📘 Algebra · Chapter MN09🎯 NDA Level : High Priority
Matrices and Determinants is one of the most predictably tested chapters in NDA Mathematics. The chapter is wide — covering matrix types, operations, determinants, inverses, and solving equations — but each individual question type is mechanical and formula-driven. A well-practised student can reliably score full marks here.
📌 What to expect in NDA (based on 2022–2024 papers): (1) Evaluating 2×2 and 3×3 determinants by expansion;
(2) Properties of determinants — row/column operations, effect of interchanging rows;
(3) Finding the inverse of a 2×2 matrix using adjoint method;
(4) Matrix multiplication of two 2×2 or 2×3 matrices;
(5) Transpose, symmetric, and skew-symmetric matrix identification;
(6) Solving 2×2 or 3×3 system of equations using Cramer's Rule;
(7) Condition for a system to have no unique solution (D = 0);
(8) Statement-based questions: rank, invertibility, properties of special matrices.
Topics at a Glance
① Matrix Types
Row, column, square, identity, diagonal, null
② Matrix Operations
Addition, scalar mult, multiplication, transpose
③ Determinants
2×2 formula, 3×3 expansion, properties
④ Adjoint & Inverse
adj A, A⁻¹ = adj A / |A|, conditions
⑤ Cramer's Rule
D, D₁, D₂, D₃ for 2×2 and 3×3 systems
⑥ Matrix Method
AX = B → X = A⁻¹B
1. Matrices — Definition & Types
1.1
Definition & Notation
A matrix is a rectangular array of numbers arranged in rows and columns
A matrix of order m×n has m rows and n columns. Each entry aᵢⱼ is in row i, column j. The order is always written as rows × columns.
⚡ Matrix Notation & Order
A = [aᵢⱼ]ₘₓₙ means m rows, n columns, element at row i column j is aᵢⱼ
Example — 2×3 matrix:
A = | a₁₁ a₁₂ a₁₃ | ← Row 1
| a₂₁ a₂₂ a₂₃ | ← Row 2
For A to equal B: same order AND aᵢⱼ = bᵢⱼ for all i, j (element-wise equality)
Square matrix: m = n (number of rows = number of columns)
Leading diagonal: elements a₁₁, a₂₂, a₃₃, … (where i = j)
A matrix is NOT a number — it is an array. Two matrices can be added/compared only if they have the same order.
Types of Matrices — Complete Reference
Row Matrix (1×n)
Has only 1 row and any number of columns.
[1 3 −2]
Column Matrix (m×1)
Has only 1 column and any number of rows.
|4| |−1| |7|
Square Matrix (n×n)
Equal number of rows and columns. Determinant is defined.
|1 2| |3 4|
Zero / Null Matrix
All elements are zero. Acts like 0 in addition.
|0 0| |0 0|
Identity Matrix (I)
Diagonal elements = 1, all others = 0. AI = IA = A.
|1 0| |0 1|
Diagonal Matrix
Square matrix with all non-diagonal elements = 0.
|3 0| |0 5|
Symmetric Matrix
A = Aᵀ (transpose equals original). aᵢⱼ = aⱼᵢ for all i,j.
|1 2| |2 3|
Skew-Symmetric Matrix
Aᵀ = −A. Diagonal elements = 0. aᵢⱼ = −aⱼᵢ.
|0 2| |−2 0|
⚠ Key Identification Tests:
• Symmetric: check aᵢⱼ = aⱼᵢ — i.e., element at (1,2) = element at (2,1), etc.
• Skew-symmetric: diagonal must all be 0, and aᵢⱼ = −aⱼᵢ.
• Every square matrix A can be written as A = P + Q where P = (A+Aᵀ)/2 is symmetric and Q = (A−Aᵀ)/2 is skew-symmetric. This decomposition is directly tested in NDA.
Multiplication is the most tested operation — order matters
⚡ Operations Summary
Addition (A + B):
Same order required. Add element-by-element: (A+B)ᵢⱼ = aᵢⱼ + bᵢⱼ
Commutative: A+B = B+A. Associative: A+(B+C) = (A+B)+C
Scalar Multiplication (kA):
Multiply every element by k: (kA)ᵢⱼ = k·aᵢⱼ
Matrix Multiplication (AB):
A must be m×n, B must be n×p → product AB is m×p
(AB)ᵢⱼ = Σ aᵢₖ·bₖⱼ (row i of A dotted with column j of B)
NOT commutative: AB ≠ BA in general
Associative: (AB)C = A(BC)
Distributive: A(B+C) = AB + AC
Transpose (Aᵀ):
Interchange rows and columns: (Aᵀ)ᵢⱼ = aⱼᵢ
(Aᵀ)ᵀ = A; (AB)ᵀ = BᵀAᵀ (reverse order!); (A+B)ᵀ = Aᵀ+Bᵀ
Matrix multiplication requires the inner dimensions to match: (m×n)(n×p) = m×p. Always check order before multiplying. The rule (AB)ᵀ = BᵀAᵀ — note the reversal — is a favourite NDA statement question.
Answer: (b) Symmetric
(A+Aᵀ)ᵀ = Aᵀ+(Aᵀ)ᵀ = Aᵀ+A = A+Aᵀ. Since (A+Aᵀ)ᵀ = A+Aᵀ, it is symmetric.
🔥 TRICKY QUESTIONS
Matrices — Non-Commutative & Decomposition Traps
🧩 T1. If A = |0 1| (skew-symmetric), verify A + Aᵀ = 0 and find A².
|−1 0|
Aᵀ = |0 −1| → A + Aᵀ = |0 0| = O ✓ (skew-symmetric property: A+Aᵀ = 0).
|1 0| |0 0|
A² = |0×0+1×(−1) 0×1+1×0| = |−1 0|
|−1×0+0×(−1) −1×1+0×0| |0 −1|
A² = −I. This is a classic result: the square of a 2×2 skew-symmetric matrix is −I (times a scalar). Tested in NDA statement questions.
🧩 T2. AB = 0 but A ≠ 0 and B ≠ 0. Give an example showing this is possible.
Let A = |1 0| and B = |0 0|.
|0 0| |0 1|
AB = |1×0+0×0 1×0+0×1| = |0 0|.
|0×0+0×0 0×0+0×1| |0 0|
AB = O, yet A ≠ O and B ≠ O. Matrix multiplication has zero divisors. This directly contradicts the intuition from real numbers where ab=0 implies a=0 or b=0. Statement-verification question type in NDA.
3. Determinants
3.1
2×2 Determinant & 3×3 Expansion
Always expand along the first row by default in NDA problems
⚡ Determinant Formulae
2×2 Determinant:
|a b|
|c d| = ad − bc ← product of main diagonal minus product of anti-diagonal
3×3 Determinant (expansion along Row 1):
|a₁ b₁ c₁|
|a₂ b₂ c₂| = a₁(b₂c₃−b₃c₂) − b₁(a₂c₃−a₃c₂) + c₁(a₂b₃−a₃b₂)
|a₃ b₃ c₃|
Sign pattern for cofactors (checkerboard):
+ − +
− + −
+ − +
For 3×3: each element of row 1 multiplies its 2×2 minor (remaining 2×2 after deleting that row and column), with alternating + and − signs.
Sarrus' Rule — Visual Mnemonic for 3×3 Determinant
Fig 1: Sarrus' Rule — green arrows (↘) give positive products; red arrows (↗) give negative products. Sum = determinant.
Worked Example — 3×3 Determinant
Find |1 2 3|
|4 5 6|
|7 8 9|
Expand along Row 1:
= 1·(5×9−6×8) − 2·(4×9−6×7) + 3·(4×8−5×7)
= 1·(45−48) − 2·(36−42) + 3·(32−35)
= 1·(−3) − 2·(−6) + 3·(−3) = −3 + 12 − 9 = 0
Result = 0 means this matrix is singular (rows are in AP — a classic NDA example of a zero determinant).
3.2
Properties of Determinants
Properties allow shortcuts — avoid expanding when a property applies
Property
Statement
Effect / Use
P1 — Transpose
|Aᵀ| = |A|
Row properties apply to columns too
P2 — Row Interchange
Swapping any two rows (or columns) multiplies determinant by −1
Explains sign change in Cramer's rule
P3 — Identical Rows
If two rows (or columns) are identical, |A| = 0
Quick check for singularity
P4 — Proportional Rows
If one row is a scalar multiple of another, |A| = 0
Rows in AP → |A| = 0
P5 — Scalar Multiple
Multiplying one row by k multiplies |A| by k
|kA| = kⁿ|A| for n×n matrix
P6 — Row Addition
Adding a multiple of one row to another leaves |A| unchanged
Simplify entries before expanding
P7 — Zero Row
If any row (or column) is all zeros, |A| = 0
Singular matrix
P8 — Product
|AB| = |A|·|B|
|Aⁿ| = |A|ⁿ, |A⁻¹| = 1/|A|
📌 Most Frequently Tested Properties in NDA:
P3 (identical rows → 0), P4 (proportional rows → 0), P5 (|kA| = kⁿ|A|), and P8 (|AB| = |A|·|B|). The P5 result for |kA| is the single most common trap — students write |kA| = k|A| (correct only for 1×1, not for 2×2 or 3×3).
📋 TOPIC-WISE PYQ
Determinants — NDA-Pattern Questions
Q4. The value of |3 2| is:
|1 4|
(a) 10 (b) 12 (c) 14 (d) 8
Answer: (a) 10
= 3×4 − 2×1 = 12 − 2 = 10.
Q5. If A is a 3×3 matrix and |A| = 4, then |3A| = ?
(a) 12 (b) 36 (c) 108 (d) 64
Answer: (c) 108
|kA| = kⁿ·|A| for an n×n matrix. Here n=3, k=3: |3A| = 3³×4 = 27×4 = 108.
Common trap: writing 3×4 = 12. Remember: each row gets multiplied by k, so |kA| = kⁿ|A|.
Q6. The value of the determinant |1 a bc| is:
|1 b ca|
|1 c ab|
(a) 0 (b) 1 (c) abc (d) a+b+c
Answer: (a) 0
Multiply Col 1 by abc: |abc a bc| → each row becomes scalar multiple of a common factor.
Actually: R1→R1−R2, R2→R2−R3 simplifies. Or note: multiply each row by (a,b,c) respectively → Col3 becomes (abc, abc, abc) — then Col3 is identical → det = 0. The standard approach for this classic matrix identity confirms |A| = 0.
🔥 TRICKY QUESTIONS
Determinants — Property Application Traps
🧩 T3. Without expanding, show that |a+b b+c c+a| = 0.
|b+c c+a a+b|
|c+a a+b b+c|
(Using the 3-row version: R1+R2+R3 = 2(a+b+c) in each column — but let's use the simpler observation.)
Apply C1→C1+C2+C3: each column sum = 2(a+b+c), so Col 1 becomes all 2(a+b+c). This makes Col1 a multiple of the all-ones vector — but more directly, note that Row1+Row2+Row3 = (a+b+c)×(1,1,1) duplicated... The key insight: The sum of all three rows is the same in each row (each row sums to a+b+c+a+b+c = 2(a+b+c)), so R1→R1−R2−R3 → first row becomes zero → determinant = 0. Actually: rows are cyclic permutations → linearly dependent → |A| = 0.
🧩 T4. If A is a 2×2 matrix with |A| = 5, find |adj A|.
For an n×n matrix: |adj A| = |A|^(n−1).
Here n = 2: |adj A| = |A|^(2−1) = |A|¹ = 5.
For n = 3: |adj A| = |A|² — this is tested more often.
Also: |A⁻¹| = 1/|A| = 1/5 (since |A·A⁻¹| = |I| = 1 → |A|·|A⁻¹| = 1).
4. Adjoint & Inverse of a Matrix
4.1
Cofactor Matrix, Adjoint & Inverse
Inverse exists only when |A| ≠ 0 — this condition is directly tested
⚡ Inverse of a Matrix — Step-by-Step
Step 1 — Cofactor Cᵢⱼ:
Cᵢⱼ = (−1)^(i+j) × Mᵢⱼ
where Mᵢⱼ = minor (determinant of matrix after deleting row i, col j)
Sign pattern: + − + / − + − / + − +
Step 2 — Adjoint (adj A):
adj A = Transpose of the cofactor matrix
[adj A]ᵢⱼ = Cⱼᵢ (note: cofactor of ROW j, COL i goes into position (i,j))
Step 3 — Inverse A⁻¹:
A⁻¹ = (adj A) / |A| provided |A| ≠ 0
Key identities:
A · (adj A) = (adj A) · A = |A| · I
|adj A| = |A|^(n−1) (for n×n matrix)
adj(AB) = (adj B)(adj A) (reversed order)
(A⁻¹)⁻¹ = A
(AB)⁻¹ = B⁻¹A⁻¹ (reversed order)
A matrix is invertible (non-singular) iff |A| ≠ 0. If |A| = 0, the matrix is singular and has no inverse. This is the key condition for a unique solution to AX = B.
⚡ Quick Inverse of a 2×2 Matrix
If A = |a b| then A⁻¹ = (1/(ad−bc)) × |d −b|
|c d| |−c a|
Steps: (1) Calculate |A| = ad − bc. (2) Swap a and d. (3) Negate b and c. (4) Divide by |A|.
Example: A = |2 3| → |A| = 2×1−3×4 = 2−12 = −10
|4 1|
A⁻¹ = (1/−10) × |1 −3| = |−0.1 0.3|
|−4 2| | 0.4 −0.2|
Verify: A·A⁻¹ = I ✓
The 2×2 inverse shortcut: swap diagonal, negate off-diagonal, divide by determinant. This is the fastest method for NDA — no cofactor computation needed for 2×2.
A² = A → A·A = A → A⁻¹·(A·A) = A⁻¹·A → A = I. A = I (the identity matrix). Any matrix satisfying A² = A is called idempotent. The only invertible idempotent is the identity. Non-invertible idempotents (like projection matrices) also exist but are not asked at NDA level.
🧩 T6. If A = |cos θ −sin θ|, find A⁻¹ and note what it equals.
|sin θ cos θ|
|A| = cos²θ + sin²θ = 1.
A⁻¹ = (1/1)|cos θ sin θ| = |cos θ sin θ|
|−sin θ cos θ| |−sin θ cos θ|
But Aᵀ = |cos θ sin θ| = A⁻¹.
|−sin θ cos θ| A⁻¹ = Aᵀ — this matrix is called an orthogonal matrix (AᵀA = I). This is a direct NDA exam result about rotation matrices.
5. Solving Linear Equations — Cramer's Rule
5.1
Cramer's Rule for 2 & 3 Variable Systems
Replace the coefficient column with the constant column to get D₁, D₂, D₃
⚡ Cramer's Rule — Full Setup
For the system: a₁x + b₁y = c₁
a₂x + b₂y = c₂
D = |a₁ b₁| (coefficient determinant)
|a₂ b₂|
D₁ = |c₁ b₁| (replace x-column with constants)
|c₂ b₂|
D₂ = |a₁ c₁| (replace y-column with constants)
|a₂ c₂|
Solution: x = D₁/D, y = D₂/D (valid only when D ≠ 0)
For 3 variables (a₁x + b₁y + c₁z = d₁ etc.):
D = coefficient det (3×3)
D₁ = replace 1st column with (d₁,d₂,d₃)
D₂ = replace 2nd column with (d₁,d₂,d₃)
D₃ = replace 3rd column with (d₁,d₂,d₃)
x = D₁/D, y = D₂/D, z = D₃/D
Condition summary: D≠0 → unique solution. D=0 and all Dᵢ=0 → infinite solutions (consistent). D=0 but any Dᵢ≠0 → no solution (inconsistent). This three-way condition is directly tested in NDA.
Answer: (a) x=1, y=2, z=3
Verify: 1+2+3=6✓, 1+4+9=14✓, 1+8+27=36✓. By Cramer's Rule, compute D (3×3 Vandermonde-type) ≠ 0, then D₁,D₂,D₃ to get x=1, y=2, z=3.
6. Matrix Method — Inverse Method (AX = B)
6.1
Setting Up and Solving AX = B
Express the system as a matrix equation, then X = A⁻¹B
⚡ Matrix Method — Complete Procedure
System of equations:
a₁x + b₁y + c₁z = d₁
a₂x + b₂y + c₂z = d₂
a₃x + b₃y + c₃z = d₃
Matrix form: AX = B where
A = |a₁ b₁ c₁| X = |x| B = |d₁|
|a₂ b₂ c₂| |y| |d₂|
|a₃ b₃ c₃| |z| |d₃|
Solution:
If |A| ≠ 0: X = A⁻¹B = (adj A / |A|) × B
If |A| = 0: No unique solution exists
For 2×2 system [a₁x+b₁y=c₁, a₂x+b₂y=c₂]:
A⁻¹ = (1/(a₁b₂−a₂b₁)) × |b₂ −b₁|
|−a₂ a₁|
X = A⁻¹B gives x and y directly
The matrix method and Cramer's Rule always give the same answer. Cramer's Rule is faster for 2×2; the matrix method is more systematic for 3×3 when A⁻¹ is already known.
Worked Example — Matrix Method for 2×2 System
Solve: 2x + y = 5, x + 3y = 10 using matrix method.
Q12. For the system AX = B, if |A| = 0, the system has:
(a) Always a unique solution (b) No solution always (c) Either no solution or infinitely many (d) Two solutions
Answer: (c) Either no solution or infinitely many
|A| = 0 means A is singular → A⁻¹ does not exist → the system either has no solution (inconsistent) or infinitely many solutions (consistent but dependent).
Q13. If A⁻¹ = |2 1|, then A is:
|1 1|
(a) |1 −1| (b) |2 −1| (c) |1 1| (d) |−1 2|
|−1 2| |−1 2| |1 2| |1 −2|
Answer: (a) |1 −1| / |−1 2|
If A⁻¹ = |2 1|/|1 1|, then A = (A⁻¹)⁻¹. |A⁻¹| = 2−1 = 1.
A = (1/1)|1 −1|/|−1 2| = |1 −1|/|−1 2|.
🔥 TRICKY QUESTIONS
Linear Equations — Condition & Consistency Traps
🧩 T7. For what value of k does kx + 2y = 3 and 3x + 6y = 9 have infinitely many solutions?
Rewrite as AX = B: A = |k 2|/|3 6|, B = |3|/|9|.
For infinite solutions: D=0 AND D₁=D₂=0.
D = 6k − 6. For D=0: k=1.
Check D₁: |3 2|/|9 6| = 18−18=0 ✓. D₂: |k 3|/|3 9| = 9k−9. At k=1: 0 ✓. k = 1 gives infinitely many solutions (equations are identical). Trap: k=1 makes equations identical (3x+6y=9 is exactly 3× the first). Students often say D=0 is sufficient, but you must verify all Dᵢ = 0 for infinite (vs no) solutions.
🧩 T8. Show that if A is both symmetric and skew-symmetric, then A = O.
Symmetric: Aᵀ = A. Skew-symmetric: Aᵀ = −A.
From both: A = −A → 2A = 0 → A = O (zero matrix). This is a classic statement-verification question. The only matrix that is simultaneously symmetric and skew-symmetric is the zero matrix.
🧩 T9. If A is a 3×3 matrix with |A| = 2, find |A · adj A|.
We know: A · adj A = |A| · I.
So |A · adj A| = ||A| · I| = |A|ⁿ · |I| = 2³ × 1 = 8.
Alternatively: |A · adj A| = |A| · |adj A| = 2 × |A|^(n−1) = 2 × 2² = 2 × 4 = 8. Same answer ✓.
📋 Master Formula Sheet — MN09 Matrices & Determinants
All critical formulae for rapid pre-exam revision.
📋 Matrix Types (Quick ID)
Symmetric: Aᵀ = A (aᵢⱼ = aⱼᵢ)
Skew-symmetric: Aᵀ = −A, diagonal = 0
Idempotent: A² = A
Involutory: A² = I (i.e., A = A⁻¹)
Orthogonal: AᵀA = I (i.e., A⁻¹ = Aᵀ)
⚙ Matrix Operations
(A+B)ᵀ = Aᵀ+Bᵀ
(AB)ᵀ = BᵀAᵀ (reversed!)
(AB)⁻¹ = B⁻¹A⁻¹ (reversed!)
AB ≠ BA in general
A·adj(A) = |A|·I
📐 Determinants
|a b|/|c d| = ad − bc
Row swap → multiply det by −1
|kA| = kⁿ|A| for n×n matrix
|AB| = |A|·|B|
|adj A| = |A|^(n−1)
🔄 Inverse
A⁻¹ = adj(A)/|A| (only if |A|≠0)
2×2 shortcut: swap diag, negate off-diag, ÷|A|
|A⁻¹| = 1/|A|
(A⁻¹)⁻¹ = A
adj(AB) = adj(B)·adj(A) (reversed)
📊 Cramer's Rule
D = coeff. det; D₁,D₂,D₃ = replace column with constants
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