Matrices & Determinants

Matrices and determinants power linear-system analysis. Advanced tests properties of determinants (row/column operations), the adjoint/inverse, special matrix types (symmetric, orthogonal, idempotent), and the consistency of linear systems via rank/Cramer. Efficient determinant manipulation avoids brute expansion.

Key results

• Determinant properties: row/column swaps flip sign; a common factor pulls out; equal rows give 0; |AB| = |A||B|; |A| = |Aᵀ|; |kA| = kⁿ|A|.
• Inverse: A⁻¹ = adj(A)/|A| (exists iff |A| ≠ 0); (AB)⁻¹ = B⁻¹A⁻¹; |adj A| = |A|^(n−1).
• Systems (AX = B): unique solution if |A| ≠ 0 (Cramer); if |A| = 0, either no solution or infinitely many (check consistency).
• Special matrices: symmetric Aᵀ=A, skew-symmetric Aᵀ=−A (odd-order det 0); orthogonal AAᵀ=I; idempotent A²=A.
• Cayley–Hamilton: every matrix satisfies its own characteristic equation.

Type	Property
Singular
Orthogonal	AAᵀ = I,
Skew-symmetric (odd n)
adj A

Exam Tricks & Tips

• 🎯 Simplify determinants with row/column operations before expanding — create zeros so cofactor expansion is trivial.
• 🎯 For a homogeneous system AX=0, a non-trivial solution exists iff |A| = 0 — the standard eigen/consistency trigger.
• 🎯 Use |adj A| = |A|^(n−1) and adj(adj A) = |A|^(n−2) A for adjoint-power questions instead of computing entries.
• 🎯 Any odd-order skew-symmetric determinant is zero — a one-line answer that students often miss.
• 🎯 Cayley–Hamilton lets you compute high powers/inverse by reducing Aⁿ using the characteristic polynomial.
• ❌ Common mistake: concluding a system is inconsistent whenever |A| = 0 — it may have infinitely many solutions; you must check the augmented rank.

Expected exam pattern

1–2 questions: a determinant-property/system-consistency problem and a special-matrix or adjoint/inverse conceptual (often multiple-correct). Cayley–Hamilton integers recur.

Quick recap

Use row/column operations; |AB|=|A||B|, |kA|=kⁿ|A|. A⁻¹ = adj A/|A|; |adj A| = |A|^(n−1). Homogeneous non-trivial solution iff |A|=0. Odd-order skew-symmetric det = 0. Cayley–Hamilton for powers/inverse.

Cover the answer, recall, then check. 12 cards on matrices and determinants for JEE Advanced.

Q1. Formula for the inverse of a matrix.
A1. A⁻¹ = adj(A)/|A|, existing iff |A| ≠ 0.

Q2. Determinant of a scalar multiple kA (n×n).
A2. |kA| = kⁿ|A|.

Q3. Determinant of a product |AB|.
A3. |A||B|.

Q4. Condition for a homogeneous system AX = 0 to have a non-trivial solution.
A4. |A| = 0.

Q5. Value of |adj A| for an n×n matrix.
A5. |A|^(n−1).

Q6. Determinant of an odd-order skew-symmetric matrix.
A6. Zero.

Q7. Defining property of an orthogonal matrix and its determinant.
A7. AAᵀ = I; |A| = ±1.

Q8. Cayley–Hamilton theorem.
A8. Every square matrix satisfies its own characteristic equation.

Q9. When AX = B has a unique solution.
A9. When |A| ≠ 0 (then X = A⁻¹B; Cramer's rule applies).

Q10. Effect of swapping two rows of a determinant.
A10. The sign of the determinant changes.

Q11. Property of a determinant with two identical rows.
A11. It equals zero.

Q12. Inverse of a product (AB)⁻¹.
A12. B⁻¹A⁻¹ (order reverses).

Matrices encode linear systems and transformations; determinants measure invertibility and scaling. JEE Advanced tests the algebra of matrices, properties of determinants (used to simplify without expanding), and the consistency of linear systems.

Core concept: a matrix represents a linear map; its determinant is nonzero exactly when the map is invertible, which governs when a system Ax = b has a unique solution.

Core ideas

Beginner — matrix algebra

Matrices add entrywise and multiply row-by-column (non-commutative: AB ≠ BA generally). Special types: symmetric (Aᵀ = A), skew-symmetric (Aᵀ = −A, zero diagonal), orthogonal (AᵀA = I), idempotent (A² = A), nilpotent (Aᵏ = 0), involutory (A² = I). Transpose rules: (AB)ᵀ = BᵀAᵀ, (AB)⁻¹ = B⁻¹A⁻¹ (order reverses).

Intermediate — determinants and inverse

The determinant is linear in each row, changes sign on row swaps, and is zero if two rows are equal/proportional — these properties let you simplify a determinant by row/column operations before expanding (the practical key). |AB| = |A||B|, |Aᵀ| = |A|, |kA| = kⁿ|A| (n×n), |A⁻¹| = 1/|A|. The inverse A⁻¹ = adj(A)/|A| exists iff |A| ≠ 0; |adj A| = |A|^(n−1), and adj(adj A) = |A|^(n−2) A.

Advanced — linear systems and Cayley–Hamilton

For Ax = b (n equations): if |A| ≠ 0, unique solution x = A⁻¹b (Cramer's rule). If |A| = 0, examine consistency: for a homogeneous system Ax = 0, |A| = 0 gives non-trivial (infinite) solutions, else only the trivial one — a favourite for "find k so the system has non-trivial solutions". For non-homogeneous, |A| = 0 gives either no solution or infinitely many (use rank/augmented reasoning). The Cayley–Hamilton theorem (every matrix satisfies its own characteristic equation) lets you compute high powers and inverses of a matrix efficiently — an advanced JEE tool.

Worked example

Find the value of λ for which the system x + y + z = 6, x + 2y + 3z = 10, x + 2y + λz = μ has (a) a unique solution, (b) infinitely many, (c) no solution. The coefficient determinant is |1 1 1; 1 2 3; 1 2 λ|. Expanding: subtract row 1 from rows 2 and 3 → |1 1 1; 0 1 2; 0 1 λ−1| = 1·(1·(λ−1) − 2·1) = λ − 3. (a) Unique iff determinant ≠ 0, i.e. λ ≠ 3. (b/c) When λ = 3, rows 2 and 3 of the coefficient matrix become proportional; the system is consistent (infinitely many) if μ = 10 and inconsistent (no solution) if μ ≠ 10. The determinant-then-consistency analysis is exactly the JEE-Advanced structure.

How JEE Advanced tests this

System-consistency conditions (unique/infinite/none) via determinant and rank; determinant evaluation using row/column operations; properties of adj/inverse (|adj A|, adj(adj A)); special-matrix identities (symmetric/skew, orthogonal, idempotent); Cayley–Hamilton for powers/inverse; and homogeneous-system non-trivial-solution conditions.

Exam tricks & shortcuts

• Simplify determinants by row/column operations before expanding.
• Homogeneous system has non-trivial solutions ⇔ |A| = 0.
• |adj A| = |A|^(n−1); adj(adj A) = |A|^(n−2)A; |kA| = kⁿ|A|.
• (AB)⁻¹ = B⁻¹A⁻¹, (AB)ᵀ = BᵀAᵀ — order reverses.

Key formulas

• (AB)ᵀ = BᵀAᵀ; (AB)⁻¹ = B⁻¹A⁻¹.
• Inverse: A⁻¹ = adj(A)/|A| (|A| ≠ 0); A·adj(A) = |A|I.
• |AB| = |A||B|; |kA| = kⁿ|A|; |adj A| = |A|ⁿ⁻¹.
• Systems: AX = B → X = A⁻¹B; Cramer x_i = Δ_i/Δ.
• Symmetric Aᵀ = A; skew Aᵀ = −A.

Matrix multiplication is non-commutative; the determinant decides invertibility and solvability of linear systems.

Worked Example

Problem: Solve the system 2x + y = 8 and x + 3y = 9 by finding the inverse of the coefficient matrix A = [[2, 1], [1, 3]] and using X = A⁻¹B.

Solution:
Write the system as AX = B with A = [[2, 1], [1, 3]], X = [x, y]ᵀ, B = [8, 9]ᵀ.

Determinant of A:
det A = (2)(3) − (1)(1) = 6 − 1 = 5.
Since det A ≠ 0, A is invertible.

Inverse of a 2×2 matrix [[a, b], [c, d]] is (1/det)[[d, −b], [−c, a]]:
A⁻¹ = (1/5)[[3, −1], [−1, 2]].

Now X = A⁻¹B:
X = (1/5)[[3, −1], [−1, 2]] [8, 9]ᵀ
= (1/5)[ (3·8 − 1·9), (−1·8 + 2·9) ]ᵀ
= (1/5)[ (24 − 9), (−8 + 18) ]ᵀ
= (1/5)[15, 10]ᵀ = [3, 2]ᵀ.

Answer: x = 3, y = 2.