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NCERT Solutions for Exercise 3.3 Class 12 Maths Chapter 3 - Matrices

NCERT Solutions for Exercise 3.3 Class 12 Maths Chapter 3 - Matrices

Edited By Komal Miglani | Updated on Apr 17, 2025 04:24 PM IST | #CBSE Class 12th

Given a matrix A such that A=[aij] is an m×n matrix, then the matrix obtained by interchanging the rows and columns of A is called the transpose of A. In this exercise, you will get NCERT solutions for Class 12 Maths Chapter 3 Exercise 3.3, which will include questions related to properties of the transpose of the matrices, symmetric and skew-symmetric matrices in Exercise 3.3, Class 12 Maths. Going through these solutions will help you to understand the concept clearly. These NCERT solutions are created by a subject matter expert at Careers360 to give a more systematic and proper approach for each question. You should try to solve these Class 12th maths chapter 3 exercise 3.3 of the NCERT on your own. You can take help from these solutions, which are prepared by experts who know how best to answer in board exams

Class 12 Maths Chapter 3 Exercise 3.3 Solutions: Download PDF

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Matrices Exercise: 3.3

Question 1(i). Find the transpose of each of the following matrices:

[512−1]

Answer:

A=[512−1]

The transpose of the given matrix is

AT=[512−1]

Question 1(ii). Find the transpose of each of the following matrices:

[1−123]

Answer:

A=[1−123]

Interchanging the rows and columns of the matrix A, we get

AT=[12−13]

Question 1(iii) Find the transpose of each of the following matrices:

[−15635623−1]

Answer:

A=[−15635623−1]

Transpose is obtained by interchanging the rows and columns of matrix

AT=[−13255366−1]

Question 2(i). If A=[−123579−211] and B=[−41−5120131], then verify

(A+B)′=A′+B′

Answer:

A=[−123579−211] and B=[−41−5120131]

(A+B)′=A′+B′

L.H.S : (A+B)′

A+B=[−123579−211] +[−41−5120131]

A+B=[−1+(−4)2+13+(−5)5+17+29+0−2+11+31+1]

A+B=[−53−2699−142]

(A+B)′=[−56−1394−292]

R.H.S : A′+B′

A′+B′=[−15−2271391] +[−411123−501]

A′+B′=[−1+(−4)5+1−2+12+17+21+33+(−5)9+01+1]

A′+B′=[−56−1394−292]

Thus we find that the LHS is equal to RHS and hence verified.

Question 2(ii). If A=[−123579−211] and B=[−41−5120131], then verify

(A−B)′=A′−B′

Answer:

A=[−123579−211] and B=[−41−5120131]

(A−B)′=A′−B′

L.H.S : (A−B)′

A−B=[−123579−211] −[−41−5120131]

A−B=[−1−(−4)2−13−(−5)5−17−29−0−2−11−31−1]

A−B=[318459−3−20]

(A−B)′=[34−315−2890]

R.H.S : A′−B′

A′−B′=[−15−2271391] −[−411123−501]

A′−B′=[−1−(−4)5−1−2−12−17−21−33−(−5)9−01−1]

A′−B′=[34−315−2890]

Hence, L.H.S = R.H.S. so verified that

(A−B)′=A′−B′.

Question 3(i). If A′=[34−1201] and B=[−121123], then verify

(A+B)′=A′+B′

Answer:

A′=[34−1201] B=[−121123]

A=(A′)′=[3−10421]

To prove: (A+B)′=A′+B′

L.H.S:(A+B)′=

A+B=[3−10421] +[−121123]

A+B=[3+(−1)−1+(−1)0+14+12+21+3]

A+B=[2−21544]

∴(A+B)′=[251414]

R.H.S: A′+B′

A′+B′=[34−1201] +[−112213]

A′+B′=[251414]

Hence, L.H.S = R.H.S i.e. (A+B)′=A′+B′.

Question 3(ii). If A=[34−1201] and B=[−121123], then verify

(A−B)′=A′−B′

Answer:

A′=[34−1201] B=[−121123]

A=(A′)′=[3−10421]

To prove: (A−B)′=A′−B′

L.H.S:(A−B)′=

A−B=[3−10421] −[−121123]

A−B=[3−(−1)−1−(2)0−14−12−21−3]

A−B=[4−3−130−2]

∴(A−B)′=[43−30−1−2]

R.H.S: A′−B′

A′−B′=[34−1201] −[−112213]

A′−B′=[43−30−1−2]

Hence, L.H.S = R.H.S i.e. (A−B)′=A′−B′.

Question 4. If A′=[−2312] and B=[−1012], then find (A+2B)′

Answer:

B=[−1012]

A′=[−2312]

A=(A′)′=[−2132]

(A+2B)′ :

A+2B=[−2132]+2[−1012]

A+2B=[−2132]+[−2024]

A+2B=[−2+(−2)1+03+22+4]

A+2B=[−4156]

Transpose is obtained by interchanging rows and columns and the transpose of A+2B is

(A+2B)′=[−4516]

Question 5(i) For the matrices A and B, verify that (AB)′=B′A′, where

A=[1−43], B=[−121]

Answer:

A=[1−43], B=[−121]

To prove : (AB)′=B′A′

L.H.S:(AB)′

AB=[1−43][−121]

AB=[−1214−8−4−363]

(AB)′=[−14−32−861−43]

R.H.S:B′A′

B′=[−121]

A′=[1−43]

B′A′=[−121][1−43]

B′A′=[−14−32−861−43]

Hence, L.H.S =R.H.S

so it is verified that (AB)′=B′A′.

Question 5(ii) For the matrices A and B, verify that (AB)′=B′A′, where

A=[012], B=[157]

Answer:

A=[012], B=[157]

To prove : (AB)′=B′A′

L.H.S:(AB)′

AB=[012][157]

AB=[00015721014]

(AB)′=[01205100714]

R.H.S:B′A′

B′=[157]

A′=[012]

B′A′=[157][012]

B′A′=[01205100714]

Hence, L.H.S =R.H.S i.e.(AB)′=B′A′.

Question 6(i). If A=[cos⁡αsin⁡α−sin⁡αcos⁡α], then verify that A′A=I

Answer:

A=[cos⁡αsin⁡α−sin⁡αcos⁡α]

By interchanging rows and columns, we get the transpose of A

A′=[cos⁡α−sin⁡αsin⁡αcos⁡α]

To prove: A′A=I

L.H.S :A′A

A′A=[cos⁡α−sin⁡αsin⁡αcos⁡α] [cos⁡αsin⁡α−sin⁡αcos⁡α]

A′A=[cos2⁡α+sin2⁡αsin⁡αcos⁡α−sin⁡α cosαsin⁡αcos⁡α−sin⁡αcos⁡α sin2α+cos2⁡α]

A′A=[1001]=I=R.H.S

Question 6(ii). If A=[sin⁡αcos⁡α−cos⁡αsin⁡α], then verify that A′A=I

Answer:

A=[sin⁡αcos⁡α−cos⁡αsin⁡α]

By interchanging columns and rows of the matrix A we get the transpose of A

A′=[sin⁡α−cos⁡αcos⁡αsin⁡α]

To prove: A′A=I

L.H.S :A′A

A′A=[sin⁡α−cos⁡αcos⁡αsin⁡α] [sin⁡αcos⁡α−cos⁡αsin⁡α]

A′A=[cos2⁡α+sin2⁡αsin⁡αcos⁡α−sin⁡α cosαsin⁡αcos⁡α−sin⁡αcos⁡α sin2α+cos2⁡α]

A′A=[1001]=I=R.H.S

Question 7(i). Show that the matrix A=[1−15−121513] is a symmetric matrix.

Answer:

A=[1−15−121513]

The transpose of A is

A′=[1−15−121513]

Since A″=A, so given matrix is a symmetric matrix.

Question 7(ii) Show that the matrix A=[01−1−1011−10] is a skew-symmetric matrix.

Answer:

A=[01−1−1011−10]

The transpose of A is

A′=[0−1110−1−110]

A′=−[01−1−1011−10]

A′=−A

Since A′=−A so given matrix is a skew-symmetric matrix.

Question 8(i). For the matrix A=[1567], verify that

(A+A′) is a symmetric matrix.

Answer:

A=[1567]

A′=[1657]

A+A′=[1567] +[1657]

A+A′=[1+15+66+57+7]

A+A′=[2111114]

(A+A′)′=[2111114]

We have A+A′=(A+A′)′

Hence, (A+A′) is a symmetric matrix.

Question 8(ii) For the matrix A=[1567], verify that

(A−A′) is a skew symmetric matrix.

Answer:

A=[1567]

A′=[1657]

A−A′=[1567] −[1657]

A−A′=[1−15−66−57−7]

A−A′=[0−110]

(A−A′)′=[01−10]=−(A−A′)

We have A−A′=−(A−A′)′

Hence, (A−A′) is a skew-symmetric matrix.

Question 9. Find 12(A+A′) and 12(A−A′), when A=[0ab−a0c−b−c0]

Answer:

A=[0ab−a0c−b−c0]

The transpose of the matrix is obtained by interchanging rows and columns

A′=[0−a−ba0−cbc0]

12(A+A′)=12([0ab−a0c−b−c0] +[0−a−ba0−cbc0])

12(A+A′)=12([0+0a+(−a)b+(−b)−a+a0+0c+(−c)−b+b−c+c0+0])

12(A+A′)=12[000000000]

12(A+A′)=[000000000]

12(A+A′)=0

12(A−A′)=12([0ab−a0c−b−c0]−[0−a−ba0−cbc0])

12(A−A′)=12([0−0a−(−a)b−(−b)−a−a0−0c−(−c)−b−b−c−c0−0])

12(A−A′)=12[02a2b−2a02c−2b−2c0]

12(A−A′)=[0ab−a0c−b−c0]

Question 10(i). Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

[351−1]

Answer:

A=[351−1]

A′=[315−1]

A+A′=[351−1]+[315−1]

A+A′=[666−2]

Let

B=12(A+A′)=12[666−2]=[333−1]

B′=[333−1]=B

Thus, 12(A+A′) is a symmetric matrix.

A−A′=[351−1]−[315−1]

A−A′=[04−40]

Let

C=12(A−A′)=12[04−40]=[02−20]

C′=[0−220]

C=−C′

Thus, 12(A−A′) is a skew symmetric matrix.

Represent A as the sum of B and C.

B+C=[333−1] +[02−20] =[351−1]=A

Question:10(ii). Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

[6−22−23−12−13]

Answer:

A=[6−22−23−12−13]

A′=[6−22−23−12−13]

A+A′=[6−22−23−12−13]+[6−22−23−12−13]

A+A′=[12−44−46−24−26]

Let

B=12(A+A′)=12[12−44−46−24−26]=[6−22−23−12−13]

B′=[6−22−23−12−13]=B

Thus, 12(A+A′) is a symmetric matrix.

A−A′=[6−22−23−12−13]−[6−22−23−12−13]

A−A′=[000000000]

Let

C=12(A−A′)=12[000000000]=[000000000]

C′=[000000000]

C=−C′

Thus, 12(A−A′) is a skew-symmetric matrix.

Represent A as the sum of B and C.

B+C=[6−22−23−12−13] +[000000000] =[6−22−23−12−13]=A

Question 10(iii). Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

[33−1−2−21−4−52]

Answer:

A=[33−1−2−21−4−52]

A′=[3−2−43−2−5−112]

A+A′=[33−1−2−21−4−52]+[3−2−43−2−5−112]

A+A′=[61−51−4−4−5−44]

Let

B=12(A+A′)=12[61−51−4−4−5−44]=[312−5212−2−2−52−22]

B′=[312−5212−2−2−52−22]=B

Thus, 12(A+A′) is a symmetric matrix.

A−A′=[33−1−2−21−4−52]−[3−2−43−2−5−112]

A−A′=[053−506−3−60]

Let

C=12(A−A′)=12[053−506−3−60]=[05232−5203−32−30]

C′=[0−52−32520−33230]

C=−C′

Thus, 12(A−A′) is a skew-symmetric matrix.

Represent A as the sum of B and C.

B+C=[312−5212−2−2−52−22] +[05232−5203−32−30] =[33−1−2−21−4−52]=A

Question 10(iv). Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

[15−12]

Answer:

A=[15−12]

A′=[1−152]

A+A′=[15−12]+[1−152]

A+A′=[2444]

Let

B=12(A+A′)=12[2444]=[1222]

B′=[1222]=B

Thus, 12(A+A′) is a symmetric matrix.

A−A′=[15−12]−[1−152]

A−A′=[06−60]

Let

C=12(A−A′)=12[06−60]=[03−30]

C′=[0−330]

C=−C′

Thus, 12(A−A′) is a skew-symmetric matrix.

Represent A as the sum of B and C.

B+C=[1222] −[0−330] =[15−12]=A

Question 11 Choose the correct answer in the Exercises 11 and 12.

If A, B are symmetric matrices of the same order, then AB – BA is a

(A) Skew-symmetric matrix
(B) Symmetric matrix
(C) Zero matrix
(D) Identity matrix

Answer:

If A, B are symmetric matrices, then

A′=A and B′=B

we have, (AB−BA)′=(AB)′−(BA)′=B′A′−A′B′

=BA−AB

=−(AB−BA)

Hence, we have (AB−BA)=−(AB−BA)′

Thus,( AB-BA)' is skew symmetric.

Option A is correct.

Question 12 Choose the correct answer in the Exercises 11 and 12.

If A=[cos⁡α−sin⁡αsin⁡αcos⁡α] and A+A′=I, then the value of α is

(A) π6

(B) π3

(C) π

(D) 3π2

Answer:

A=[cos⁡α−sin⁡αsin⁡αcos⁡α]

A′=[cos⁡αsin⁡α−sin⁡αcos⁡α]

A+A′=[cos⁡α−sin⁡αsin⁡αcos⁡α]+[cos⁡αsin⁡α−sin⁡αcos⁡α]=[1001]

A+A′=[2cos⁡α002cos⁡α]=[1001]

2cosα=1

cosα=12

α=π3

Option B is correct.


Also Read,

Topics covered in Chapter 3: Matrices: Exercise 3.3

  • Intoduction
  • Transpose of a matrix: If A=[aij] is an m×n matrix, then the matrix obtained by interchanging the rows and columns of A is called the transpose of A.
  • Properties of the transpose of the matrices:
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JEE Main high scoring chapters and topics

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  1. (A′)′=A,
  2. (k A)′=k A′ (where k is any constant)
  3. (A+B)′=A′+B′
  4. (AB)′=B′A′
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  • Symmetric Matrix: A square matrix A=[aij] is said to be symmetric if A′=A, that is, [aij]=[aji] for all possible values of i and j.
  • Skew Symmetric Matrix: A square matrix A=[aij] is said to be skew symmetric matrix if A′=−A, that is aji=−aij for all possible values of i and j. Now, if we put i=j, we have aii=−aii. Therefore 2aii=0 or aii=0 for all i 's.
  • Theorem 1: For any square matrix A with real number entries, A+A′ is a symmetric matrix and A−A′ is a skew-symmetric matrix.
  • Theorem 2: Any square matrix can be expressed as the sum of a symmetric and a skew-symmetric matrix.
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Also, read,

NCERT Solutions Subject Wise

These links lead to NCERT textbook solutions for other subjects. Students can check and analyse these well-structured solutions for a deeper understanding.

Subject-wise NCERT Exemplar solutions

Students may visit these NCERT exemplar links for additional practice.

Frequently Asked Questions (FAQs)

1. How these NCERT textbook solutions are helpful in board exam ?

NCERT solutions will help you to solve NCERT problems when you are not able to solve them on your own. For more questions NCERT exemplar problems will be useful. For CBSE board exam NCERT syllabus will be useful for exam preparation. Practice class 12 ex 3.3 to command the concepts.

2. What is the definition of order of a Matrix ?

The order of matrix having m rows and n columns is m x n. 

3. What is symmetric matrix ?

If the transpose of matrix A is equal to matrix A then matrix A is a symmetric matrix.

4. What is skew-symmetric matrix ?

If the transpose of matrix A is equal to the negative of matrix A then matrix A is a skew-symmetric matrix.

5. What are the diagonal elements of skew symmetric matrix ?

All the diagonal elements of a skew-symmetric matrix are zero.

6. What is the transpose of A' ?

(A')' = A

Hence the transpose of A' is matrix A.

7. If A is symmetric matrix then A' ?

If A is a symmetric matrix then A' = A.

8. If A is symmetric matrix and k is a constant then (kA) ' ?

If A is a symmetric matrix and k is a constant then (kA) ' = k (A)'

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Manasvini Gupta 30 January,2025
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Kanishka kaushikii 17 September,2024
i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

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Yuvan 30 August,2024
i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

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Ashvadha 12 July,2024
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 Option 4)

20,000 \, \, J - 50,000 \, \, J
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A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

 Option 2)

\; K\;
Option 3)

zero\;

 Option 4)

K/4
Read More

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

 Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced
Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

 Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .
Read More

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

 Option 2)

3.125 × 10-2
Option 3)

1.25 × 10-2

 Option 4)

2.5 × 10-2
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If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

 Option 2)

increase two fold
Option 3)

remain unchanged

 Option 4)

be a function of the molecular mass of the substance.
Read More

With increase of temperature, which of these changes?

Option 1)

Molality

 Option 2)

Weight fraction of solute
Option 3)

Fraction of solute present in water

 Option 4)

Mole fraction.
Read More

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

 Option 2)

6.023 × 1022
Option 3)

half that in 8 g He

 Option 4)

558.5 × 6.023 × 1023
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A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

 Option 2)

more than 3 but less than 6
Option 3)

more than 6 but less than 9

 Option 4)

more than 9
Read More

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