Matrices & Determinants — vidhyapath 2027

Concept Core

From matrix types to Cramer's Rule — structured for exam speed.

Matrix Types — Quick Reference

    
      TypeDefinitionExample
Row matrix1 × n matrix[1 2 3]
Column matrixn × 1 matrixSingle column
Square matrixn × n2×2, 3×3
DiagonalOff-diagonal elements = 0diag(a,b,c)
Identity (I)Diagonal = 1, rest = 0AI = IA = A
SymmetricA = Aᵀaᵢⱼ = aⱼᵢ
Skew-symmetricA = −Aᵀaᵢⱼ = −aⱼᵢ; diagonal = 0
OrthogonalAAᵀ = I → A⁻¹ = AᵀRotation matrices

  

Type	Definition	Example
Row matrix	1 × n matrix	[1 2 3]
Column matrix	n × 1 matrix	Single column
Square matrix	n × n	2×2, 3×3
Diagonal	Off-diagonal elements = 0	diag(a,b,c)
Identity (I)	Diagonal = 1, rest = 0	AI = IA = A
Symmetric	A = Aᵀ	aᵢⱼ = aⱼᵢ
Skew-symmetric	A = −Aᵀ	aᵢⱼ = −aⱼᵢ; diagonal = 0
Orthogonal	AAᵀ = I → A⁻¹ = Aᵀ	Rotation matrices

Matrix Operations

Addition: Same order only. Add element-by-element. Commutative.

Multiplication: (m×n)(n×p) = (m×p). NOT commutative: AB ≠ BA in general.

Transpose: (Aᵀ)ᵢⱼ = Aⱼᵢ. (AB)ᵀ = BᵀAᵀ

Trace: tr(A) = sum of diagonal elements = sum of eigenvalues

Determinant of 2×2 and 3×3

2×2:

|A| = ad − bc for A = [[a,b],[c,d]]

3×3 (cofactor expansion along row 1):

|A| = a₁₁C₁₁ + a₁₂C₁₂ + a₁₃C₁₃

where Cᵢⱼ = (−1)^(i+j) × Mᵢⱼ (Mᵢⱼ = minor)

Properties of Determinants

1. |Aᵀ| = |A| — transposing doesn't change determinant

2. Swapping two rows/cols: sign of det changes

3. Row of zeros → det = 0

4. Two identical rows/cols → det = 0

5. |kA| = kⁿ|A| for n×n matrix

6. |AB| = |A|·|B|

Inverse of a Matrix

A⁻¹ exists only when |A| ≠ 0 (A is non-singular).

A⁻¹ = adj(A) / |A|

adj(A) = transpose of cofactor matrix

Properties: AA⁻¹ = I | (AB)⁻¹ = B⁻¹A⁻¹ | (A⁻¹)ᵀ = (Aᵀ)⁻¹

Cramer's Rule — Solving Linear Systems

For AX = B where A is n×n, solution: xᵢ = Dᵢ/D

where D = |A| and Dᵢ = |A with i-th column replaced by B|

Consistency:

D ≠ 0: unique solution

D = 0, all Dᵢ = 0: infinitely many solutions

D = 0, any Dᵢ ≠ 0: no solution (inconsistent)

Eigenvalues — Characteristic Equation

Eigenvalues λ satisfy: det(A − λI) = 0 (characteristic equation)

For 2×2 matrix: λ² − (tr A)λ + |A| = 0

λ₁ + λ₂ = tr(A) = a₁₁ + a₂₂ λ₁λ₂ = |A|

Cayley-Hamilton Theorem: every matrix satisfies its own characteristic equation. Use to find powers of A.

Formula Vault

All matrix and determinant formulas in one place.

2×2 Determinant

|A| = ad − bc

A = [[a,b],[c,d]]

2×2 Inverse

A⁻¹ = (1/|A|)[[d,−b],[−c,a]]

Swap diagonal, negate off-diagonal

Adjoint

adj(A) = (Cofactor matrix)ᵀ

A·adj(A) = |A|·I

Inverse Formula

A⁻¹ = adj(A) / |A|

Only when |A| ≠ 0

Determinant of Product

|AB| = |A| · |B|

|Aⁿ| = |A|ⁿ

Determinant of Scalar Multiple

|kA| = kⁿ|A|

n = order of matrix

Cramer's Rule

x = D₁/D, y = D₂/D

D = |coeff matrix|

Eigenvalue Sum & Product

Σλ = tr(A), Πλ = |A|

For 2×2 matrix

Transpose Properties

(AB)ᵀ = BᵀAᵀ; (A+B)ᵀ = Aᵀ+Bᵀ

Reverse order for products

Rank

ρ(A) = max order of non-zero minor

Full rank n×n: ρ=n → A invertible

Worked Examples

5 problems covering the full EAPCET spectrum.

EasyEvaluate the determinant of [[3,1],[−2,4]]▾

Find |A| for A = [[3, 1], [−2, 4]].

|A| = ad − bc = (3)(4) − (1)(−2) = 12 + 2 = 14

✓ |A| = 14

EasyFind the inverse of A = [[2,1],[5,3]]▾

Find A⁻¹ for A = [[2,1],[5,3]].

|A| = (2)(3) − (1)(5) = 6 − 5 = 1

A⁻¹ = (1/1) × [[3,−1],[−5,2]] = [[3,−1],[−5,2]]

✓ A⁻¹ = [[3,−1],[−5,2]]

MediumSolve using Cramer's Rule: 2x+y=5, 3x−y=0▾

Solve the system: 2x + y = 5 and 3x − y = 0 using Cramer's Rule.

Coefficient matrix: A = [[2,1],[3,−1]]. D = |A| = (2)(−1)−(1)(3) = −2−3 = −5

D₁ (replace col 1 by constants [5,0]): |[[5,1],[0,−1]]| = −5−0 = −5

D₂ (replace col 2 by [5,0]): |[[2,5],[3,0]]| = 0−15 = −15

x = D₁/D = −5/−5 = 1; y = D₂/D = −15/−5 = 3

✓ x = 1, y = 3

EAPCET LevelFind k such that the system has no solution▾

For what value of k does the system x+2y=3, 3x+ky=9 have no solution?

D = |[[1,2],[3,k]]| = k − 6

For no solution: D = 0 AND not all Dᵢ = 0. So k − 6 = 0 → k = 6

Verify D₁: |[[3,2],[9,6]]| = 18−18 = 0. D₂: |[[1,3],[3,9]]| = 9−9 = 0.

Both Dᵢ = 0 when k=6 → infinitely many solutions, NOT no solution. Try: equations become x+2y=3 and 3x+6y=9 → 3(x+2y)=9 → same line → infinitely many solutions

For no solution we need D=0 but at least one Dᵢ≠0. Change problem: if RHS was 9 and 10 instead of 3 and 9, then k=6 gives no solution. For this problem: no unique k gives no solution — the system is always consistent.

✓ For k = 6: infinitely many solutions. For k ≠ 6: unique solution. No solution is not possible here.

Trap QuestionIf A is a 3×3 matrix with |A| = 5, find |3A| and |adj A|▾

A is a 3×3 matrix with |A| = 5. Find (i) |3A| and (ii) |adj A|.

|3A|: Use |kA| = kⁿ|A| where n = 3. So |3A| = 3³ × 5 = 27 × 5 = 135

|adj A|: Use formula |adj A| = |A|^(n−1) = 5^(3−1) = 5² = 25

✓ |3A| = 135, |adj A| = 25

Mistake DNA

5 common errors that cost marks in EAPCET matrix questions.

✖️

Assuming Matrix Multiplication is Commutative

AB = BA is NOT generally true. This error appears in simplification problems.

❌ Wrong

AB = BA assumed
→ wrong simplification
of (AB)ᵀ = AᵀBᵀ ✗

✓ Correct

(AB)ᵀ = BᵀAᵀ ✓
(AB)⁻¹ = B⁻¹A⁻¹ ✓
Order reverses for products

For any product of matrices, transpose or inverse reverses the order. This is one of the most frequently tested properties.

🔢

|kA| = k|A| instead of kⁿ|A|

The scalar factor k is raised to the power n (order of matrix), not left as k.

❌ Wrong

For 3×3, |A|=4:
|2A| = 2 × 4 = 8 ✗

✓ Correct

|2A| = 2³ × 4 = 32 ✓
Each row gets factor k,
so kⁿ overall

Each row of the matrix can be scaled by k, contributing k to the determinant. For n rows, that's kⁿ. This is NOT the same as multiplying a single row.

🔑

Computing Inverse When |A| = 0

A singular matrix (|A|=0) has no inverse. Students still try to compute A⁻¹.

❌ Wrong

|A| = 0 but still compute
A⁻¹ = adj(A)/0 ✗
(division by zero)

✓ Correct

If |A| = 0: A is singular
→ A⁻¹ does not exist ✓
First check |A| ≠ 0

Always compute |A| first. If it's zero, stop — the matrix is singular. This also means the linear system either has infinitely many or no solutions.

📊

Cofactor Sign Errors in 3×3 Determinant

Forgetting the (−1)^(i+j) checkerboard sign pattern when computing cofactors.

❌ Wrong

All cofactors positive ✗
C₁₂ = +M₁₂ instead
of −M₁₂

✓ Correct

Sign pattern: + − +
− + −
+ − +

Cofactor Cᵢⱼ = (−1)^(i+j) × Mᵢⱼ. Corner elements are +, edge centers are −. Memorise the checkerboard.

➗

D = 0 → Automatically Claiming "No Solution"

When D = 0, the system might have infinitely many solutions (not necessarily no solution).

❌ Wrong

D = 0 → "system has
no solution" ✗

✓ Correct

D = 0: check all Dᵢ
All Dᵢ = 0 → ∞ solutions
Any Dᵢ ≠ 0 → no solution ✓

D = 0 is the starting point, not the conclusion. You must check D₁, D₂, etc. to determine if the system is inconsistent or has infinitely many solutions.

Chapter Intelligence

Matrices is one of the highest-scoring chapters if you know exactly what to practise.

EAPCET Topic Weightage (2019–2024)

Inverse of a matrix

3×3 determinant evaluation

Cramer's Rule

|kA|, |adj A| properties

Matrix multiplication

Eigenvalues / char. equation

High-Yield PYQ Patterns

Find A⁻¹ using adj/det |adj A| and |kA| Solve 2×2 system using Cramer's Consistency of 3-equation system Symmetric / skew-symmetric Eigenvalue of 2×2 matrix

Exam Strategy

For 2×2 inverse: use the swap-negate formula directly — don't compute adjoint through cofactors. It's 4× faster.
Before solving any system: compute D first. If D ≠ 0, unique solution exists — use Cramer's. If D = 0, investigate further.
Properties questions (|kA|, |adj A|): memorise the two formulas |kA|=kⁿ|A| and |adj A|=|A|^(n−1). These appear as 30-second MCQs every year.
3×3 determinant: expand along the row or column with the most zeros — it reduces the number of 2×2 determinants you need to compute.
Check: (AB)ᵀ = BᵀAᵀ, (AB)⁻¹ = B⁻¹A⁻¹. Order reversal is tested very frequently in EAPCET.