Matrices Class 12 NCERT Chapter 3: Official PDF and Guide

Studying matrices class 12 from the Mathematics Part I textbook? This page holds the official Chapter 3 Matrices PDF, together with a plain-language guide to everything the chapter teaches — matrix order and notation, the types of matrices, addition and multiplication, the transpose, symmetric and skew symmetric matrices, and invertible matrices.

The official PDF is right below; the guide that follows walks through each section in the order the book itself presents them.

Download the official NCERT Class 12 Maths Chapter 3 Matrices PDF

Download the NCERT Class 12 Maths Chapter 3 Matrices PDF from the official NCERT website to read the chapter exactly as published, with every worked example and exercise printed as in the book itself. This is the file this page describes: the PDF carries the full text and exercises, and the sections below carry the thinking behind them.


What the chapter holds Count Where it is used
Printed pages 42
Sections in the chapter 14
Tables 8
Worked examples 2 solved step by step in our NCERT Solutions
Exercise questions 44 answered in our NCERT Solutions
Official NCERT PDF Download the chapter PDF the chapter exactly as NCERT publishes it



Chapter 3 Matrices Class 12 at a Glance

Matrices is the chapter the rest of Class 12 maths builds on. It defines what a matrix is, how matrices are added and multiplied, and which special matrices matter — and Chapter 4, Determinants, assumes every part of it.

The table below shows how much the chapter holds — its sections, worked examples and exercise questions — so you know what is in the file before you open it.

The chapter’s arc is one straight line. Section 3.2 defines a matrix and its order. Section 3.3 lists the seven types of matrices and what equality means. Section 3.4 builds addition, scalar multiplication and matrix multiplication with their properties. Section 3.5 defines the transpose. Section 3.6 uses it for symmetric and skew symmetric matrices. Section 3.7 closes with invertible matrices.

Miscellaneous Examples and the closing Summary round it off.

What Chapter 3 Matrices Covers, Section by Section

Use this map to jump straight to the part you are stuck on. Each row names the section as the book numbers it and what that section teaches.

Section What it teaches
3.1 Introduction Why matrices exist: compact, simple methods for systems of linear equations, spreadsheets, rotations and reflections, cryptography.
3.2 Matrix and 3.2.1 Order of a matrix Definition 1: an ordered rectangular array of numbers or functions; order \(m \times n\), \(mn\) elements, notation \(A = [a_{ij}]_{m \times n}\).
3.3 Types of Matrices The seven types — column, row, square, diagonal, scalar, identity, zero — then equality of matrices.
3.4 Operations on Matrices Addition, scalar multiplication, difference, then matrix multiplication with its properties.
3.5 Transpose of a Matrix Definition 3, and the four properties including \((AB)’ = B’A’\).
3.6 Symmetric and Skew Symmetric Matrices The two special square matrices defined through \(A’\), and the splitting theorem.
3.7 Invertible Matrices Inverse via \(AB = BA = I\), uniqueness, and \((AB)^{-1} = B^{-1}A^{-1}\).
Miscellaneous Examples and Exercise; Summary Mixed problems and proofs, then the chapter’s own closing summary.

Matrix, Order and Notation: The Starting Definitions

Think of a matrix as a rectangular box of numbers. The whole chapter is about what you are allowed to do with that box — and the first step is learning to read it.

A matrix is an ordered rectangular array of numbers or functions, called its elements or entries (NCERT Section 3.2). Matrices are named with capital letters.

The position of each entry matters: in \(a_{ij}\), the first subscript \(i\) names the row and the second subscript \(j\) names the column, so \(a_{23}\) is the entry in the second row and third column (NCERT Section 3.2.1).

If the matrix has \(m\) rows and \(n\) columns, its order is \(m \times n\) and it holds \(m \times n\) entries. This chapter works only with matrices whose elements are real numbers or functions taking real values.

Build one to see the notation work. Construct a \(2 \times 3\) matrix with \(a_{ij} = i + 2j\):

  1. Step 1: Write the positions you need: \(i = 1, 2\) and \(j = 1, 2, 3\).
  2. Step 2: Substitute each pair into \(a_{ij} = i + 2j\).

\[ a_{11} = 1 + 2(1) = 3, \quad a_{12} = 1 + 2(2) = 5, \quad a_{13} = 1 + 2(3) = 7 \]

\[ a_{21} = 2 + 2(1) = 4, \quad a_{22} = 2 + 2(2) = 6, \quad a_{23} = 2 + 2(3) = 8 \]

Result: \(A = \begin{bmatrix} 3 & 5 & 7 \\ 4 & 6 & 8 \end{bmatrix}\).

Why learn this vocabulary now? Because every later section — addition, multiplication, transpose, inverse — is written using \(a_{ij}\), order and \([a_{ij}]_{m \times n}\). Many Exercise 3.1 questions ask you to read entries directly from this notation (NCERT Section 3.2.1).

Matrix Addition and Scalar Multiplication: The Same Order Rule

Addition and scalar multiplication are entry-by-entry operations. The only hard rule is that addition requires both matrices to have the same shape.

If \(A = [a_{ij}]\) and \(B = [b_{ij}]\) are both of order \(m \times n\), then \(A + B\) is the matrix \([a_{ij} + b_{ij}]\) (NCERT Section 3.4.1). The book motivates this with two factories producing shoes in matching price categories: each total is a corresponding entry added — \(80 + 90\), \(75 + 70\), and so on.

The matching rows and columns are what make the addition sensible.

The book is explicit about the trap: if \(A\) and \(B\) are not of the same order, \(A + B\) is not defined (NCERT Section 3.3.1 note). Check order first, always.

Scalar multiplication is simpler. For a scalar \(k\), the matrix \(kA\) multiplies every entry of \(A\) by \(k\): \(kA = k[a_{ij}]_{m \times n} = [ka_{ij}]\) (NCERT Section 3.4.2). The negative of a matrix is \(-A = (-1)A\), and the difference is \(A – B = A + (-1)B\).

Addition obeys four laws (NCERT Section 3.4.3): it is commutative (\(A + B = B + A\)), associative (\((A + B) + C = A + (B + C)\)), the zero matrix \(O\) is the additive identity (\(A + O = A\)), and \(-A\) is the additive inverse (\(A + (-A) = O\)). Scalar multiplication distributes: \(k(A + B) = kA + kB\) and \((k + l)A = kA + lA\) (NCERT Section 3.4.4).

Use these laws to solve for an unknown matrix. Take

Given: \(2X + A = 3B\) with \(A = \begin{bmatrix} 2 & 6 \\ -4 & 8 \end{bmatrix}\), \(B = \begin{bmatrix} 4 & 0 \\ 2 & 2 \end{bmatrix}\).

Find \(X\).

Step 1: Add \(-A\) (the additive inverse) to both sides, then use the additive identity \(O\) to simplify.

\[ 2X = 3B – A = 3\begin{bmatrix} 4 & 0 \\ 2 & 2 \end{bmatrix} – \begin{bmatrix} 2 & 6 \\ -4 & 8 \end{bmatrix} = \begin{bmatrix} 12 & 0 \\ 6 & 6 \end{bmatrix} + \begin{bmatrix} -2 & -6 \\ 4 & -8 \end{bmatrix} = \begin{bmatrix} 10 & -6 \\ 10 & -2 \end{bmatrix} \]

Step 2: Multiply by \(\frac{1}{2}\), the scalar that undoes the \(2\).

\[ X = \frac{1}{2}\begin{bmatrix} 10 & -6 \\ 10 & -2 \end{bmatrix} = \begin{bmatrix} 5 & -3 \\ 5 & -1 \end{bmatrix} \]

Step 3 (check): \(2X + A = \begin{bmatrix} 10 & -6 \\ 10 & -2 \end{bmatrix} + \begin{bmatrix} 2 & 6 \\ -4 & 8 \end{bmatrix} = \begin{bmatrix} 12 & 0 \\ 6 & 6 \end{bmatrix} = 3B\).

The solution is correct.

Matrix Multiplication: The Compatibility Rule and a Worked Example

Unlike addition, multiplication does not pair up matching positions. Each entry of the product is a whole row of the first matrix multiplied by a whole column of the second.

The rule: the product \(AB\) is defined when the number of columns of \(A\) equals the number of rows of \(B\).

If \(A\) is \(m \times n\) and \(B\) is \(n \times p\), then \(AB\) is \(m \times p\), and (NCERT Section 3.4.5) \[ c_{ik} = a_{i1}b_{1k} + a_{i2}b_{2k} + \cdots + a_{in}b_{nk} = \sum_{j=1}^{n} a_{ij}b_{jk} \]

The book’s shopping example shows the point: pen and story-book requirements times the price column gives the money each person needs. Row of requirements, column of prices, sum of products — that is the entire operation.

Work one product with fresh numbers. Take \(A = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}\) (order \(2 \times 2\)) and \(B = \begin{bmatrix} 5 & 2 \\ 1 & 6 \end{bmatrix}\) (order \(2 \times 2\)). Since columns of \(A\) = rows of \(B\), the product exists and is \(2 \times 2\).

Entry (1,1): first row of \(A\) times first column of \(B\).

\[ c_{11} = 1(5) + 3(1) = 8 \]

Entry (1,2): first row of \(A\) times second column of \(B\).

\[ c_{12} = 1(2) + 3(6) = 20 \]

Entry (2,1): second row of \(A\) times first column of \(B\).

\[ c_{21} = 2(5) + 4(1) = 14 \]

Entry (2,2): second row of \(A\) times second column of \(B\).

\[ c_{22} = 2(2) + 4(6) = 28 \]

Result: \(AB = \begin{bmatrix} 8 & 20 \\ 14 & 28 \end{bmatrix}\).

Note that \(AB\) being defined never guarantees \(BA\) is defined — the compatibility check runs both ways. Multiplication is associative, distributive over addition, and the identity matrix \(I\) of matching order acts as the multiplicative identity (NCERT Section 3.4.6).

When Matrix Rules Break: Non Commutativity and Zero Products

For ordinary numbers, \(ab = 0\) forces \(a = 0\) or \(b = 0\), and order of multiplication never matters. Both intuitions fail for matrices — this section is where most conceptual errors start.

First, \(AB\) is generally not equal to \(BA\), even when both products are defined and have the same order. The book shows this with worked examples and adds the same warning (NCERT Section 3.4.5). With the pair above, both products exist and both are \(2 \times 2\):

Compute \(BA\) with the same matrices: \(B = \begin{bmatrix} 5 & 2 \\ 1 & 6 \end{bmatrix}\), \(A = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}\).

\[ BA = \begin{bmatrix} 5 \times 1 + 2 \times 2 & 5 \times 3 + 2 \times 4 \\ 1 \times 1 + 6 \times 2 & 1 \times 3 + 6 \times 4 \end{bmatrix} = \begin{bmatrix} 9 & 23 \\ 13 & 27 \end{bmatrix} \]

Compare: \(AB = \begin{bmatrix} 8 & 20 \\ 14 & 28 \end{bmatrix} \neq \begin{bmatrix} 9 & 23 \\ 13 & 27 \end{bmatrix} = BA\). The order of multiplication changes the answer.

One honest exception: diagonal matrices of the same order do commute (NCERT Section 3.4.5). But that is a special case, not the rule.

Second, a zero product does not mean a zero factor. Invent a fresh pair of non-zero matrices:

Take \(A = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}\) and \(B = \begin{bmatrix} 2 & -2 \\ -1 & 1 \end{bmatrix}\), both non-zero.

\[ AB = \begin{bmatrix} 1(2) + 2(-1) & 1(-2) + 2(1) \\ 2(2) + 4(-1) & 2(-2) + 4(1) \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \]

Conclusion: two non-zero matrices can multiply to the zero matrix. A zero product tells you nothing about the factors.

Transpose of a Matrix: Four Properties and the Reversal Rule

The transpose flips a matrix across its main diagonal, turning rows into columns. Nearly every later definition in this chapter — symmetric, skew symmetric, inverse checks — is written in terms of \(A’\).

If \(A = [a_{ij}]\) is \(m \times n\), its transpose \(A’\) (also written \(A^T\)) is \([a_{ji}]\) of order \(n \times m\) (NCERT Section 3.5). Example: \(A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{bmatrix}\) gives \(A’ = \begin{bmatrix} 1 & 3 & 5 \\ 2 & 4 & 6 \end{bmatrix}\).

The four properties of the transpose (NCERT Section 3.5.1):

  • \((A’)’ = A\) — transposing twice returns the original matrix.
  • \((kA)’ = kA’\) — a scalar passes through.
  • \((A + B)’ = A’ + B’\) — a sum transposes entry-wise.
  • \((AB)’ = B’A’\) — and this one is the easy one to misremember: the order reverses.

Why does the order reverse? The entry \(c_{ik}\) of \(AB\) comes from row \(i\) of \(A\) and column \(k\) of \(B\). After transposing, that entry sits in row \(k\), column \(i\) — which is exactly what row \(k\) of \(B’\) times column \(i\) of \(A’\) produces. Rows and columns swapped, so the factors swap.

Verify it on the pair from the multiplication section. We already have \(AB = \begin{bmatrix} 8 & 20 \\ 14 & 28 \end{bmatrix}\), so \((AB)’ = \begin{bmatrix} 8 & 14 \\ 20 & 28 \end{bmatrix}\). On the other side, \(B’ = \begin{bmatrix} 5 & 1 \\ 2 & 6 \end{bmatrix}\) and \(A’ = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\), giving \(B’A’ = \begin{bmatrix} 5(1) + 1(3) & 5(2) + 1(4) \\ 2(1) + 6(3) & 2(2) + 6(4) \end{bmatrix} = \begin{bmatrix} 8 & 14 \\ 20 & 28 \end{bmatrix}\). Both sides match.

Symmetric and Skew Symmetric Matrices: The Splitting Trick

These are the two special square matrices the chapter defines through the transpose. A square matrix is symmetric if \(A’ = A\), meaning \(a_{ij} = a_{ji}\) for every pair — the matrix reads the same above and below the diagonal. It is skew symmetric if \(A’ = -A\), meaning \(a_{ji} = -a_{ij}\) (NCERT Section 3.6).

The skew condition forces every diagonal entry to be zero: putting \(i = j\) gives \(a_{ii} = -a_{ii}\), so \(2a_{ii} = 0\) and \(a_{ii} = 0\). A skew symmetric matrix always has an all-zero diagonal.

Two theorems carry this section (NCERT Section 3.6). Theorem 1: for any square matrix \(A\), the matrix \(A + A’\) is symmetric and \(A – A’\) is skew symmetric. Theorem 2: any square matrix can be written as the sum of a symmetric and a skew symmetric matrix:

\[ A = \frac{1}{2}(A + A’) + \frac{1}{2}(A – A’) \]

Why does the \(\frac{1}{2}\) appear? Because \(A + A’\) doubles the symmetric part while cancelling the skew part, and \(A – A’\) does the opposite — halving each puts the original \(A\) back together.

Split \(A = \begin{bmatrix} 4 & 7 \\ 2 & 6 \end{bmatrix}\). First find \(A’ = \begin{bmatrix} 4 & 2 \\ 7 & 6 \end{bmatrix}\).

Step 1: Form the symmetric part \(P = \frac{1}{2}(A + A’)\).

\[ P = \frac{1}{2}\begin{bmatrix} 8 & 9 \\ 9 & 12 \end{bmatrix} = \begin{bmatrix} 4 & \frac{9}{2} \\ \frac{9}{2} & 6 \end{bmatrix} \]

Step 2: Verify \(P’ = P\).

The matrix reads the same mirrored, so it is symmetric.

Step 3: Form the skew part \(Q = \frac{1}{2}(A – A’)\).

\[ Q = \frac{1}{2}\begin{bmatrix} 0 & 5 \\ -5 & 0 \end{bmatrix} = \begin{bmatrix} 0 & \frac{5}{2} \\ -\frac{5}{2} & 0 \end{bmatrix} \]

Step 4: Check \(Q’ = -Q\), and note the diagonal is all zeros — the skew signature.

Rebuild: \(P + Q = \begin{bmatrix} 4 & 7 \\ 2 & 6 \end{bmatrix} = A\). The split works.

Invertible Matrices: What Inverse Means and When It Exists

The inverse is the matrix analogue of a reciprocal. The definition demands that the product work in both directions.

A square matrix \(A\) is invertible if there exists another square matrix \(B\) of the same order such that \(AB = BA = I\). That \(B\) is called the inverse of \(A\) and is written \(A^{-1}\) (NCERT Section 3.7).

The book’s note is blunt: a rectangular matrix has no inverse, because both products \(AB\) and \(BA\) must be defined and equal, which forces both matrices to be square of the same order (NCERT Section 3.7 note).

Verify an inverse pair with fresh numbers.

Let \(A = \begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix}\) and \(B = \begin{bmatrix} 2 & -1 \\ -5 & 3 \end{bmatrix}\).

\[ AB = \begin{bmatrix} 3(2) + 1(-5) & 3(-1) + 1(3) \\ 5(2) + 2(-5) & 5(-1) + 2(3) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]

\[ BA = \begin{bmatrix} 2(3) + (-1)(5) & 2(1) + (-1)(2) \\ -5(3) + 3(5) & -5(1) + 3(2) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \]

Conclusion: \(AB = BA = I\), so \(B = A^{-1}\). Both directions had to check out.

The inverse is unique: if both \(B\) and \(C\) are inverses of \(A\), then \(B = BI = B(AC) = (BA)C = IC = C\) (NCERT Section 3.7, Theorem 3). And the product rule reverses the order: \((AB)^{-1} = B^{-1}A^{-1}\) — to undo \(AB\), you first undo \(B\), then undo \(A\) (Theorem 4).

Remember: this chapter only defines the inverse; computing it by a formula comes in the next chapter.

Reading Matrices: The Arrays and Tables That Carry This Chapter

This chapter has no diagrams. Its ideas live in matrix arrays, data tables and worked layouts — so knowing how to read those arrangements is the skill that unlocks it.

The general array shows every position at once (NCERT Section 3.2.1):
\[ A = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix}_{m \times n} \]
The first subscript is the row, the second is the column, and the total count is \(m \times n\).

Matrices also store geometry. A point \((x, y)\) in the plane can be written as the column matrix \(\begin{bmatrix} x \\ y \end{bmatrix}\) or the row matrix \([x\ \ y]\).

The vertices of a closed figure follow the same idea: a quadrilateral with vertices \(P(2, 1), Q(4, 3), R(1, 4), S(-2, 2)\) becomes \[ \begin{bmatrix} 2 & 4 & 4 & -2 \\ 1 & 3 & 4 & 2 \end{bmatrix}_{2 \times 3 + 1} \quad \text{or} \quad \begin{bmatrix} 2 & 1 \\ 4 & 3 \\ 1 & 4 \\ -2 & 2 \end{bmatrix}_{4 \times 2} \]

The first arrangement stores one point per column; the second stores one point per row. Both carry the same figure.

The chapter opens exactly this way. The possession counts of notebooks and pens for three people sit in a table whose row and column labels — who owns, what item — tell you what each entry means. Example tables for factory production and farmers’ sales work the same way: rows and columns are labelled by category, and each entry is one quantity.

When you read any data table in the chapter, ask two questions first: what do the rows store, and what do the columns store?

Every Definition in Chapter 3, One Table

Keep this table open while doing the exercises — it is the whole chapter’s terminology in one place, with fresh examples throughout.

Term Condition that defines it Quick example
Matrix Ordered rectangular array of numbers or functions; entries are called elements (NCERT Section 3.2, Definition 1) \(\begin{bmatrix} 1 & 0 \\ 2 & 5 \\ 3 & -4 \end{bmatrix}\)
Order \(m \times n\) \(m\) rows, \(n\) columns; holds \(mn\) elements \([a_{ij}]_{2 \times 3}\) holds 6 elements
Column matrix Only one column: order \(m \times 1\) \(\begin{bmatrix} 7 \\ -2 \\ 8 \end{bmatrix}\)
Row matrix Only one row: order \(1 \times n\) \(\begin{bmatrix} 4 & 6 & 9 \end{bmatrix}\)
Square matrix Rows equal columns: \(m = n\) \(\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\)
Diagonal matrix All non-diagonal elements zero \(\begin{bmatrix} 2 & 0 \\ 0 & 5 \end{bmatrix}\)
Scalar matrix Diagonal matrix with all diagonal elements equal to one constant \(k\) \(\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}\)
Identity matrix Diagonal entries all 1, rest all 0; written \(I_n\) \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)
Zero matrix Every element is 0; written \(O\) \(\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)
Equality of matrices Same order and \(a_{ij} = b_{ij}\) for all \(i, j\) (NCERT Section 3.3.1, Definition 2) \(\begin{bmatrix} 2 & 3 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 3 \\ 0 & 1 \end{bmatrix}\)
Negative of a matrix \(-A = (-1)A\); every entry changes sign \(-\begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} -2 & 1 \\ -3 & -4 \end{bmatrix}\)
Transpose \(A’\) Rows and columns interchanged; \([a_{ji}]_{n \times m}\) (NCERT Section 3.5) \(A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{bmatrix}\) gives \(A’ = \begin{bmatrix} 1 & 3 & 5 \\ 2 & 4 & 6 \end{bmatrix}\)
Symmetric matrix \(A’ = A\), so \(a_{ij} = a_{ji}\) (NCERT Section 3.6, Definition 4) \(\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}\)
Skew symmetric matrix \(A’ = -A\), so \(a_{ji} = -a_{ij}\); all diagonal entries zero (NCERT Section 3.6, Definition 5) \(\begin{bmatrix} 0 & 4 \\ -4 & 0 \end{bmatrix}\)
Invertible matrix Square \(A\) with some \(B\) such that \(AB = BA = I\); \(B = A^{-1}\), unique (NCERT Section 3.7) \(A = \begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix}\), \(A^{-1} = \begin{bmatrix} 3 & -1 \\ -5 & 2 \end{bmatrix}\)

The Mistakes Students Make With Matrices

Every mistake below is a place where matrix rules differ from number rules. Read each one before you meet it in the exercises.

Mistake Correct rule How to check your answer
Adding matrices of different orders \(A + B\) is defined only when both have the same order (NCERT Section 3.4.1) Write both orders first; if they differ, the sum does not exist
Assuming \(AB = BA\) Multiplication is not commutative even when both products are defined and same-sized (NCERT Section 3.4.5) Compute both products once; diagonal matrices of same order are the exception
Assuming \(AB = O\) forces \(A = O\) or \(B = O\) Two non-zero matrices can give a zero product (NCERT Section 3.4.5) Check each factor entry-by-entry; a zero product identifies no zero factor
Writing \((AB)’ = A’B’\) The transpose reverses order: \((AB)’ = B’A’\) (NCERT Section 3.5.1) Test on a \(2 \times 2\) pair; the reversal is exactly why the order swaps
Forgetting skew symmetric diagonals are zero \(a_{ii} = -a_{ii}\) forces \(2a_{ii} = 0\), so every diagonal entry is 0 (NCERT Section 3.6) Look at the diagonal of your skew matrix — it must be all zeros
Hunting an inverse for a rectangular matrix Inverses exist only for square matrices, and are unique when they exist; both products must equal \(I\) (NCERT Section 3.7) Verify \(AB = BA = I\) before you call \(B\) the inverse

Which Exercises Test Which Skill

The exercises are arranged in the same order as the sections, so a missed question tells you exactly which section to reread.

Exercise Skills tested Where
Exercise 3.1 Order of a matrix, counting elements, reading entries like \(a_{13}\), constructing matrices from a formula such as \(a_{ij} = \frac{(i+j)^2}{2}\), equality of matrices by solving for unknowns, plus MCQs on square matrices and possible orders End of Section 3.3
Exercise 3.2 Every operation at once: addition, subtraction, scalar multiples, matrix products, equations with unknown matrices like \(2X + Y = \dots\), verifying \(AB \neq BA\), evaluating expressions like \(A^2 – 5A + 6I\), and word problems (trust fund, bookshop, production) End of Section 3.4
Exercise 3.3 Transposes, verifying the four transpose properties, symmetric and skew symmetric checks, and splitting matrices into symmetric plus skew symmetric parts End of Section 3.6
Exercise 3.4 The definition of inverse: the condition \(AB = BA = I\) that must hold for one matrix to be the inverse of another End of Section 3.7
Miscellaneous Exercise on Chapter 3 Proofs about symmetric matrices, \(A’A = I\) verifications, and applied sales and revenue problems using matrix multiplication After the Miscellaneous Examples

What a full-marks answer shows: state the order check before any operation; when equating matrices, write the corresponding-element equations explicitly; for skew symmetric answers, point out the zero diagonal; and only call a matrix an inverse after showing both \(AB\) and \(BA\).

One honest note: textbook contents and the examinable syllabus are not always identical — check the current official CBSE syllabus before deciding what to skim.

The Chapter in One Chain: Last Minute Review

This is the chapter’s own closing summary, rebuilt as one connected chain (NCERT Summary). Follow it in order and you have covered the chapter.

  1. A matrix is an ordered rectangular array of numbers or functions.
  2. \(m\) rows and \(n\) columns make order \(m \times n\), holding \(mn\) elements.
  3. One column → column matrix; one row → row matrix; equal rows and columns → square matrix.
  4. Diagonal: all non-diagonal entries zero. Scalar: diagonal entries all equal \(k\). Identity: diagonal ones. Zero: everything zero.
  5. Equality needs the same order and \(a_{ij} = b_{ij}\) for all \(i, j\).
  6. \(kA\) multiplies every entry by \(k\); \(-A = (-1)A\); \(A – B = A + (-1)B\).
  7. \(A + B\) needs the same order, adds entry-wise, and is commutative, associative, with identity \(O\) and inverse \(-A\). Also \(k(A + B) = kA + kB\) and \((k + l)A = kA + lA\).
  8. \(AB\) is defined when columns of \(A\) = rows of \(B\); then \(c_{ik} = \sum a_{ij}b_{jk}\), the product is associative and distributive, but not commutative, and a zero product does not mean a zero factor.
  9. Transpose \(A’\) swaps rows and columns: \((A’)’ = A\), \((kA)’ = kA’\), \((A + B)’ = A’ + B’\), \((AB)’ = B’A’\).
  10. Symmetric means \(A’ = A\); skew symmetric means \(A’ = -A\) with zero diagonal; every square matrix is \(\frac{1}{2}(A + A’) + \frac{1}{2}(A – A’)\).
  11. The inverse exists only for square matrices, satisfies \(AB = BA = I\), is unique, and \((AB)^{-1} = B^{-1}A^{-1}\).

Chapter 4, Determinants, picks up right here: it takes the square matrices defined in this chapter and assigns each one a single number.

Continue in the same textbook and site from these pages.

Sources and Data Verification

This page describes NCERT Class 12 Mathematics Part I, Chapter 3 (Matrices), in its official edition on ncert.nic.in. It covers the NCERT textbook chapter only — not the CBSE syllabus, curriculum or sample material — and it is maintained for the 2026-27 session.

NCERT settles textbooks, editions and PDFs; CBSE settles curriculum, syllabus and examinations.

Frequently Asked Questions on Matrices Class 12

When is the product of two matrices defined and what is the order of the result?

The product \(AB\) exists when the number of columns of \(A\) equals the number of rows of \(B\). If \(A\) is \(m \times n\) and \(B\) is \(n \times p\), then \(AB\) is \(m \times p\) (NCERT Section 3.4.5). The middle dimension \(n\) must match; the outer dimensions \(m\) and \(p\) become the result’s order.

Is AB always equal to BA in matrix multiplication?

No. Matrix multiplication is not commutative. Even when both \(AB\) and \(BA\) are defined and have the same order, the two products can differ — the worked pair above gives \(AB = \begin{bmatrix} 8 & 20 \\ 14 & 28 \end{bmatrix}\) but \(BA = \begin{bmatrix} 9 & 23 \\ 13 & 27 \end{bmatrix}\) (NCERT Section 3.4.5). Diagonal matrices of the same order are the exception.

Can two non zero matrices multiply to a zero matrix?

Yes. The product \(\begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} \begin{bmatrix} 2 & -2 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\) shows two non-zero matrices with a zero product (NCERT Section 3.4.5). So a zero product does not imply a zero factor.

How do I split any square matrix into a symmetric and a skew symmetric matrix?

Use \(A = \frac{1}{2}(A + A’) + \frac{1}{2}(A – A’)\). The first part is symmetric, the second is skew symmetric, and adding them returns \(A\) (NCERT Section 3.6, Theorem 2). Verify your work by transposing each part: \(P’ = P\) and \(Q’ = -Q\).

When does a matrix have an inverse and is it unique?

A square matrix \(A\) has an inverse when there is a matrix \(B\) of the same order with \(AB = BA = I\); then \(B = A^{-1}\). Rectangular matrices never have inverses, and when an inverse exists it is unique (NCERT Section 3.7)

What does the row and column position tell you in the entry aij?

In \(a_{ij}\), the first subscript \(i\) is the row number and the second subscript \(j\) is the column number. So \(a_{23}\) sits in the second row, third column (NCERT Section 3.2.1). Reading this notation correctly is the first skill Exercise 3.1 tests.

Reference: NCERT Class 12 Mathematics textbook, chapter 3, official edition on ncert.nic.in.


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