanding: identical or proportional rows, a zero row, or rows in arithmetic progression (the middle row is the average of its neighbours) all force the determinant to vanish.
- Product rules: |AB| = |A||B| with order irrelevant for the scalar result, |A^n| = |A|^n, |A^T| = |A| — but |A + B| ≠ |A| + |B| in general.
- Adjoint family: A(adj A) = |A| I, |adj A| = |A|^(n−1), |A^(−1)| = 1/|A|, and adj(AB) = (adj B)(adj A).
- Skew-symmetric result: every odd-order skew-symmetric determinant equals zero — a free fact worth a full mark.
- Vandermonde pattern: rows 1; a b c; a² b² c² evaluate to (a − b)(b − c)(c − a), the engine behind many factorisation and divisibility problems.
How to work through it
Evaluate D = |1 1 1; a b c; a² b² c²| the way examiners expect. Send C2 → C2 − C1 and C3 → C3 − C1; the first row becomes 1, 0, 0, so expanding along it leaves (b − a)(c² − a²) − (c − a)(b² − a²). Factor each difference of squares: (b − a)(c − a)(c + a) − (c − a)(b − a)(b + a) = (b − a)(c − a)(c − b), which rearranges to (a − b)(b − c)(c − a). Now the payoff question: can three distinct points (a, a²), (b, b²), (c, c²) on the parabola y = x² ever be collinear? The area determinant with rows (a, a², 1), (b, b², 1), (c, c², 1) is, up to sign, the same Vandermonde object — nonzero for distinct a, b, c. So no three distinct points of a parabola are collinear, a two-line proof that expansion from scratch would bury under arithmetic.
Where marks leak
JEE Main asks property-application MCQs: given |A| = 5 for a 3 × 3 matrix, report |2A| (40), |adj A| (25), |A²| (25) — candidates who write 10 for the first have scaled one row instead of the whole matrix. JEE Advanced prefers determinant equations and cyclic patterns, such as showing the symmetric-looking determinant with (b + c)², a², a² down the first row equals 2abc(a + b + c)³ — verified instantly at a = b = c = 1, where both sides give 54. Two habits protect marks: check any claimed identity at one numerical triple before trusting the algebra, and never split |A + B| into |A| + |B|; the distributive instinct is exactly what option-writers plant. Finally, transpose invariance means any column property quoted for rows is automatically true — quoting it as a row fact is safer than improvising.
Frequently asked questions
What is |kA| for an n × n matrix?
k^n |A|, because every one of the n rows carries the factor k — |3A| = 27|A| for a 3 × 3 matrix.
When is a determinant zero without expanding?
Identical or proportional rows, a zero row, rows in arithmetic progression, or an odd-order skew-symmetric matrix.
What is |adj A| for a 3 × 3 matrix with |A| = 4?
|A|^(n−1) = |A|² = 16, and the chain continues: |adj(adj A)| = |A|^((n−1)²) = 4^4.
Does |AB| = |A||B| hold in both orders?
Yes — both equal |A||B| since these are scalars, even though AB ≠ BA as matrices.
What does the Vandermonde determinant equal?
(a − b)(b − c)(c − a) for rows 1; a b c; a² b² c², zero exactly when two of a, b, c coincide.