If AB = 0 for two non-zero square matrices A and B, then:

At least one of A, B must be the zero matrix

The points (1,2), (3,6) and (5,10) are:

Vertices of a triangle of area 2

Matrices and Determinants

Matrices are among the most reliable scoring topics in JEE, because the questions are procedural and the procedures are short. The one idea that unlocks the whole chapter is that the determinant tells you whether a matrix can be undone: if |A| ≠ 0 an inverse exists and a linear system has exactly one solution, and if |A| = 0 it does not and the system either fails or has infinitely many solutions. Almost every question here is a disguised way of asking whether some determinant vanishes. Beyond that, most lost marks come from one habit: assuming matrix multiplication behaves like ordinary multiplication, which it does not.

Matrix Algebra and Where the Analogy Breaks

⚠️ Matrix multiplication is not commutative, and AB = 0 does not mean A or B is zero

Three familiar rules of numbers fail for matrices, and the paper tests all three. AB ≠ BA in general, so (A + B)² = A² + AB + BA + B² and not A² + 2AB + B² unless A and B happen to commute. AB = 0 does not imply A = 0 or B = 0 — a matrix of ones times a matrix with rows summing to zero can vanish while neither factor does. And AB = AC does not imply B = C unless A is invertible, because cancelling requires multiplying by A⁻¹, which needs |A| ≠ 0. Every one of these is a standard trap, and the safe habit is never to cancel or expand a matrix expression without first asking whether the matrix you are cancelling is invertible.

Special matrices and what defines them
Type	Definition	Consequence
Symmetric	Aᵗ = A	A + Aᵗ is always symmetric
Skew-symmetric	Aᵗ = −A	Diagonal entries are all zero; odd order gives \|A\| = 0
Orthogonal	AAᵗ = I	\|A\| = ±1
Idempotent	A² = A	\|A\| is 0 or 1
Nilpotent	Aᵏ = 0 for some k	\|A\| = 0 necessarily
Involutory	A² = I	A is its own inverse

Every square matrix splits uniquely as the sum of a symmetric and a skew-symmetric part: A = ½(A + Aᵗ) + ½(A − Aᵗ). Two order facts are worth holding: a skew-symmetric matrix of odd order always has determinant zero, so it can never be invertible, and the transpose reverses a product, (AB)ᵗ = BᵗAᵗ, exactly as the inverse does, (AB)⁻¹ = B⁻¹A⁻¹.

Determinants and Their Properties

Properties that let you avoid expanding
Operation	Effect on the determinant
Swap two rows or columns	Sign changes
Two rows identical or proportional	Determinant is zero
Multiply one row by k	Determinant multiplied by k
Multiply the whole n × n matrix by k	kⁿ\|A\| — not k\|A\|
Add a multiple of one row to another	Determinant unchanged — the key simplification tool
Transpose	Determinant unchanged

🧠 Look for a row operation that creates a zero row before expanding

Because adding a multiple of one row to another leaves the determinant unchanged, a determinant whose rows have a hidden linear relationship can be reduced to zero in a single step. If the entries of each row sum to the same value, adding all the columns into the first makes that column constant, which often exposes a proportional pair immediately. This is why determinants that look formidable — three rows of quadratics in a, b, c, say — frequently turn out to be zero, and why the first move should always be to hunt for structure rather than to expand. Expanding a 3 × 3 costs six products; spotting a proportional pair costs nothing.

Two multiplicative facts recur. |AB| = |A||B|, so |Aⁿ| = |A|ⁿ, and for an n × n matrix |adj A| = |A|ⁿ⁻¹. The inverse is A⁻¹ = adj(A)/|A|, which exists precisely when |A| ≠ 0. And the area of a triangle with vertices (x₁,y₁), (x₂,y₂), (x₃,y₃) is half the absolute value of the determinant with rows (x₁, y₁, 1) and so on — so the three points are collinear exactly when that determinant vanishes, which is the quickest collinearity test available.

Solving Linear Systems

Cramer's rule and the three cases
Case	Condition	Outcome
Consistent, unique solution	Δ ≠ 0	x = Δ₁/Δ, y = Δ₂/Δ, z = Δ₃/Δ
Inconsistent	Δ = 0 and some Δᵢ ≠ 0	No solution — the planes have no common point
Infinitely many solutions	Δ = 0 and every Δᵢ = 0	A line or plane of solutions
Homogeneous, trivial only	Δ ≠ 0	x = y = z = 0 is the only solution
Homogeneous, non-trivial	Δ = 0	Infinitely many solutions exist

🎯 Why a homogeneous system never has "no solution"

A homogeneous system has all constant terms zero, so x = y = z = 0 always satisfies it — the trivial solution is free of charge. That means a homogeneous system can never be inconsistent, and the only question worth asking about it is whether a non-trivial solution also exists, which happens exactly when Δ = 0. This is why questions on homogeneous systems always take the form "find the value of k for which the system has a non-trivial solution": they are asking you to solve Δ = 0 and nothing else. Recognising that immediately removes two of the three cases from consideration and turns a case analysis into one determinant.

⚠️ Δ = 0 does not by itself mean infinitely many solutions

This is the single most common error in the chapter. When Δ = 0 the system has either no solution or infinitely many, and which one it is depends entirely on the numerator determinants: if any Δᵢ is non-zero the system is inconsistent, and only if all of them vanish are there infinitely many solutions. Geometrically, three planes with Δ = 0 may form a triangular prism with no common point, or they may all contain a common line. So a question that gives you Δ = 0 and asks for the number of solutions is incomplete until you have checked at least one Δᵢ — and answering "infinitely many" straight from Δ = 0 is exactly the response the option list is built to catch.

Key takeaways

AB ≠ BA in general, AB = 0 does not force A or B to vanish, and cancelling A needs |A| ≠ 0.
A skew-symmetric matrix has zero diagonal, and one of odd order always has determinant zero.
Multiplying an n × n matrix by k multiplies the determinant by kⁿ, not by k.
Adding a multiple of one row to another leaves the determinant unchanged — the main simplification tool.
|AB| = |A||B|, |Aⁿ| = |A|ⁿ and |adj A| = |A|ⁿ⁻¹ for an n × n matrix.
Three points are collinear exactly when the area determinant vanishes.
Δ ≠ 0 gives a unique solution; Δ = 0 gives either no solution or infinitely many — check the Δᵢ.
A homogeneous system is never inconsistent; it has a non-trivial solution exactly when Δ = 0.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

If A is a 3 × 3 matrix with |A| = 2, then |3A| equals:
1. 6
2. 18
3. 54
4. 2
Show answer
Answer: C — 54
Multiplying a matrix by a scalar multiplies every one of its three rows by that scalar, and each row contributes a factor, so the determinant is multiplied by 3³ = 27, giving 27 × 2 = 54. The answer 6 comes from treating the scalar as if it multiplied the determinant just once, which is the standard slip — the exponent must match the order of the matrix.
A system of three linear equations has Δ = 0 and Δ₁ ≠ 0. The system:
1. Has a unique solution
2. Has infinitely many solutions
3. Has no solution
4. Cannot be classified
Show answer
Answer: C — Has no solution
With Δ = 0 the unique-solution case is ruled out, and infinitely many solutions require every numerator determinant to vanish as well. Since Δ₁ is non-zero, the system is inconsistent and has no solution — geometrically the three planes enclose a prism with no common point. Jumping from Δ = 0 straight to "infinitely many" without checking the Δᵢ is the error this question is designed to catch.
The determinant of a skew-symmetric matrix of order 3 is:
1. Always zero
2. Always 1
3. Always positive
4. Any real number
Show answer
Answer: A — Always zero
For a skew-symmetric matrix Aᵗ = −A, so taking determinants gives |A| = |Aᵗ| = |−A| = (−1)³|A| = −|A|, which forces |A| = 0 for odd order. The argument fails for even order, where (−1)ⁿ is +1 and the determinant can be non-zero — which is why the order named in the question is the whole point.
A homogeneous system of three equations in three unknowns has a non-trivial solution if and only if:
1. Δ ≠ 0
2. Δ = 0
3. Δ = 1
4. All Δᵢ ≠ 0
Show answer
Answer: B — Δ = 0
A homogeneous system always admits the trivial solution, so it is never inconsistent; the only issue is whether a second solution exists, and that requires the coefficient matrix to be singular, Δ = 0. If Δ ≠ 0 the matrix is invertible and the trivial solution is the only one. This is why such questions reduce to solving a single determinant equation for the parameter.
For a 3 × 3 matrix A with |A| = 4, |adj A| equals:
1. 4
2. 12
3. 16
4. 64
Show answer
Answer: C — 16
For an n × n matrix, |adj A| = |A|ⁿ⁻¹, so with n = 3 the answer is 4² = 16. The relation follows from A·adj A = |A|I, whose determinant gives |A|·|adj A| = |A|ⁿ. Quoting |adj A| = |A| is the usual slip, and it happens to be correct only for a 2 × 2 matrix, where n − 1 = 1.
If AB = 0 for two non-zero square matrices A and B, then:
1. This is impossible
2. Both A and B must be singular
3. At least one of A, B must be the zero matrix
4. A must equal −B
Show answer
Answer: B — Both A and B must be singular
Taking determinants gives |A||B| = 0, so at least one determinant is zero; and in fact if either were invertible you could multiply through by its inverse and force the other to be the zero matrix, contradicting the premise — so both must be singular. Unlike with numbers, a product of matrices can vanish without either factor vanishing, which is exactly why matrix cancellation is illegitimate without invertibility.
The points (1,2), (3,6) and (5,10) are:
1. Vertices of a triangle of area 2
2. Collinear
3. Vertices of a right triangle
4. Concyclic
Show answer
Answer: B — Collinear
Forming the determinant with rows (1,2,1), (3,6,1) and (5,10,1) gives zero, so the enclosed area is zero and the three points lie on one line — here y = 2x. The determinant test is the fastest available check for collinearity and needs no slope comparison, which makes it useful even when one of the segments happens to be vertical.
For square matrices A and B, (AB)⁻¹ equals:
1. A⁻¹B⁻¹
2. B⁻¹A⁻¹
3. (BA)⁻¹
4. AB
Show answer
Answer: B — B⁻¹A⁻¹
Check by multiplying: (AB)(B⁻¹A⁻¹) = A(BB⁻¹)A⁻¹ = AA⁻¹ = I, so the order must reverse. Writing A⁻¹B⁻¹ instead would leave A⁻¹ and B adjacent with nothing to cancel, and since matrix multiplication is not commutative the two expressions are genuinely different. The transpose behaves the same way, with (AB)ᵗ = BᵗAᵗ.