Matrices and determinants for NDA

Matrices and determinants is a short, self-contained unit, and one of the most scoring in the NDA paper. Most questions are designed so that a property gives the answer in two lines while brute-force expansion takes two minutes. The skill to build here is noticing the property first: identical rows, a triangular matrix, a scalar pulled out, a known value of det A.

A note on notation, since this page has no matrix typesetting: [[1, 2], [3, 4]] means the matrix with rows (1, 2) and (3, 4). The determinant of that matrix is written |1 2; 3 4|, with rows separated by semicolons.

Typical question types are:

• products and powers of small matrices;
• identifying symmetric, skew-symmetric, singular or triangular matrices;
• evaluating a determinant, often by a property rather than expansion;
• det(kA), det(adj A) and det(A⁻¹) from det A;
• the inverse of a 2 × 2 matrix, or of A from a matrix equation;
• solving or testing the consistency of a system of linear equations.

Types of matrices

Every square matrix is the sum of a symmetric and a skew-symmetric matrix: A = ½(A + Aᵀ) + ½(A − Aᵀ).

Operations

• Addition needs matrices of the same order.
• Multiplication AB needs the number of columns of A to equal the number of rows of B. An m × n matrix times an n × p matrix gives an m × p matrix. Each entry is a row of A multiplied term by term with a column of B, then added.
• In general AB ≠ BA, and AB = O does not mean A = O or B = O.
• (AB)ᵀ = BᵀAᵀ and (AB)⁻¹ = B⁻¹A⁻¹: the order reverses.

Determinants

For a 2 × 2 matrix, |a b; c d| = ad − bc.

For a 3 × 3 matrix, expand along any row or column with the sign pattern + − + (then − + −, then + − +). Expanding |a₁ b₁ c₁; a₂ b₂ c₂; a₃ b₃ c₃| along the first row gives a₁(b₂c₃ − b₃c₂) − b₁(a₂c₃ − a₃c₂) + c₁(a₂b₃ − a₃b₂). Choose the row or column with the most zeros.

Properties that save time

Adjoint and inverse

The adjoint of A is the transpose of its matrix of cofactors. For A = [[a, b], [c, d]], adj A = [[d, −b], [−c, a]]: swap the diagonal, change the sign of the other two.

The key identity is A (adj A) = (adj A) A = (det A) I. From it:

• A⁻¹ = adj A ÷ det A, which exists only when det A ≠ 0;
• for an n × n matrix, det(adj A) = (det A)ⁿ⁻¹;
• det(A⁻¹) = 1 ÷ det A.

Solving linear equations

Write the system as AX = B.

Matrix method: if det A ≠ 0, the unique solution is X = A⁻¹B.

Cramer's rule: with D = det A, and Dx, Dy, Dz formed by replacing the x, y or z column of A with the constants, x = Dx/D, y = Dy/D, z = Dz/D.

Consistency:

• D ≠ 0: a unique solution.
• D = 0 and at least one of Dx, Dy, Dz is not 0: no solution.
• D = 0 and all of them are 0: infinitely many solutions or none; check the equations further.
• A homogeneous system (all constants 0) always has the solution x = y = z = 0. It has other solutions only if D = 0.

Area of a triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃) is half the absolute value of |x₁ y₁ 1; x₂ y₂ 1; x₃ y₃ 1|. The points are collinear when this determinant is 0.

Worked NDA-style MCQs

Q1. If A = [[2, 3], [1, 2]], then A² − 4A + I equals:
(a) I (b) O (c) A (d) 2I

A² = [[4 + 3, 6 + 6], [2 + 2, 3 + 4]] = [[7, 12], [4, 7]]. 4A = [[8, 12], [4, 8]]. A² − 4A = [[−1, 0], [0, −1]] = −I, so A² − 4A + I = O. Answer: (b).

A bonus: A(4I − A) = I, so A⁻¹ = 4I − A = [[2, −3], [−1, 2]].

Q2. The value of |1 a b+c; 1 b c+a; 1 c a+b| is:
(a) a + b + c (b) abc (c) 0 (d) 1

Add the second column to the third. Every entry of the third column becomes a + b + c, which makes it (a + b + c) times the first column. Proportional columns give 0. Answer: (c).

Q3. A is a 3 × 3 matrix with det A = 5. Then det(adj A) and det(2A) are:
(a) 25 and 40 (b) 25 and 10 (c) 125 and 40 (d) 5 and 10

det(adj A) = 5³⁻¹ = 25. det(2A) = 2³ × 5 = 40. Answer: (a).

Q4. By Cramer's rule, the solution of 2x + 3y = 8 and x − y = −1 is:
(a) x = 2, y = 1 (b) x = 1, y = 2 (c) x = −1, y = 2 (d) x = 1, y = −2

D = |2 3; 1 −1| = −2 − 3 = −5. Dx = |8 3; −1 −1| = −8 + 3 = −5. Dy = |2 8; 1 −1| = −2 − 8 = −10. So x = 1 and y = 2. Check: 2 + 6 = 8 and 1 − 2 = −1. Answer: (b).

Q5. For what value of λ does the system x + y + z = 0, x + 2y + 3z = 0, x + 3y + λz = 0 have a non-zero solution?
(a) 3 (b) 4 (c) 5 (d) 6

The system is homogeneous, so we need D = 0. D = 1(2λ − 9) − 1(λ − 3) + 1(3 − 2) = λ − 5. So λ = 5. Answer: (c).

Q6. The area of the triangle with vertices (1, 2), (4, 6) and (7, 2) is:
(a) 6 (b) 12 (c) 24 (d) 18

Half of |1(6 − 2) + 4(2 − 2) + 7(2 − 6)| = half of |4 + 0 − 28| = 12. Check: the base from (1, 2) to (7, 2) is 6 and the height is 4, so the area is 12. Answer: (b).

Common mistakes

• Assuming AB = BA. Always keep the order in products and in (AB)⁻¹ = B⁻¹A⁻¹.
• det(kA) = k det A. For an n × n matrix it is kⁿ det A.
• Forgetting the sign pattern when expanding along the second row or column.
• Inverting a singular matrix. Check det A ≠ 0 first.
• Declaring "infinitely many solutions" too soon when D = 0. Check the other determinants and the equations.

Practice set

1. If A is 3 × 4 and B is 4 × 2, the order of AB is: (a) 3 × 2 (b) 4 × 4 (c) 2 × 3 (d) not defined
2. |2 −3; 4 5| equals: (a) −2 (b) 22 (c) 2 (d) −22
3. The determinant of [[3, 0, 0], [5, −2, 0], [7, 1, 4]] is: (a) 24 (b) −24 (c) 5 (d) 0
4. If A = [[1, 2], [2, 1]], then A² − 2A equals: (a) I (b) 3I (c) O (d) 2A
5. The inverse of [[3, 5], [1, 2]] is: (a) [[2, −5], [−1, 3]] (b) [[3, −5], [−1, 2]] (c) [[2, 5], [1, 3]] (d) [[−2, 5], [1, −3]]
6. [[k, 4], [1, k]] is singular when k is: (a) ±4 (b) ±2 (c) 0 (d) 4 only
7. If A is 3 × 3 with det A = 4, then det(3A) is: (a) 12 (b) 36 (c) 108 (d) 64
8. The determinant of any 3 × 3 skew-symmetric matrix is: (a) 1 (b) −1 (c) 0 (d) the product of the diagonal

Answers:

1. (a) 3 × 2. The inner orders (4 and 4) match.
2. (b) 22. 2 × 5 − (−3)(4) = 10 + 12.
3. (b) −24. Lower triangular: 3 × (−2) × 4.
4. (b) 3I. A² = [[5, 4], [4, 5]], and subtracting 2A = [[2, 4], [4, 2]] leaves [[3, 0], [0, 3]].
5. (a). det = 6 − 5 = 1, so the inverse is the adjoint.
6. (b) ±2. k² − 4 = 0.
7. (c) 108. 3³ × 4.
8. (c) 0. Skew-symmetric of odd order.

What to do next

• Write the properties table from memory, and for each property one example that it solves.
• Solve 30 questions from old NDA papers on this unit, timed at 75 seconds each.
• See where determinants reappear in straight lines (area and collinearity) and vectors (the scalar triple product).