Matrices and Determinants — Summary
Matrices and determinants are the entry point to GATE EC Engineering Mathematics (which carries ~13 marks / paper). They power circuit-equation solving (nodal/mesh), state-space, and appear as direct 1–2 mark questions on rank, determinant value, and inverse.
Key results
- det(AB) = det(A)·det(B); det(Aᵀ) = det(A); det(A⁻¹) = 1/det(A); det(kA) = kⁿ·det(A) for n×n.
- A is invertible ⇔ det(A) ≠ 0 ⇔ rank = n ⇔ columns independent.
- (AB)⁻¹ = B⁻¹A⁻¹; (AB)ᵀ = BᵀAᵀ.
- Rank = number of non-zero rows in row-echelon form = order of the largest non-zero minor.
- A⁻¹ = adj(A)/det(A); adj(A) = transpose of cofactor matrix.
| Matrix type | Defining property | Determinant |
|---|---|---|
| Symmetric | A = Aᵀ | real eigenvalues |
| Skew-symmetric (odd n) | A = −Aᵀ | 0 |
| Orthogonal | AᵀA = I | ±1 |
| Triangular | zeros one side | product of diagonal |
| Singular | det = 0 | not invertible |
Exam Tricks & Tips
- 🎯 Determinant = product of eigenvalues; trace = sum of eigenvalues — use to cross-check without expansion.
- 🎯 For a triangular (or diagonal) matrix, det is just the diagonal product — spot this instantly.
- 🎯 Row/column operations don't change rank; swapping two rows flips det sign, scaling a row by k scales det by k.
- 🎯 Skew-symmetric matrix of odd order always has det = 0 (a common trap answer).
- 🎯 For 2×2, A⁻¹ = (1/det)·[[d,−b],[−c,a]] — memorise the swap-and-negate pattern.
- 🎯 Rank of a product: rank(AB) ≤ min(rank A, rank B).
- ❌ Common mistake: assuming det(A+B) = det(A)+det(B) — determinant is NOT additive; only multiplicative.
Expected exam pattern
1-mark: find det/rank/inverse of a small numeric matrix, or identify singular condition (solve for a parameter making det = 0). 2-mark: rank after operations, or det via eigenvalue/trace shortcuts.
Quick recap
Determinant is multiplicative not additive; det ≠ 0 ⇔ full rank ⇔ invertible. Rank = largest non-zero minor. Use trace/det = sum/product of eigenvalues to verify. Skew-symmetric odd order → det 0.
Matrices and Determinants — Flashcards
Cover the answer, recall, then check. 12 cards on matrices and determinants for GATE EC.
Q1. When is a square matrix A invertible?
A1. iff det(A) ≠ 0, equivalently rank(A) = n, equivalently its columns (rows) are linearly independent.
Q2. State det(AB) and det(kA) for an n×n matrix.
A2. det(AB) = det(A)·det(B); det(kA) = kⁿ·det(A).
Q3. How is det related to eigenvalues and trace?
A3. det(A) = product of all eigenvalues; trace(A) = sum of diagonal = sum of eigenvalues.
Q4. What is the determinant of any odd-order skew-symmetric matrix?
A4. Zero. (Because det(A) = det(−Aᵀ) = (−1)ⁿ det(A), and for odd n this forces det = 0.)
Q5. Give the inverse formula for a general non-singular matrix.
A5. A⁻¹ = adj(A)/det(A), where adj(A) is the transpose of the cofactor matrix.
Q6. How do you find the rank of a matrix?
A6. Reduce to row-echelon form; rank = number of non-zero rows = order of the largest non-vanishing minor.
Q7. What is det of a triangular or diagonal matrix?
A7. The product of its diagonal entries.
Q8. State (AB)⁻¹ and (AB)ᵀ.
A8. (AB)⁻¹ = B⁻¹A⁻¹ and (AB)ᵀ = BᵀAᵀ (order reverses).
Q9. For a 2×2 matrix [[a,b],[c,d]], write the inverse.
A9. (1/(ad−bc))·[[d,−b],[−c,a]].
Q10. How does swapping two rows affect the determinant?
A10. It multiplies the determinant by −1 (sign flip). Adding a multiple of one row to another leaves det unchanged.
Q11. What property makes AᵀA = I?
A11. A is orthogonal; then det(A) = ±1 and A⁻¹ = Aᵀ.
Q12. Bound on rank of a product AB.
A12. rank(AB) ≤ min(rank(A), rank(B)).
Matrices and Determinants
Matrices are the language linear systems, circuits, and transforms all speak, and the determinant is the single number that tells you whether a system has a unique solution, whether a matrix is invertible, and how it scales volume. GATE Engineering Mathematics reliably delivers 1–2 marks here, often as a quick rank or determinant computation that gates a larger problem.
Core concept: A matrix encodes a linear system; its determinant tests invertibility, and its rank counts independent equations, deciding whether solutions are unique, infinite, or absent.
The theory
Beginner — determinant essentials
For A (n×n), det(A) = 0 ⇔ A is singular (non-invertible, rows/columns linearly dependent). Useful properties:
- det(AB) = det(A)·det(B); det(Aᵀ) = det(A); det(A^(−1)) = 1/det(A).
- det(kA) = kⁿ·det(A) for an n×n matrix.
- Swapping two rows flips the sign; adding a multiple of one row to another leaves det unchanged.
- Triangular/diagonal matrix: det = product of diagonal entries.
Intermediate — rank and solvability
The rank is the number of linearly independent rows (= independent columns), found by row-reducing to echelon form and counting non-zero rows. For A·x = b with A being m×n:
- Unique solution: rank(A) = rank([A|b]) = n.
- Infinite solutions: rank(A) = rank([A|b]) < n (free variables = n − rank).
- No solution: rank(A) < rank([A|b]) (inconsistent).
Homogeneous A·x = 0 has a non-trivial solution ⇔ det(A) = 0 (rank < n).
Advanced — inverse and special forms
- Inverse: A^(−1) = adj(A)/det(A); adj(A) is the transpose of the cofactor matrix.
- Cayley-Hamilton: every matrix satisfies its own characteristic equation — a fast route to A^(−1) and high powers of A.
- Special matrices: orthogonal (AᵀA = I, det = ±1), symmetric (A = Aᵀ), skew-symmetric (Aᵀ = −A, so det = 0 for odd n), unitary (complex analogue of orthogonal).
- Trace = sum of diagonal = sum of eigenvalues; det = product of eigenvalues.
Worked example
For what value of k does the system x + 2y + 3z = 0, 2x + ky + 6z = 0, 3x + 6y + 9z = 0 have non-trivial solutions?
Non-trivial ⇔ det = 0. Note row3 = 3·row1, so the matrix is already singular for any k — det = 0 always. So non-trivial solutions exist for all k. (If instead only rows 1,2 were given independently, solve det = 0.)
Sanity: rank ≤ 2 < 3 always ⇒ at least one free variable. ✔
GATE relevance
1–2 mark NAT/MCQ: evaluate a determinant, find rank, decide consistency/number of solutions, compute an inverse, or use det properties (det(kA), det(AB)). Also the entry point to eigenvalue problems and state-space controllability.
Exam tricks and shortcuts
- Spot dependent rows/columns instantly ⇒ det = 0, skip expansion.
- det(kA) = kⁿ det(A) — the exponent n trips people up.
- Product of eigenvalues = det; sum = trace — cross-check without solving.
- Mnemonic: "Rank matches ⇒ consistent; equals n ⇒ unique."
Writing det(kA) = k·det(A). For an n×n matrix every one of the n rows scales by k, so det(kA) = kⁿ·det(A). Forgetting the exponent is the single most common determinant error.
- ✓- det(A) = 0 ⇔ singular ⇔ rows/cols dependent ⇔ non-trivial Ax=0.
- ✓- det(AB)=det A·det B; det(kA)=kⁿ det A; det(Aᵀ)=det A.
- ✓- Solutions: unique (rank=rank[A|b]=n), infinite (<n), none (rank A<rank[A|b]).
- ✓- A^(−1) = adj(A)/det(A); trace = Σλ, det = Πλ.
- ✓- Orthogonal: AᵀA = I, det = ±1.
- ✓The determinant flags invertibility and the rank counts independent equations; together they classify every linear system as unique, infinite, or inconsistent. Use property shortcuts (dependent rows, kⁿ scaling, trace/det = sum/product of eigenvalues) rather than brute-force expansion.
Matrices and Determinants — Formula Sheet
Key formulas
- Determinant: |A| ≠ 0 ⇔ invertible; A⁻¹ = adj(A)/|A|.
- Properties: |AB| = |A||B|; |kA| = kⁿ|A|; |Aᵀ| = |A|.
- Rank: number of linearly independent rows/columns; system consistent if rank(A) = rank(A|B).
- Unique solution ⇔ rank = number of unknowns; infinite if rank < unknowns; none if ranks differ.
- Cayley–Hamilton: every matrix satisfies its own characteristic equation.
- ✓- A⁻¹ = adj(A)/|A|; |AB| = |A||B|.
- ✓- Consistency: rank(A) = rank(A|B).
- ✓- Unique solution ⇔ rank = number of unknowns.
Rank determines the solvability of AX = B; Cayley–Hamilton lets you compute powers/inverses of a matrix.
Matrices and Determinants — Worked Example
Worked Example
Problem: Using Cramer's rule, solve for x in the linear system
x + y + z = 6, x + 2y + 3z = 14, x + 4y + 9z = 36.
Solution:
Write the coefficient matrix and its determinant D:
A = [[1, 1, 1], [1, 2, 3], [1, 4, 9]].
The nodes 1, 2, 3 form a Vandermonde structure, so
D = (2−1)(3−1)(3−2) = 1 × 2 × 1 = 2.
Since D = 2 ≠ 0, the system has a unique solution and Cramer's rule applies.
For x, replace the first column of A with the right-hand side (6, 14, 36):
Dx = |[[6, 1, 1], [14, 2, 3], [36, 4, 9]]|.
Expand along row 1:
Dx = 6(2×9 − 3×4) − 1(14×9 − 3×36) + 1(14×4 − 2×36)
Dx = 6(18 − 12) − 1(126 − 108) + 1(56 − 72)
Dx = 6(6) − 1(18) + 1(−16) = 36 − 18 − 16 = 2.
Therefore x = Dx / D = 2 / 2 = 1.
Back-substituting confirms the full solution (x, y, z) = (1, 2, 3):
check row 3 → 1 + 4(2) + 9(3) = 1 + 8 + 27 = 36 ✓.
Answer: x = 1 (complete solution (1, 2, 3)).
- ✓- A unique solution exists only when the coefficient determinant D ≠ 0; then Cramer's rule gives each variable as Dᵢ / D.
- ✓- Dᵢ is formed by replacing the i-th column of A with the constants column — everything else stays fixed.
- ✓- For matrices built from powers of nodes, the Vandermonde formula ∏(xⱼ − xᵢ) evaluates the determinant instantly.