EAPCET Mathematics · Intermediate first year · Paper IA

Matrices

Matrices covers matrix algebra, determinants, adjoint and inverse, rank, and solving systems of linear equations. It is one of the most formula-driven chapters in Paper IA.

Subject
Mathematics
Taught in
Intermediate first year · Paper IA
Mathematics in the paper
80 of 160 questions

Key formulas and results

  • (AB)T=BTAT(AB)^{T} = B^{T}A^{T} and (AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1}
  • A−1=adj⁡(A)∣A∣A^{-1} = \dfrac{\operatorname{adj}(A)}{|A|}, defined only when ∣A∣≠0|A| \ne 0
  • ∣adj⁡A∣=∣A∣n−1|\operatorname{adj} A| = |A|^{n-1} for an n×nn \times n matrix
  • ∣kA∣=kn∣A∣|kA| = k^{n}|A| for an n×nn \times n matrix
  • Every square matrix A=A+AT2+A−AT2A = \dfrac{A + A^{T}}{2} + \dfrac{A - A^{T}}{2}, a symmetric plus a skew-symmetric part
  • AX=BAX = B has a unique solution when ∣A∣≠0|A| \ne 0
  • Rank of an n×nn \times n matrix with ∣A∣≠0|A| \ne 0 is nn; otherwise the rank is the order of the largest square submatrix with non-zero determinant
  • Orthogonal: AAT=IAA^{T} = I; idempotent: A2=AA^{2} = A; involutory: A2=IA^{2} = I (so A−1=AA^{-1} = A); nilpotent: Ak=OA^{k} = O for some kk
  • Symmetric: AT=AA^{T} = A; skew-symmetric: AT=−AA^{T} = -A

What this chapter covers

  • Types of matrices: symmetric, skew-symmetric, orthogonal
  • Transpose and its properties
  • Determinants and their properties
  • Adjoint and inverse of a matrix
  • Rank of a matrix
  • Consistency and solution of linear equations (Cramer’s rule, matrix inversion, Gauss–Jordan)

How it is usually asked

  • Evaluate a determinant using its properties instead of expanding
  • Find |adj A|, |A⁻¹| or |kA| from a given |A|
  • Find the inverse of a 2 × 2 or 3 × 3 matrix
  • Decide whether a system of equations is consistent, or find its solution

Understanding Matrices

A matrix is a rectangular array of numbers, and matrix algebra follows rules that differ from ordinary numbers. Addition is entry by entry, but multiplication combines rows of the first matrix with columns of the second, so AB is defined only when the column count of A equals the row count of B, and in general AB ≠ BA. The transpose swaps rows and columns, and it reverses the order in a product: (AB)ᵀ = BᵀAᵀ.

The determinant is a single number attached to a square matrix. If |A| ≠ 0, A has an inverse given by adj A divided by |A|, where adj A is the transpose of the matrix of cofactors. Results such as |kA| = kⁿ|A|, |adj A| = |A|ⁿ⁻¹ and |A⁻¹| = 1/|A| let you find determinants of related matrices without computing them from scratch.

For a system of linear equations written as AX = B, a non-zero determinant means a unique solution, found by Cramer’s rule or by X = A⁻¹B. When |A| = 0 the system has either no solution or infinitely many, and the rank of the matrices decides which. EAPCET questions mostly test these determinant shortcuts, inverses of small matrices and the consistency of a system.

Worked example

EAPCET-style question

AA is a 3×33 \times 3 matrix with ∣A∣=4|A| = 4. The value of ∣adj⁡(2A)∣|\operatorname{adj}(2A)| is

  1. (1)64
  2. (2)128
  3. (3)256
  4. (4)1024

Solution

  1. Let B=2AB = 2A. Since AA is 3×33 \times 3, ∣B∣=23∣A∣=8×4=32|B| = 2^{3}|A| = 8 \times 4 = 32.
  2. For any 3×33 \times 3 matrix BB, ∣adj⁡B∣=∣B∣3−1=∣B∣2|\operatorname{adj} B| = |B|^{3-1} = |B|^{2}.
  3. So ∣adj⁡(2A)∣=322=1024|\operatorname{adj}(2A)| = 32^{2} = 1024.

Answer: 1024

Shortcuts and results to remember

  • For AX=BAX = B: ∣A∣≠0|A| \ne 0 gives a unique solution; ∣A∣=0|A| = 0 gives either infinitely many solutions or none, so check consistency (compare ranks).
  • If every element is zero the rank is 0; if the matrix is non-zero but every 2×22 \times 2 minor is zero, the rank is 1.

More practice questions

Practice question 1

The rank of the matrix [123246369]\begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{bmatrix} is

  1. (1)1
  2. (2)2
  3. (3)3
  4. (4)0

Solution

  1. Rows 2 and 3 are 2 and 3 times row 1, so the determinant is 0 and the rank is less than 3
  2. Every 2×22 \times 2 minor is 0 (e.g. 1⋅4−2⋅2=01 \cdot 4 - 2 \cdot 2 = 0), so the rank is less than 2
  3. The matrix is non-zero, so the rank is 1

Answer: 1

Common mistakes to avoid

  • Writing |kA| = k|A|; for an n × n matrix every row is multiplied by k, so |kA| = kⁿ|A|.
  • Writing (AB)⁻¹ = A⁻¹B⁻¹; the order reverses, so (AB)⁻¹ = B⁻¹A⁻¹, and likewise (AB)ᵀ = BᵀAᵀ.
  • Using the matrix of cofactors itself as adj A; the adjoint is the transpose of the cofactor matrix.
  • Forgetting the sign pattern (−1)ⁱ⁺ʲ when computing cofactors, which flips the sign of half the entries.

How to prepare it

Results like |adj A| = |A|ⁿ⁻¹ turn many questions into one-line calculations — learn them before practising long expansions.

Before expanding a determinant, look for row or column operations that create zeros.

Once the formulas feel familiar, test them the way the exam does: practise Matrices questions chapter-wise, then attempt a full previous year paper to see whether it holds up under a 3-hour timer.

Related Mathematics chapters

See every chapter in the EAPCET Mathematics syllabus.

Practise Matrices chapter-wise

Chapter-wise quizzes show exactly which topics in this chapter you still get wrong — before a full paper does.

Frequently asked questions

Is Matrices part of the EAPCET Mathematics syllabus?
Yes. Matrices is taught in Intermediate first year · Paper IA Mathematics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.
How should I prepare Matrices for EAPCET?
Results like |adj A| = |A|ⁿ⁻¹ turn many questions into one-line calculations — learn them before practising long expansions. Before expanding a determinant, look for row or column operations that create zeros.
How many Mathematics questions are there in EAPCET?
80 of the 160 questions in the Engineering paper are Mathematics, spread across 37 chapters.
What mistakes do students make in Matrices?
Writing |kA| = k|A|; for an n × n matrix every row is multiplied by k, so |kA| = kⁿ|A|. Writing (AB)⁻¹ = A⁻¹B⁻¹; the order reverses, so (AB)⁻¹ = B⁻¹A⁻¹, and likewise (AB)ᵀ = BᵀAᵀ. Using the matrix of cofactors itself as adj A; the adjoint is the transpose of the cofactor matrix. Forgetting the sign pattern (−1)ⁱ⁺ʲ when computing cofactors, which flips the sign of half the entries.