MathsCBSE Class 12 · Chapter 3

Matrices Class 12: Notes, Methods and Solved Questions

Updated 14 October 2026 · Fact-checked

Matrices is Chapter 3 of CBSE Class 12 Maths. It covers the order and types of matrices, equality, addition, scalar multiplication and multiplication with their properties, non-commutativity and zero products for 2 × 2 matrices, the transpose, symmetric and skew-symmetric matrices, and invertible matrices with the proof that an inverse is unique.

What Matrices is about

A matrix is a rectangular array of numbers arranged in rows and columns. It is a compact way to store data (marks, prices, quantities) and, as you will see in the next chapter, to solve systems of linear equations. You learn the order of a matrix, how to build a matrix from a rule for $a_{ij}$ , the special types (row, column, square, diagonal, scalar, identity, zero), and when two matrices are equal.

The operations are addition, scalar multiplication and matrix multiplication. Multiplication needs the number of columns of the first matrix to equal the number of rows of the second, and it is not commutative: AB and BA can differ, and AB can be the zero matrix even when neither A nor B is zero. The transpose A′ swaps rows and columns; a matrix is symmetric if A′ = A and skew-symmetric if A′ = −A, and every square matrix splits uniquely into the sum of one of each. Finally, a square matrix A is invertible if there is B with AB = BA = I; such a B is unique.

Elementary row operations and finding an inverse by them are not in the 2026-27 syllabus. Matrices and Determinants together form Unit II (10 marks). In our question bank this chapter has 56 questions (only Integrals and Continuity and Differentiability have more), mostly 1-mark MCQs, together with 5-mark questions on inverses and systems of equations.

Weightage: This chapter is part of Unit II, Algebra, which carries 10 of the 80 theory marks together with Determinants.Source: CBSE Mathematics syllabus 2026-27, Unit II Algebra

In real board papers: asked in 8 of the 8 board papers we hold (2023 to 2026), on average 8.5 marks per paper. See the questions

Key concepts

Matrix, order and constructing a matrix

NCERT section 3.2

A matrix with m rows and n columns has order m × n and mn elements. The element in the ith row and jth column is $a_{ij}$ , and the matrix is written $A = [a_{ij}]_{m \times n}$ .

To construct a matrix from a rule, substitute each pair (i, j). For example, for a 2 × 3 matrix with $a_{ij} = 2 i - j$ : $a_{11} = 1$ , $a_{12} = 0$ , $a_{13} = - 1$ , $a_{21} = 3$ , $a_{22} = 2$ , $a_{23} = 1$ .

If a matrix has k elements, its possible orders are the factor pairs of k: a matrix with 12 elements can be 1 × 12, 2 × 6, 3 × 4, 4 × 3, 6 × 2 or 12 × 1 (six orders).

Worked example

Construct a 2 × 2 matrix $A = [a_{ij}]$ with $a_{ij} = \frac{( i + 2 j ) ^{2}}{2}$ . [2 marks]

Try it first, then show the solutionSolution

$a_{11} = \frac{( 1 + 2 ) ^{2}}{2} = \frac{9}{2}$ ; $a_{12} = \frac{( 1 + 4 ) ^{2}}{2} = \frac{25}{2}$
$a_{21} = \frac{( 2 + 2 ) ^{2}}{2} = 8$ ; $a_{22} = \frac{( 2 + 4 ) ^{2}}{2} = 18$
$A = [\frac{9}{2} 8 \frac{25}{2} 18]$

Answer: $A = [9/2 8 25/2 18]$

Types of matrices and equality

NCERT section 3.3 and 3.3.1

Row matrix: order 1 × n. Column matrix: order m × 1.
Square matrix: m = n; its diagonal elements are $a_{11}, a_{22}, \dots$ .
Diagonal matrix: square, all non-diagonal elements zero.
Scalar matrix: diagonal with all diagonal elements equal.
Identity matrix I: diagonal elements 1, others 0.
Zero (null) matrix O: all elements 0.

Equality: A = B if they have the same order and every corresponding element is equal. Equating elements gives equations for unknowns.

Worked example

Find x and y if $[x + y - 1 4 x - y] = [7 - 1 41]$ . [2 marks]

Try it first, then show the solutionSolution

Equating corresponding elements: x + y = 7 and x − y = 1
Adding: 2x = 8, x = 4; then y = 3

Answer: x = 4, y = 3

Addition and scalar multiplication

NCERT section 3.4.1 to 3.4.4

Matrices of the same order are added element by element: $A + B = [a_{ij} + b_{ij}]$ . Scalar multiplication multiplies every element: $k A = [k a_{ij}]$ , and $- A = (- 1) A$ , $A - B = A + (- B)$ .

Properties (A, B, C of the same order, k, l scalars):

$A + B = B + A$ (commutative); $(A + B) + C = A + (B + C)$ (associative)
$A + O = A$ ; $A + (- A) = O$
$k (A + B) = k A + k B$ ; $(k + l) A = k A + l A$

Matrix equations such as $2 X + A = B$ are solved like ordinary equations: $X = \frac{1}{2} (B - A)$ .

Worked example

If $A = [31 - 2 4]$ and $B = [5 - 3 02]$ , find X such that $2 X + A = B$ . [2 marks]

Try it first, then show the solutionSolution

$2 X = B - A = [5 - 3 - 3 - 1 0 + 2 2 - 4] = [2 - 4 2 - 2]$
$X = [1 - 2 1 - 1]$

Answer: $X = [1 - 2 1 - 1]$

Matrix multiplication and non-commutativity

NCERT section 3.4.5

If A is m × n and B is n × p, then AB is defined and has order m × p; its (i, k) element is the sum of products of the ith row of A with the kth column of B: $c_{ik} = \sum_{j} a_{ij} b_{j k}$ .

Key facts:

AB may be defined while BA is not; and if both are defined they may have different orders.
Even for square matrices, in general $A B \neq = B A$ (non-commutative).
AB can be O although $A \neq = O$ and $B \neq = O$ . For example, with $A = [1111]$ and $B = [1 - 1 - 1 1]$ , $A B = O$ .
Multiplication is associative, $(A B) C = A (B C)$ , and distributive, $A (B + C) = A B + A C$ ; $A I = I A = A$ .
Because $A B \neq = B A$ in general, $(A + B)^{2} = A^{2} + A B + B A + B^{2}$ , not $A^{2} + 2 A B + B^{2}$ .

Worked example

For $A = [1021]$ and $B = [1301]$ , find AB and BA. Are they equal? [2 marks]

Try it first, then show the solutionSolution

$A B = [1 + 6 0 + 3 0 + 2 0 + 1] = [7321]$
$B A = [1 + 0 3 + 0 2 + 0 6 + 1] = [1327]$
AB ≠ BA.

Answer: $A B = [7321]$ , $B A = [1327]$ ; not equal

Transpose and its properties

NCERT section 3.5

The transpose A′ (or $A^{T}$ ) is obtained by interchanging rows and columns: if $A = [a_{ij}]_{m \times n}$ then $A^{'} = [a_{j i}]_{n \times m}$ .

Properties:

$(A^{'})^{'} = A$
$(k A)^{'} = k A^{'}$
$(A + B)^{'} = A^{'} + B^{'}$
$(A B)^{'} = B^{'} A^{'}$ (the order reverses)

Worked example

Verify $(A B)^{'} = B^{'} A^{'}$ for $A = 1 - 2 3$ and $B = [21]$ . [2 marks]

Try it first, then show the solutionSolution

$A B = 2 - 4 6 1 - 2 3$ , so $(A B)^{'} = [21 - 4 - 2 63]$
$B^{'} = [21]$ , $A^{'} = [1 - 2 3]$ ; $B^{'} A^{'} = [21 - 4 - 2 63]$
The two are equal.

Answer: Both equal $[21 - 4 - 2 63]$

Symmetric and skew-symmetric matrices

NCERT section 3.6

A square matrix A is symmetric if $A^{'} = A$ ( $a_{ij} = a_{j i}$ ) and skew-symmetric if $A^{'} = - A$ ( $a_{ij} = - a_{j i}$ ). In a skew-symmetric matrix every diagonal element is zero, because $a_{ii} = - a_{ii}$ .

Results (with proofs from the transpose rules):

For any square A, $A + A^{'}$ is symmetric and $A - A^{'}$ is skew-symmetric. Proof: $(A + A^{'})^{'} = A^{'} + A$ , and $(A - A^{'})^{'} = A^{'} - A = - (A - A^{'})$ .
Every square matrix is the sum of a symmetric and a skew-symmetric matrix, uniquely:

$A = \frac{1}{2} (A + A^{'}) + \frac{1}{2} (A - A^{'})$

Worked example

Express $A = [31 5 - 1]$ as the sum of a symmetric and a skew-symmetric matrix. [3 marks]

Try it first, then show the solutionSolution

$A^{'} = [35 1 - 1]$
$P = \frac{1}{2} (A + A^{'}) = \frac{1}{2} [66 6 - 2] = [33 3 - 1]$ (symmetric)
$Q = \frac{1}{2} (A - A^{'}) = \frac{1}{2} [0 - 4 40] = [0 - 2 20]$ (skew-symmetric)
Check: $P + Q = [31 5 - 1] = A$

Answer: $A = [33 3 - 1] + [0 - 2 20]$

Invertible matrices and uniqueness of the inverse

NCERT section 3.7

A square matrix A of order m is invertible if there is a square matrix B of the same order with $A B = B A = I$ . Then B is the inverse, written $A^{- 1}$ .

Uniqueness (proof): suppose B and C are both inverses of A. Then $A B = B A = I$ and $A C = C A = I$ . So $B = B I = B (A C) = (B A) C = I C = C$ . Hence the inverse, if it exists, is unique.

Inverse of a product: if A and B are invertible of the same order, $(A B)^{- 1} = B^{- 1} A^{- 1}$ , since $(A B) (B^{- 1} A^{- 1}) = A (B B^{- 1}) A^{- 1} = A A^{- 1} = I$ .

A matrix equation can give the inverse directly: if $A^{2} - 5 A + 7 I = O$ , multiply by $A^{- 1}$ to get $A - 5 I + 7 A^{- 1} = O$ , so $A^{- 1} = \frac{1}{7} (5 I - A)$ . (Finding the inverse by elementary operations is not in the syllabus; the adjoint method is in the next chapter.)

Worked example

Show that $A = [2 - 1 13]$ satisfies $A^{2} - 5 A + 7 I = O$ , and hence find $A^{- 1}$ . [3 marks]

Try it first, then show the solutionSolution

$A^{2} = [4 - 1 - 2 - 3 2 + 3 - 1 + 9] = [3 - 5 58]$
$A^{2} - 5 A + 7 I = [3 - 10 + 7 - 5 + 5 + 0 5 - 5 + 0 8 - 15 + 7] = O$
Multiply by $A^{- 1}$ : $A - 5 I + 7 A^{- 1} = O \Rightarrow A^{- 1} = \frac{1}{7} (5 I - A) = \frac{1}{7} [31 - 1 2]$

Answer: $A^{- 1} = \frac{1}{7} [31 - 1 2]$

Maths in real life: Matrices

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Matrix multiplication

Suppose a stationery shop sells notebooks at ₹40, pens at ₹10 and erasers at ₹5, and on Monday sells 20 notebooks, 50 pens and 30 erasers, and on Tuesday 15 notebooks, 40 pens and 25 erasers.

Use matrix multiplication to find the sales revenue on each day.

Try it first, then show the solutionSolution

Quantity matrix Q (rows = days): $[201550403025]$ ; price column P: $40105$
Monday: $20 \times 40 + 50 \times 10 + 30 \times 5 = 800 + 500 + 150 = 1450$
Tuesday: $15 \times 40 + 40 \times 10 + 25 \times 5 = 600 + 400 + 125 = 1125$
$QP = [14501125]$

Answer: ₹1450 on Monday and ₹1125 on Tuesday.

Memory hook: Quantities times prices, row by column: that is what matrix multiplication was made for.

Addition and scalar multiplication

Imagine a bakery's daily output of three items from two branches is stored in matrices, and the owner plans a 10% increase in production next month.

Suppose branch outputs are $A = [1006040]$ and $B = [805030]$ (cakes, breads, pies). Find the combined output and the planned combined output after a 10% increase.

Try it first, then show the solutionSolution

$A + B = [18011070]$
Planned output $= 1.1 (A + B) = [19812177]$

Answer: Combined [180 110 70]; planned [198 121 77].

Memory hook: Add matching entries; scale every entry.

Non-commutativity

Suppose a photo app applies two transformations to a picture: first a stretch S, then a shear H, and the result is compared with doing them in the other order.

Suppose $S = [2001]$ and $H = [1011]$ . Compare HS and SH.

Try it first, then show the solutionSolution

$H S = [2011]$
$S H = [2021]$
They differ, so the order of the two transformations changes the result.

Answer: HS ≠ SH: the order matters.

Memory hook: Socks then shoes is not shoes then socks: matrix products care about order.

Topics in this chapter

Matrices formula sheet

General element and order: $A = [a_{ij}]_{m \times n}$
Addition and scalar multiple: $A + B = [a_{ij} + b_{ij}], k A = [k a_{ij}]$
Product: $(A B)_{ik} = \sum_{j = 1}^{n} a_{ij} b_{j k}, (m \times n) (n \times p) = m \times p$
Square of a sum: $(A + B)^{2} = A^{2} + A B + B A + B^{2}$
Transpose rules: $(A^{'})^{'} = A, (k A)^{'} = k A^{'}, (A + B)^{'} = A^{'} + B^{'}, (A B)^{'} = B^{'} A^{'}$
Symmetric and skew-symmetric: $A^{'} = A (symmetric), A^{'} = - A (skew-symmetric)$
Decomposition: $A = \frac{1}{2} (A + A^{'}) + \frac{1}{2} (A - A^{'})$
Inverse: $A A^{- 1} = A^{- 1} A = I$
Inverse of a product: $(A B)^{- 1} = B^{- 1} A^{- 1}$
Inverse of a transpose: $(A^{'})^{- 1} = (A^{- 1})^{'}$

How the board asks Matrices

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Order, types and construction	1-mark MCQs: number of possible orders, type of a given matrix, constructing $a_{ij}$ from a rule, counting matrices with entries from a set.	Factor pairs for orders; definitions for types; substitute each (i, j); for counting, (choices)^(number of entries).
Equality and operations	Find unknowns from equal matrices; solve $2 X + A = B$ ; conformability of products.	Equate corresponding entries; treat matrix equations like algebra; check columns of the first = rows of the second.
Multiplication and identities	Show $A^{2} - 5 A + 7 I = O$ ; find AB and BA; zero products; $(A + B)^{2}$ .	Multiply row by column carefully; never assume AB = BA.
Symmetric and skew-symmetric	Find unknowns making a matrix symmetric or skew-symmetric; express a matrix as a sum of the two; MCQs on products of such matrices.	Use $a_{ij} = \pm a_{j i}$ and zero diagonal for skew; $P = \frac{1}{2} (A + A^{'})$ , $Q = \frac{1}{2} (A - A^{'})$ .
Invertible matrices	Prove uniqueness of inverse; find $A^{- 1}$ from a matrix equation; $(A B)^{- 1}$ ; MCQs on true or false statements about inverses.	Use $A B = B A = I$ ; multiply both sides of the given equation by $A^{- 1}$ .

Tricks to remember Matrices

Short memory aids. Say each one aloud once, then use it on a question the same day.

Inside numbers must match
(m × n)(n × p): the inside numbers (n) must match; the outside numbers give the order of the product (m × p).
Use it when: Conformability and order of products.
Row into column
Run along the row of the first matrix and down the column of the second, multiplying and adding.
Use it when: Computing products.
Transpose reverses the order
(AB)′ = B′A′ and (AB)⁻¹ = B⁻¹A⁻¹: taking off socks and shoes happens in reverse order.
Use it when: Transpose and inverse of products.
Skew has a zero spine
Every diagonal entry of a skew-symmetric matrix is zero; off-diagonal entries are mirror images with opposite signs.
Use it when: Finding unknowns in skew-symmetric matrices.
Half-sum and half-difference
Symmetric part = half of (A + A′); skew part = half of (A − A′).
Use it when: Decomposition questions.

Matrices: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Number of possible orders · 1 mark · MCQBoard level

A matrix has 24 elements. The number of possible orders it can have is:
(a) 6
(b) 8
(c) 4
(d) 12

Try it first, then show the solutionSolution

24 has divisors 1, 2, 3, 4, 6, 8, 12, 24: eight ordered factor pairs (1 × 24, …, 24 × 1).

Answer: (b) 8

Example 2: Why AA′ is symmetric · 1 mark · Assertion-reasonBoard level

Assertion (A): If A is a square matrix, then $A A^{'}$ is symmetric.
Reason (R): $(A B)^{'} = B^{'} A^{'}$ for matrices of suitable orders.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

Try it first, then show the solutionSolution

$(A A^{'})^{'} = (A^{'})^{'} A^{'} = A A^{'}$ , using R. Both true; R explains A.

Answer: (a)

Example 3: Equal matrices · 2 marks · Very short answerHarder board level

Find x and y if $[2 x - 1 0 3 y x + y] = [x + 3 0 128]$ . [2 marks]

Try it first, then show the solutionSolution

$2 x - 1 = x + 3 \Rightarrow x = 4$ ; $3 y = 12 \Rightarrow y = 4$
Check: $x + y = 8$ ✓

Answer: x = 4, y = 4

Example 4: Zero product · 2 marks · Very short answerHarder board level

Let $A = [1111]$ and $B = [1 - 1 - 1 1]$ . Show that AB = O although A ≠ O and B ≠ O. What does this tell you about cancelling in matrix equations? [2 marks]

Try it first, then show the solutionSolution

$A B = [1 - 1 1 - 1 - 1 + 1 - 1 + 1] = O$
So from AB = O we cannot conclude A = O or B = O, and from AB = AC we cannot cancel A unless A is invertible.

Answer: AB = O; cancellation is not allowed in general.

Example 5: A matrix identity and the inverse · 3 marks · Short answerHarder board level

If $A = [1324]$ , show that $A^{2} - 5 A - 2 I = O$ and hence find $A^{- 1}$ . [3 marks]

Try it first, then show the solutionSolution

$A^{2} = [7151022]$ ; $5 A = [5151020]$
$A^{2} - 5 A - 2 I = [7 - 5 - 2 0 0 22 - 20 - 2] = O$
Multiply by $A^{- 1}$ : $A - 5 I - 2 A^{- 1} = O \Rightarrow A^{- 1} = \frac{1}{2} (A - 5 I) = \frac{1}{2} [- 4 3 2 - 1] = [- 2 \frac{3}{2} 1 - \frac{1}{2}]$

Answer: $A^{- 1} = [- 2 3/2 1 - 1/2]$

Example 6: Symmetric-skew decomposition of a 3 × 3 matrix · 5 marks · Long answerHarder board level

Express $A = 204436 - 6 21$ as the sum of a symmetric and a skew-symmetric matrix. [5 marks]

Try it first, then show the solutionSolution

$A^{'} = 24 - 6 032461$
$A + A^{'} = 44 - 2 468 - 2 82$ , so $P = 22 - 1 234 - 1 41$ (P′ = P)
$A - A^{'} = 0 - 4 10 404 - 10 - 4 0$ , so $Q = 0 - 2 5 202 - 5 - 2 0$ (Q′ = −Q)
Check: $P + Q = 204436 - 6 21 = A$

Answer: A = P + Q with P and Q as above.

Example 7: When does (A + B)² = A² + B²? · 3 marks · Short answerHOTS / competency

If A and B are square matrices of the same order with $(A + B)^{2} = A^{2} + B^{2}$ , what can you conclude? Check with $A = [1000]$ and $B = [0001]$ . [3 marks]

Try it first, then show the solutionSolution

$(A + B)^{2} = A^{2} + A B + B A + B^{2}$ , so the condition means $A B + B A = O$ , i.e. $A B = - B A$ .
For the given matrices: $A B = O$ and $B A = O$ , so $A B + B A = O$ and the identity holds.
Check: $A + B = I$ , $(A + B)^{2} = I$ , and $A^{2} + B^{2} = A + B = I$ ✓

Answer: AB = −BA; the given pair satisfies it (AB = BA = O).

Example 8: Inverse of a product · 3 marks · Short answerHOTS / competency

If A and B are invertible matrices of the same order, prove that $(A B)^{- 1} = B^{- 1} A^{- 1}$ . Hence show that if A is invertible and AB = AC, then B = C. [3 marks]

Try it first, then show the solutionSolution

$(A B) (B^{- 1} A^{- 1}) = A (B B^{- 1}) A^{- 1} = A I A^{- 1} = A A^{- 1} = I$ ; similarly $(B^{- 1} A^{- 1}) (A B) = I$ .
So $B^{- 1} A^{- 1}$ is the (unique) inverse of AB.
If AB = AC, multiply on the left by $A^{- 1}$ : $A^{- 1} A B = A^{- 1} A C \Rightarrow B = C$ .

Answer: $(A B)^{- 1} = B^{- 1} A^{- 1}$ ; cancellation works when A is invertible.

Case-study practice (Section E style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 2 marks.

Sales at two fruit stalls

Suppose two fruit stalls record sales of apples, bananas and oranges (in kg) as $S = [1012151086]$ (row 1: stall X, row 2: stall Y), and the prices per kg are given by $P = 1205080$ (₹).

(i) · 1 mark
What is the order of SP?
Answer
Answer: 2 × 1
(2 × 3)(3 × 1)
(ii) · 1 mark
Find the revenue of stall X.
Answer
Answer: ₹2590
10(120) + 15(50) + 8(80) = 1200 + 750 + 640
(iii) · 2 marks
Find the revenue of stall Y and the total revenue of both stalls.
Answer
Answer: ₹2420; total ₹5010
12(120) + 10(50) + 6(80) = 1440 + 500 + 480 = 2420; 2590 + 2420 = 5010

A matrix that undoes itself

Suppose a code uses the matrix $A = [0110]$ to swap the two letters of each pair in a message: a pair written as a column $[x y]$ is sent as $A [x y]$ .

(i) · 1 mark
Find $A^{2}$ .
Answer
Answer: I
$[0110]^{2} = [1001]$
(ii) · 1 mark
What is $A^{- 1}$ ?
Answer
Answer: A itself
$A \cdot A = I$ , and the inverse is unique.
(iii) · 2 marks
Is A symmetric? Find $A^{2027}$ and explain.
Answer
Answer: Yes, A′ = A; $A^{2027} = A$ .
$A^{2} = I$ , so even powers give I and odd powers give A; 2027 is odd.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

If $A = 02 b a 01 - 3 - 1 0$ is skew-symmetric, find a and b. [2 marks] (2 marks)

Step	Marks
Skew-symmetric: $a_{ij} = - a_{j i}$ ; $a_{12} = - a_{21} \Rightarrow a = - 2$	1
$a_{31} = - a_{13} \Rightarrow b = 3$ (check: $a_{32} = 1 = - a_{23}$ )	1

Express $A = [47 - 1 2]$ as the sum of a symmetric and a skew-symmetric matrix. [3 marks] (3 marks)

Step	Marks
$A^{'} = [4 - 1 72]$	1
$P = \frac{1}{2} (A + A^{'}) = [4332]$ , shown symmetric	1
$Q = \frac{1}{2} (A - A^{'}) = [04 - 4 0]$ , shown skew-symmetric; $P + Q = A$	1

(a) Prove that the inverse of a square matrix, if it exists, is unique. (b) For $A = [34 - 2 - 2]$ , show that $A^{2} - A + 2 I = O$ and hence find $A^{- 1}$ . (c) Verify your answer. [5 marks] (5 marks)

Step	Marks
(a) If B and C are inverses: $B = B I = B (A C) = (B A) C = I C = C$	1
(b) $A^{2} = [9 - 8 12 - 8 - 6 + 4 - 8 + 4] = [14 - 2 - 4]$	1
$A^{2} - A + 2 I = [1 - 3 + 2 4 - 4 - 2 + 2 - 4 + 2 + 2] = O$	1
Multiply by $A^{- 1}$ : $A - I + 2 A^{- 1} = O$ , so $A^{- 1} = \frac{1}{2} (I - A) = \frac{1}{2} [- 2 - 4 23] = [- 1 - 2 1 \frac{3}{2}]$	1
(c) $A A^{- 1} = [- 3 + 4 - 4 + 4 3 - 3 4 - 3] = I$	1

Where students lose marks in Matrices

The slips examiners see most often, and what to write instead.

Mistake: Assuming AB = BA
Habit from numbers.
Wrong working
$(A + B)^{2} = A^{2} + 2 A B + B^{2}$
Right working
$(A + B)^{2} = A^{2} + A B + B A + B^{2}$
Do this: Keep products in the order they appear.
Usually costs: 1 mark in MCQs; method marks elsewhere.
Mistake: Multiplying matrices element by element
Confusing with addition.
Do this: Use row-by-column: each entry of AB is a sum of products.
Usually costs: Full marks of the step.
Mistake: Writing $(A B)^{'} = A^{'} B^{'}$
Forgetting the order reverses.
Do this: $(A B)^{'} = B^{'} A^{'}$ ; similarly $(A B)^{- 1} = B^{- 1} A^{- 1}$ .
Usually costs: 1 mark.
Mistake: Arithmetic slips in 3 × 3 products
Rushing.
Do this: Write each entry as a sum before adding; check one entry by recomputing.
Usually costs: Final answer marks.
Mistake: Using elementary row operations to find the inverse
Older textbooks.
Do this: Elementary operations are not in the 2026-27 syllabus; use the adjoint method (Determinants) or a given matrix equation.
Usually costs: Possibly the method marks.

NCERT exercise guide: Matrices

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercise 3.1 (order, construction, types)

What it tests: Orders, elements, possible orders for a given number of elements, constructing matrices from rules, and equality.

How to approach it: Factor pairs; substitute (i, j); equate entries.

Same kind as Exercise 3.1

If a matrix has 18 elements, what are its possible orders? What if it has 7 elements?

Show the solutionSolution

Factor pairs of 18: 1 × 18, 2 × 9, 3 × 6, 6 × 3, 9 × 2, 18 × 1 (six orders)
7 is prime: 1 × 7 and 7 × 1 only

Answer: Six orders for 18; two orders for 7

Exercise 3.2 (operations)

What it tests: Addition, subtraction, scalar multiples, products, matrix equations, verifying identities like $A^{2} - 5 A + 7 I = O$ , and word problems.

How to approach it: Row-by-column products; solve matrix equations algebraically; never commute.

Same kind as Exercise 3.2

If $A = [1324]$ , find $A^{2} - 5 A$ .

Show the solutionSolution

$A^{2} = [1 + 6 3 + 12 2 + 8 6 + 16] = [7151022]$
$A^{2} - 5 A = [7 - 5 15 - 15 10 - 10 22 - 20] = [2002] = 2 I$

Answer: 2I

Exercise 3.3 (transpose, symmetric and skew-symmetric)

What it tests: Transposes, verifying transpose rules, showing matrices symmetric or skew-symmetric, and decomposing a square matrix.

How to approach it: Use the transpose rules; for decomposition use half-sum and half-difference.

Same kind as Exercise 3.3

Show that $A = 0 - 2 1 20 - 4 - 1 40$ is skew-symmetric.

Show the solutionSolution

$A^{'} = 02 - 1 - 2 04 1 - 4 0$
This equals −A, so A is skew-symmetric.

Answer: A′ = −A

Matrices: previous year questions

Asked in 8 of the 8 board papers we hold (2023 to 2026), on average 8.5 marks per paper; 40 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Matrices

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

2026, set 65/1/1, Q2 · 1 mark · MCQ

Which of the following cannot be the order of a row-matrix ?

(A) $2 \times 1$
(B) $1 \times 2$
(C) $1 \times 1$
(D) $1 \times n$

Try it first, then show our solutionSolution

A row matrix has exactly one row, so its order is $1 \times n$ . $2 \times 1$ has two rows (a column matrix), so it cannot be a row matrix. Option (A).
1 mark

Answer: $2 \times 1$ , option (A)

Where the marks are

Identifies that a row matrix must have order $1 \times n$ (1 mark)

See it in the full 2026 board paper, set 65/1/1

2026, set 65/1/1, Q33(a) · 5 marks · Long answer

If $A = 0 - 2 1 2 - 1 - 1 1 - 2 0$ , find $A^{- 1}$ and use it to solve the following system of equations :
$- 2 y + z = 7$ , $2 x - y - z = 8$ , $x - 2 y = 10$

Try it first, then show our solutionSolution

$∣ A ∣ = 0 (0 - 2) - 2 (0 + 2) + 1 (2 + 1) = - 4 + 3 = - 1 \neq = 0$ , so $A^{- 1}$ exists.
1 mark
Cofactors: $C_{11} = - 2, C_{12} = - 2, C_{13} = 3; C_{21} = - 1, C_{22} = - 1, C_{23} = 2; C_{31} = - 3, C_{32} = - 2, C_{33} = 4$ .
$adj A = - 2 - 2 3 - 1 - 1 2 - 3 - 2 4$
1½ marks
$A^{- 1} = \frac{1}{∣ A ∣} adj A = 22 - 3 11 - 2 32 - 4$
½ mark
The coefficient matrix of the system is $021 - 2 - 1 - 2 1 - 1 0 = A^{T}$ . So the system is $A^{T} X = B$ with $X = x y z$ , $B = 7810$ , giving $X = (A^{T})^{- 1} B = (A^{- 1})^{T} B$ .
½ mark
$x y z = 213212 - 3 - 2 - 4 7810 = 14 + 16 - 30 7 + 8 - 20 21 + 16 - 40 = 0 - 5 - 3$
$∴ x = 0, y = - 5, z = - 3$
1½ marks

Answer: $A^{- 1} = 22 - 3 11 - 2 32 - 4$ ; $x = 0, y = - 5, z = - 3$

Where the marks are

Finds $∣ A ∣ = - 1$ (1 mark)
Correct adj A (1½ marks)
Correct $A^{- 1}$ (½ mark)
Recognises system as $A^{T} X = B$ so $X = (A^{- 1})^{T} B$ (½ mark)
Solves to get $x = 0, y = - 5, z = - 3$ (1½ marks)

See it in the full 2026 board paper, set 65/1/1

2025, set 65/7/1, Q22 · 2 marks · Very short answer

If $A = [2 - 1 32]$ , then show that $A^{2} - 4 A + 7 I = 0$ .

Try it first, then show our solutionSolution

$A^{2} = [2 - 1 32] [2 - 1 32] = [4 - 3 - 2 - 2 6 + 6 - 3 + 4] = [1 - 4 121]$
1 mark
$A^{2} - 4 A + 7 I = [1 - 4 121] - [8 - 4 128] + [7007] = [1 - 8 + 7 - 4 + 4 + 0 12 - 12 + 0 1 - 8 + 7] = [0000] = O$ . Hence proved.
1 mark

Answer: $A^{2} - 4 A + 7 I = O$ (verified)

Where the marks are

Correctly computes $A^{2} = [1 - 4 121]$ (1 mark)
Substitutes and shows the result is the zero matrix (1 mark)

See it in the full 2025 board paper, set 65/7/1

Every Matrices question, by year

For 2027

Sample paper 2026-27: Q4 (1 mark, solved), Q5 (1 mark, solved), Q6 (1 mark, solved), Q26 (3 marks, solved), Q38 (4 marks, solved)

2026

Set 65/1/1: Q2 (1 mark, solved), Q3 (1 mark, solved), Q6 (1 mark, solved), Q33(a) (5 marks, solved)

Set 65/5/1: Q1 (1 mark, solved), Q2 (1 mark, solved), Q4 (1 mark, solved), Q5 (1 mark, solved), Q32 (5 marks, solved)

Sample paper 2025-26: Q2 (1 mark, solved), Q3 (1 mark, solved), Q5 (1 mark, solved), Q32 (5 marks, solved)

2025

Set 65/1/1: Q1 (1 mark, solved), Q9 (1 mark, solved), Q13 (1 mark, solved), Q15 (1 mark, solved), Q35 (5 marks, solved)

Set 65/7/1: Q3 (1 mark, solved), Q4 (1 mark, solved), Q7 (1 mark, solved), Q22 (2 marks, solved), Q27(a) (3 marks, solved), Q27(b) (3 marks, solved)

Sample paper 2024-25: Q2 (1 mark, solved), Q7 (1 mark, solved), Q33 (5 marks, solved)

2024

Set 65/1/1: Q2 (1 mark, solved), Q5 (1 mark, solved), Q18 (1 mark, solved), Q32(a) (5 marks, solved), Q32(b) (5 marks, solved)

Set 65/5/1: Q2 (1 mark, solved), Q4 (1 mark, solved), Q5 (1 mark, solved), Q6 (1 mark, solved), Q32(a) (5 marks, solved), Q32(b) (5 marks, solved)

Sample paper 2023-24: Q1 (1 mark, solved), Q2 (1 mark, solved), Q13 (1 mark, solved), Q34 (5 marks, solved)

2023

Set 65/1/1: Q1 (1 mark, solved), Q4 (1 mark, solved), Q14 (1 mark, solved), Q33(a) (5 marks, solved), Q33(b) (5 marks, solved)

Set 65/5/1: Q3 (1 mark, solved), Q4 (1 mark, solved), Q26 (3 marks, solved), Q37 (4 marks, solved)

Practice questions on Matrices

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
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
Hint
Outside numbers.
Answer
(a) 3 × 2
Q2 · 1 mark · MCQBoard level
The number of all possible 2 × 2 matrices with each entry 0 or 1 is:
(a) 4
(b) 8
(c) 16
(d) 32
Hint
Two choices for each of 4 entries.
Answer
(c) 16
Q3 · 1 mark · Very short answerBoard level
If $A = [25 3 - 1]$ , write A′ and check whether A is symmetric.
Hint
Swap rows and columns.
Answer
$A^{'} = [23 5 - 1] \neq = A$ : not symmetric.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): For any square matrix A, $A - A^{'}$ is skew-symmetric.
Reason (R): $(A - A^{'})^{'} = A^{'} - A$ .
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
$A^{'} - A = - (A - A^{'})$ .
Answer
(a)
Q5 · 2 marks · Very short answerHarder board level
If $A = [42 - 3 - 1]$ , find k such that $A^{2} = k A - 2 I$ .
Hint
Compute A² and compare one entry.
Answer
$A^{2} = [106 - 9 - 5]$ ; $k A - 2 I = [4 k - 2 2 k - 3 k - k - 2]$ ; $- 3 k = - 9 \Rightarrow k = 3$ (all entries agree).
Q6 · 2 marks · Very short answerHarder board level
Find x if $[1 x 1] 101110 10 - 1 12 x = O$ .
Hint
Multiply the first two matrices first.
Answer
Row product: $[1 + 0 + 1, 1 + x + 0, 1 + 0 - 1] = [2, 1 + x, 0]$ . Times the column: $2 + 2 (1 + x) + 0 = 4 + 2 x = 0$ , so $x = - 2$ .
Q7 · 3 marks · Short answerHOTS / competency
If A is a symmetric matrix and B a skew-symmetric matrix of the same order, show that AB − BA is symmetric.
Hint
Use $(A B)^{'} = B^{'} A^{'}$ .
Answer
$(A B - B A)^{'} = B^{'} A^{'} - A^{'} B^{'} = (- B) A - A (- B) = - B A + A B = A B - B A$ . So it is symmetric.
Q8 · 2 marks · Very short answerHarder board level
If $A = [0010]$ , show that $A^{2} = O$ but $A \neq = O$ . Use this to explain why $(A + I)^{2} = 2 A + I$ .
Hint
A and I commute.
Answer
$A^{2} = [0000]$ . Since AI = IA, $(A + I)^{2} = A^{2} + 2 A + I = 2 A + I$ .
Q9 · 2 marks · Very short answerHarder board level
Express $6 - 2 2 - 2 3 - 1 2 - 1 3$ as a sum of symmetric and skew-symmetric matrices. What do you notice?
Hint
Check whether A′ = A first.
Answer
A′ = A, so the symmetric part is A itself and the skew-symmetric part is the zero matrix.
Q10 · 3 marks · Short answerHOTS / competency
If $A = [cos α sin α - sin α cos α]$ , show that $A^{'} A = I$ and hence write $A^{- 1}$ .
Hint
Use $cos^{2} α + sin^{2} α = 1$ .
Answer
$A^{'} A = [cos^{2} α + sin^{2} α - sin α cos α + cos α sin α - cos α sin α + sin α cos α sin^{2} α + cos^{2} α] = I$ ; similarly $A A^{'} = I$ , so $A^{- 1} = A^{'} = [cos α - sin α sin α cos α]$ .
Q11 · 2 marks · Very short answerHarder board level
A matrix $A = [a_{ij}]_{3 \times 3}$ has $a_{ij} = i - j$ . Show that A is skew-symmetric.
Hint
Compare $a_{ij}$ and $a_{j i}$ .
Answer
$a_{j i} = j - i = - (i - j) = - a_{ij}$ for all i, j, so A′ = −A. Explicitly $A = 012 - 1 01 - 2 - 1 0$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.When is the product AB defined?Show answer
When the number of columns of A equals the number of rows of B.
2.Order of AB if A is 2 × 3 and B is 3 × 4?Show answer
2 × 4.
3.What is a scalar matrix?Show answer
A diagonal matrix with all diagonal elements equal.
4.Write $(A B)^{'}$ in terms of A′ and B′.Show answer
$B^{'} A^{'}$ .
5.What are the diagonal elements of a skew-symmetric matrix?Show answer
All zero.
6.Give two non-zero 2 × 2 matrices whose product is O.Show answer
$[1111] [1 - 1 - 1 1] = O$ .
7.If AB = BA = I, what is B?Show answer
The (unique) inverse of A.
8.Number of 2 × 2 matrices with each entry 0 or 1?Show answer
$2^{4} = 16$ .

Best order to study it

Learn notation, order and constructing matrices from rules; then the types of matrices and equality.
Practise addition, scalar multiplication and matrix equations.
Master multiplication: conformability, row-by-column calculation, non-commutativity and zero products.
Learn the transpose and its four properties.
Study symmetric and skew-symmetric matrices, the two proofs and the decomposition.
Finish with invertible matrices: definition, uniqueness proof, $(A B)^{- 1}$ and inverses from matrix equations.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you construct a matrix from a rule such as $a_{ij} = 2 i - j$ ?
Can you list all possible orders of a matrix with a given number of elements?
Can you find unknowns from equal matrices?
Can you multiply 2 × 2 and 3 × 3 matrices accurately?
Can you give an example showing AB ≠ BA and one with AB = O?
Can you verify $(A B)^{'} = B^{'} A^{'}$ ?
Can you prove that $A + A^{'}$ is symmetric and $A - A^{'}$ is skew-symmetric?
Can you express a matrix as a sum of symmetric and skew-symmetric parts?
Can you prove the inverse of a matrix is unique?
Can you find $A^{- 1}$ from an equation such as $A^{2} - 5 A + 7 I = O$ ?

Matrices: frequently asked questions

Are elementary row operations in the 2026-27 syllabus?

No. Elementary operations (transformations) of a matrix and the inverse by elementary operations (NCERT 3.7 and 3.8.1 in older numbering) are not in the 2026-27 syllabus.

How many marks do Matrices and Determinants carry?

Unit II (Algebra: Matrices and Determinants) carries 10 of the 80 marks.

What kinds of questions come from this chapter?

In our question bank this chapter has 56 questions, a total beaten only by Integrals and Continuity and Differentiability, mostly 1-mark MCQs on types, orders, products, transposes and symmetric or skew-symmetric matrices, plus 5-mark questions on finding an inverse and solving equations (which uses the next chapter's adjoint method).

Is matrix multiplication ever commutative?

Sometimes, for particular matrices (for example A with I, or A with itself), but not in general. Never assume it.

Can a non-square matrix have an inverse?

No. Only square matrices can be invertible, because we need AB = BA = I with both products of the same order.

Are diagonal matrices symmetric?

Yes. All off-diagonal entries are zero, so a_{ij} = a_{ji} trivially.

Based on: CBSE Mathematics (041) syllabus 2026-27; NCERT Mathematics Part I, Textbook for Class XII, Chapter 3 (Reprint 2026-27)