Careers360 Logo
School
Search Colleges, Exams, Schools & more
NCERT Solutions for Exercise 4.5 Class 12 Maths Chapter 4 - Determinants

NCERT Solutions for Exercise 4.5 Class 12 Maths Chapter 4 - Determinants

Edited By Ramraj Saini | Updated on Dec 03, 2023 04:11 PM IST | #CBSE Class 12th

NCERT Solutions For Class 12 Maths Chapter 4 Exercise 4.5

NCERT Solutions for Exercise 4.5 Class 12 Maths Chapter 4 Determinants are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. In this article, you will get NCERT solutions for Class 12 Maths chapter 4 Determinants Exercise 4.5 consists of questions related to adjoint, cofactor, and inverse of a matrix. The adjoint of a square matrix A is defined as the transpose of the matrix of the cofactors of matrix A. The adjoint of matrix A is denoted by adj (A) and the inverse of matrix A is defined by A-1. It is a very important concept used in the solving system of linear equations, statistics, and scientific research.

Exercise 4.5 Class 12 Maths is very important for the board exams as generally one question is asked from this exercise. You are advised to solve the questions from Class 12 Maths ch 4 ex 4.5 including examples given before the exercise. You can go through Class 12 Maths chapter 4 exercise 4.5 solutions to understand the concept. 12th class Maths exercise 4.5 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

Also, see

VMC VIQ Scholarship Test

Register for Vidyamandir Intellect Quest. Get Scholarship and Cash Rewards.

Apply
Pearson | PTE

Register now for PTE & Unlock 10% OFF : Use promo code: 'C360SPL10'. Limited Period Offer! Trusted by 3,500+ universities globally

Apply

Assess NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5

Download PDF

Determinants Exercise: 4.5

Question:1 Find adjoint of each of the matrices.

\small \begin{bmatrix} 1 &2 \\ 3 & 4 \end{bmatrix}

Answer:

Given matrix: \small \begin{bmatrix} 1 &2 \\ 3 & 4 \end{bmatrix}= A

Then we have,

A_{11} = 4, A_{12}=-(1)3, A_{21} = -(1)2,\ and\ A_{22}= 1

Hence we get:

adjA = \begin{bmatrix} A_{11} & A_{12} \\ A_{21} &A_{22} \end{bmatrix}^T = \begin{bmatrix} A_{11} & A_{21} \\ A_{12} &A_{22} \end{bmatrix} = \begin{bmatrix} 4 & -2 \\ -3 &1 \end{bmatrix}

Question:2 Find adjoint of each of the matrices

\small \begin{bmatrix} 1 &-1 &2 \\ 2 & 3 &5 \\ -2 & 0 &1 \end{bmatrix}

Answer:

Given the matrix: \small A = \begin{bmatrix} 1 &-1 &2 \\ 2 & 3 &5 \\ -2 & 0 &1 \end{bmatrix}

Then we have,

A_{11} = (-1)^{1+1}\begin{vmatrix} 3 &5 \\ 0& 1 \end{vmatrix} =(3-0)= 3

A_{12} = (-1)^{1+2}\begin{vmatrix} 2 &5 \\ -2& 1 \end{vmatrix} =-(2+10)= -12

A_{13} = (-1)^{1+3}\begin{vmatrix} 2 &3 \\ -2& 0 \end{vmatrix} =0+6= 6

A_{21} = (-1)^{2+1}\begin{vmatrix} -1 &2 \\ 0& 1 \end{vmatrix} =-(-1-0)= 1

A_{22} = (-1)^{2+2}\begin{vmatrix} 1 &2 \\ -2& 1 \end{vmatrix} =(1+4)= 5

A_{23} = (-1)^{2+3}\begin{vmatrix} 1 &-1 \\-2& 0 \end{vmatrix} =-(0-2)= 2

A_{31} = (-1)^{3+1}\begin{vmatrix} -1 &2 \\ 3& 5 \end{vmatrix} =(-5-6)= -11

A_{32} = (-1)^{3+2}\begin{vmatrix} 1 &2 \\2& 5\end{vmatrix} =-(5-4)= -1

A_{33} = (-1)^{3+3}\begin{vmatrix} 1 &-1 \\ 2& 3 \end{vmatrix} =(3+2)= 5

Hence we get:

adjA = \begin{bmatrix} A_{11} &A_{21} &A_{31} \\ A_{12}&A_{22} &A_{32} \\ A_{13}&A_{23} &A_{33} \end{bmatrix} = \begin{bmatrix} 3 &1 &-11 \\ -12&5 &-1 \\ 6&2 &5 \end{bmatrix}

Question:3 Verify \small A (adj A)=(adj A)A=|A|I.

\small \begin{bmatrix} 2 &3 \\ -4 & -6 \end{bmatrix}

Answer:

Given the matrix: \small \begin{bmatrix} 2 &3 \\ -4 & -6 \end{bmatrix}

Let \small A = \begin{bmatrix} 2 &3 \\ -4 & -6 \end{bmatrix}

Calculating the cofactors;

\small A_{11} = (-1)^{1+1}(-6) = -6

\small A_{12} = (-1)^{1+2}(-4) = 4

\small A_{21} = (-1)^{2+1}(3) = -3

\small A_{22} = (-1)^{2+2}(2) = 2

Hence, \small adjA = \begin{bmatrix} -6 &-3 \\ 4& 2 \end{bmatrix}

Now,

\small A (adj A) = \begin{bmatrix} 2 &3 \\ -4&-6 \end{bmatrix}\left ( \begin{bmatrix} -6 &-3 \\ 4 &2 \end{bmatrix} \right )

\small \begin{bmatrix} -12+12 &-6+6 \\ 24-24 & 12-12 \end{bmatrix} = \begin{bmatrix} 0 &0 \\ 0 & 0 \end{bmatrix}

aslo,

\small (adjA)A = \begin{bmatrix} -6 &-3 \\ 4 & 2 \end{bmatrix}\begin{bmatrix} 2 &3 \\ -4& -6 \end{bmatrix}

\small = \begin{bmatrix} -12+12 &-18+18 \\ 8-8 & 12-12 \end{bmatrix} = \begin{bmatrix} 0 &0 \\ 0& 0 \end{bmatrix}

Now, calculating |A|;

\small |A| = -12-(-12) = -12+12 = 0

So, \small |A|I = 0\begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix} = \begin{bmatrix} 0 &0 \\ 0& 0 \end{bmatrix}

Hence we get

\small A (adj A)=(adj A)A=|A|I

Question:4 Verify \small A (adj A)=(adjA)A=|A| I.

\small \begin{bmatrix} 1 &-1 & 2\\ 3 &0 &-2 \\ 1 & 0 &3 \end{bmatrix}

Answer:

Given matrix: \small \begin{bmatrix} 1 &-1 & 2\\ 3 &0 &-2 \\ 1 & 0 &3 \end{bmatrix}

Let \small A= \begin{bmatrix} 1 &-1 & 2\\ 3 &0 &-2 \\ 1 & 0 &3 \end{bmatrix}

Calculating the cofactors;

\small A_{11} = (-1)^{1+1} \begin{vmatrix} 0 &-2 \\ 0& 3 \end{vmatrix} = 0

\small A_{12} = (-1)^{1+2} \begin{vmatrix} 3 &-2 \\1& 3 \end{vmatrix} = -(9+2) =-11

\small A_{13} = (-1)^{1+3} \begin{vmatrix} 3 &0 \\ 1& 0 \end{vmatrix} = 0

\small A_{21} = (-1)^{2+1} \begin{vmatrix} -1 &2 \\ 0& 3 \end{vmatrix} = -(-3-0)= 3

\small A_{22} = (-1)^{2+2} \begin{vmatrix} 1 &2 \\ 1& 3 \end{vmatrix} = 3-2=1

\small A_{23} = (-1)^{2+3} \begin{vmatrix} 1 &-1 \\ 1& 0 \end{vmatrix} = -(0+1) = -1

\small A_{31} = (-1)^{3+1} \begin{vmatrix} -1 &2 \\ 0& -2 \end{vmatrix} = 2

\small A_{32} = (-1)^{3+2} \begin{vmatrix} 1 &2 \\ 3& -2 \end{vmatrix} = -(-2-6) = 8

\small A_{33} = (-1)^{3+3} \begin{vmatrix} 1 &-1 \\ 3& 0 \end{vmatrix} = 0+3 =3

Hence, \small adjA = \begin{bmatrix} 0 &3 &2 \\ -11 & 1& 8\\ 0 &-1 & 3 \end{bmatrix}

Now,

\small A (adj A) =\begin{bmatrix} 1 &-1 &2 \\ 3& 0 & -2\\ 1 & 0 & 3 \end{bmatrix}\begin{bmatrix} 0 &3 &2 \\ -11& 1& 8\\ 0& -1 &3 \end{bmatrix}

\small =\begin{bmatrix} 0+11+0 &3-1-2 &2-8+6 \\ 0+0+0 & 9+0+2 & 6+0-6 \\ 0+0+0 &3+0-3 & 2+0+9 \end{bmatrix} = \begin{bmatrix} 11 & 0 &0 \\ 0& 11&0 \\ 0 & 0 & 11 \end{bmatrix}

also,

\small A (adj A) =\begin{bmatrix} 0 &3 &2 \\ -11& 1& 8\\ 0& -1 &3 \end{bmatrix}\begin{bmatrix} 1 &-1 &2 \\ 3& 0 & -2\\ 1 & 0 & 3 \end{bmatrix}

\small =\begin{bmatrix} 0+9+2 &0+0+0 &0-6+6 \\ -11+3+8 & 11+0+0 & -22-2+24 \\ 0-3+3 &0+0+0 & 0+2+9 \end{bmatrix} = \begin{bmatrix} 11 & 0 &0 \\ 0& 11&0 \\ 0 & 0 & 11 \end{bmatrix}

Now, calculating |A|;

\small |A| = 1(0-0) +1(9+2) +2(0-0) = 11

So, \small |A|I = 11\begin{bmatrix} 1 &0&0 \\ 0& 1&0 \\ 0&0&1 \end{bmatrix} = \begin{bmatrix} 11 &0&0 \\ 0& 11&0\\ 0&0&11 \end{bmatrix}

Hence we get,

\small A (adj A)=(adj A)A=|A|I.

Question:5 Find the inverse of each of the matrices (if it exists).

\small \begin{bmatrix} 2 &-2 \\ 4 & 3 \end{bmatrix}

Answer:

Given matrix : \small \begin{bmatrix} 2 &-2 \\ 4 & 3 \end{bmatrix}

To find the inverse we have to first find adjA then as we know the relation:

A^{-1} = \frac{1}{|A|}adjA

So, calculating |A| :

|A| = (6+8) = 14

Now, calculating the cofactors terms and then adjA.

A_{11} = (-1)^{1+1} (3) = 3

A_{12} = (-1)^{1+2} (4) = -4

A_{21} = (-1)^{2+1} (-2) = 2

A_{22} = (-1)^{2+2} (2) = 2

So, we have adjA = \begin{bmatrix} 3 &2 \\ -4& 2 \end{bmatrix}

Therefore inverse of A will be:

A^{-1} = \frac{1}{|A|}adjA

= \frac{1}{14}\begin{bmatrix} 3 &2 \\ -4& 2 \end{bmatrix} = \begin{bmatrix} \frac{3}{14} &\frac{1}{7} \\ \\ \frac{-2}{7} & \frac{1}{7} \end{bmatrix}

Question:6 Find the inverse of each of the matrices (if it exists).

\small \begin{bmatrix} -1 &5 \\ -3 &2 \end{bmatrix}

Answer:

Given the matrix : \small \begin{bmatrix} -1 &5 \\ -3 &2 \end{bmatrix} = A

To find the inverse we have to first find adjA then as we know the relation:

A^{-1} = \frac{1}{|A|}adjA

So, calculating |A| :

|A| = (-2+15) = 13

Now, calculating the cofactors terms and then adjA.

A_{11} = (-1)^{1+1} (2) = 2

A_{12} = (-1)^{1+2} (-3) = 3

A_{21} = (-1)^{2+1} (5) =-5

A_{22} = (-1)^{2+2} (-1) = -1

So, we have adjA = \begin{bmatrix} 2 &-5 \\ 3& -1 \end{bmatrix}

Therefore inverse of A will be:

A^{-1} = \frac{1}{|A|}adjA

= \frac{1}{13}\begin{bmatrix} 2 &-5 \\ 3& -1 \end{bmatrix} = \begin{bmatrix} \frac{2}{13} &\frac{-5}{13} \\ \\ \frac{3}{13} & \frac{-1}{13} \end{bmatrix}

Question:7 Find the inverse of each of the matrices (if it exists).

\small \begin{bmatrix} 1 &2 &3 \\ 0 &2 &4 \\ 0 &0 &5 \end{bmatrix}

Answer:

Given the matrix : \small \begin{bmatrix} 1 &2 &3 \\ 0 &2 &4 \\ 0 &0 &5 \end{bmatrix}= A

To find the inverse we have to first find adjA then as we know the relation:

A^{-1} = \frac{1}{|A|}adjA

So, calculating |A| :

|A| = 1(10-0)-2(0-0)+3(0-0) = 10

Now, calculating the cofactors terms and then adjA.

A_{11} = (-1)^{1+1} (10) = 10 A_{12} = (-1)^{1+2} (0) = 0

A_{13} = (-1)^{1+3} (0) =0 A_{21} = (-1)^{2+1} (10) = -10

A_{22} = (-1)^{2+2} (5-0) = 5 A_{23} = (-1)^{2+1} (0-0) = 0

A_{31} = (-1)^{3+1} (8-6) = 2 A_{32} = (-1)^{3+2} (4-0) =-4

A_{33} = (-1)^{3+3} (2-0) = 2

So, we have adjA = \begin{bmatrix} 10 &-10 &2 \\ 0& 5 &-4 \\ 0& 0 &2 \end{bmatrix}

Therefore inverse of A will be:

A^{-1} = \frac{1}{|A|}adjA

= \frac{1}{10}\begin{bmatrix} 10 &-10 &2 \\ 0 & 5& -4\\ 0 &0 &2 \end{bmatrix}

Question:8 Find the inverse of each of the matrices (if it exists).

\small \begin{bmatrix} 1 &0 &0 \\ 3 &3 &0 \\ 5 &2 &-1 \end{bmatrix}

Answer:

Given the matrix : \small \begin{bmatrix} 1 &0 &0 \\ 3 &3 &0 \\ 5 &2 &-1 \end{bmatrix} = A

To find the inverse we have to first find adjA then as we know the relation:

A^{-1} = \frac{1}{|A|}adjA

So, calculating |A| :

|A| = 1(-3-0)-0(-3-0)+0(6-15) = -3

Now, calculating the cofactors terms and then adjA.

A_{11} = (-1)^{1+1} (-3-0) = -3 A_{12} = (-1)^{1+2} (-3-0) = 3

A_{13} = (-1)^{1+3} (6-15) =-9 A_{21} = (-1)^{2+1} (0-0) = 0

A_{22} = (-1)^{2+2} (-1-0) = -1 A_{23} = (-1)^{2+1} (2-0) = -2

A_{31} = (-1)^{3+1} (0-0) = 0 A_{32} = (-1)^{3+2} (0-0) =0

A_{33} = (-1)^{3+3} (3-0) = 3

So, we have adjA = \begin{bmatrix} -3 &0 &0 \\ 3& -1 &0 \\ -9& -2 &3 \end{bmatrix}

Therefore inverse of A will be:

A^{-1} = \frac{1}{|A|}adjA

= \frac{-1}{3}\begin{bmatrix} -3 &0 &0 \\ 3 & -1& 0\\ -9 &-2 &3 \end{bmatrix}

Question:9 Find the inverse of each of the matrices (if it exists).

\small \begin{bmatrix} 2 &1 &3 \\ 4 &-1 &0 \\ -7 &2 &1 \end{bmatrix}

Answer:

Given the matrix : \small \begin{bmatrix} 2 &1 &3 \\ 4 &-1 &0 \\ -7 &2 &1 \end{bmatrix} =A

To find the inverse we have to first find adjA then as we know the relation:

A^{-1} = \frac{1}{|A|}adjA

So, calculating |A| :

|A| = 2(-1-0)-1(4-0)+3(8-7) =-2-4+3 = -3

Now, calculating the cofactors terms and then adjA.

A_{11} = (-1)^{1+1} (-1-0) = -1 A_{12} = (-1)^{1+2} (4-0) = -4

A_{13} = (-1)^{1+3} (8-7) =1 A_{21} = (-1)^{2+1} (1-6) = 5

A_{22} = (-1)^{2+2} (2+21) = 23 A_{23} = (-1)^{2+1} (4+7) = -11

A_{31} = (-1)^{3+1} (0+3) = 3 A_{32} = (-1)^{3+2} (0-12) =12

A_{33} = (-1)^{3+3} (-2-4) = -6

So, we have adjA = \begin{bmatrix} -1 &5 &3 \\ -4& 23 &12 \\ 1& -11 &-6 \end{bmatrix}

Therefore inverse of A will be:

A^{-1} = \frac{1}{|A|}adjA

A^{-1} = \frac{1}{-3} \begin{bmatrix} -1 &5 &3 \\ -4& 23 &12 \\ 1& -11 &-6 \end{bmatrix}

Question:10 Find the inverse of each of the matrices (if it exists).

\small \begin{bmatrix} 1 & -1 & 2\\ 0 & 2 &-3 \\ 3 &-2 &4 \end{bmatrix}

Answer:

Given the matrix : \small \begin{bmatrix} 1 & -1 & 2\\ 0 & 2 &-3 \\ 3 &-2 &4 \end{bmatrix} = A

To find the inverse we have to first find adjA then as we know the relation:

A^{-1} = \frac{1}{|A|}adjA

So, calculating |A| :

|A| = 1(8-6)+1(0+9)+2(0-6) =2+9-12 = -1

Now, calculating the cofactors terms and then adjA.

A_{11} = (-1)^{1+1} (8-6) = 2 A_{12} = (-1)^{1+2} (0+9) = -9

A_{13} = (-1)^{1+3} (0-6) =-6 A_{21} = (-1)^{2+1} (-4+4) = 0

A_{22} = (-1)^{2+2} (4-6) = -2 A_{23} = (-1)^{2+1} (-2+3) = -1

A_{31} = (-1)^{3+1} (3-4) = -1 A_{32} = (-1)^{3+2} (-3-0) =3

A_{33} = (-1)^{3+3} (2-0) = 2

So, we have adjA = \begin{bmatrix} 2 &0 &-1 \\ -9& -2 &3 \\ -6& -1 &2 \end{bmatrix}

Therefore inverse of A will be:

A^{-1} = \frac{1}{|A|}adjA

A^{-1} = \frac{1}{-1} \begin{bmatrix} 2 &0 &-1 \\ -9& -2 &3 \\ -6& -1 &2 \end{bmatrix}

A^{-1} = \begin{bmatrix} -2 &0 &1 \\ 9& 2 &-3 \\ 6& 1 &-2 \end{bmatrix}

Question:11 Find the inverse of each of the matrices (if it exists).

\small \begin{bmatrix} 1 & 0&0 \\ 0 &\cos \alpha &\sin \alpha \\ 0 &\sin \alpha &-\cos \alpha \end{bmatrix}

Answer:

Given the matrix : \small \begin{bmatrix} 1 & 0&0 \\ 0 &\cos \alpha &\sin \alpha \\ 0 &\sin \alpha &-\cos \alpha \end{bmatrix} =A

To find the inverse we have to first find adjA then as we know the relation:

A^{-1} = \frac{1}{|A|}adjA

So, calculating |A| :

|A| = 1(-\cos^2 \alpha-\sin^2 \alpha)+0(0-0)+0(0-0)

=-(\cos^2 \alpha + \sin^2 \alpha) = -1

Now, calculating the cofactors terms and then adjA.

A_{11} = (-1)^{1+1} (-\cos^2 \alpha - \sin^2 \alpha) = -1 A_{12} = (-1)^{1+2} (0-0) = 0

A_{13} = (-1)^{1+3} (0-0) =0 A_{21} = (-1)^{2+1} (0-0) = 0

A_{22} = (-1)^{2+2} (-\cos \alpha-0) = -\cos \alpha A_{23} = (-1)^{2+1} (\sin \alpha-0) = -\sin \alpha

A_{31} = (-1)^{3+1} (0-0) = 0 A_{32} = (-1)^{3+2} (\sin \alpha-0) =-\sin \alpha

A_{33} = (-1)^{3+3} (\cos \alpha - 0) = \cos \alpha

So, we have adjA = \begin{bmatrix} -1 &0 &0 \\ 0& -\cos \alpha &-\sin \alpha \\ 0& -\sin \alpha &\cos \alpha \end{bmatrix}

Therefore inverse of A will be:

A^{-1} = \frac{1}{|A|}adjA

A^{-1} = \frac{1}{-1}\begin{bmatrix} -1 &0 &0 \\ 0& -\cos \alpha &-\sin \alpha \\ 0& -\sin \alpha &\cos \alpha \end{bmatrix} = \begin{bmatrix}1 &0 &0 \\ 0&\cos \alpha &\sin \alpha \\ 0& \sin \alpha &-\cos \alpha \end{bmatrix}

Question:12 Let \small A=\begin{bmatrix} 3 &7 \\ 2 & 5 \end{bmatrix} and \small B=\begin{bmatrix} 6 &8 \\ 7 & 9 \end{bmatrix}. Verify that\small (AB)^-^1=B^{-1}A^{-1}.

Answer:

We have \small A=\begin{bmatrix} 3 &7 \\ 2 & 5 \end{bmatrix} and \small B=\begin{bmatrix} 6 &8 \\ 7 & 9 \end{bmatrix}.

then calculating;

AB = \begin{bmatrix} 3 &7 \\ 2& 5 \end{bmatrix}\begin{bmatrix} 6 &8 \\ 7& 9 \end{bmatrix}

=\begin{bmatrix} 18+49 &24+63 \\ 12+35 & 16+45 \end{bmatrix} = \begin{bmatrix} 67 &87 \\ 47& 61 \end{bmatrix}

Finding the inverse of AB.

Calculating the cofactors fo AB:

AB_{11}=(-1)^{1+1}(61) = 61 AB_{12}=(-1)^{1+2}(47) = -47

AB_{21}=(-1)^{2+1}(87) = -87 AB_{22}=(-1)^{2+2}(67) = 67

Then we have adj(AB):

adj(AB) = \begin{bmatrix} 61 &-87 \\ -47& 67 \end{bmatrix}

and |AB| = 61(67) - (-87)(-47) = 4087-4089 = -2

Therefore we have inverse:

(AB)^{-1}=\frac{1}{|AB|}adj(AB) = -\frac{1}{2} \begin{bmatrix} 61 &-87 \\ -47 & 67 \end{bmatrix}

= \begin{bmatrix} \frac{-61}{2} &\frac{87}{2} \\ \\ \frac{47}{2} & \frac{-67}{2} \end{bmatrix} .....................................(1)

Now, calculating inverses of A and B.

|A| = 15-14 = 1 and |B| = 54- 56 = -2

adjA = \begin{bmatrix} 5 &-7 \\ -2 & 3 \end{bmatrix} and adjB = \begin{bmatrix} 9 &-8 \\ -7 & 6 \end{bmatrix}

therefore we have

A^{-1} = \frac{1}{|A|}adjA= \frac{1}{1} \begin{bmatrix} 5&-7 \\ -2& 3 \end{bmatrix} and B^{-1} = \frac{1}{|B|}adjB= \frac{1}{-2} \begin{bmatrix} 9&-8 \\ -7& 6 \end{bmatrix}= \begin{bmatrix} \frac{-9}{2} & 4 \\ \\ \frac{7}{2} & -3 \end{bmatrix}

Now calculatingB^{-1}A^{-1}.

B^{-1}A^{-1} =\begin{bmatrix} \frac{-9}{2} & 4 \\ \\ \frac{7}{2} & -3 \end{bmatrix}\begin{bmatrix} 5&-7 \\ -2& 3 \end{bmatrix}

=\begin{bmatrix} \frac{-45}{2}-8 && \frac{63}{2}+12 \\ \\ \frac{35}{2}+6 && \frac{-49}{2}-9 \end{bmatrix} = \begin{bmatrix} \frac{-61}{2} && \frac{87}{2} \\ \\ \frac{47}{2} && \frac{-67}{2} \end{bmatrix}........................(2)

From (1) and (2) we get

\small (AB)^-^1=B^{-1}A^{-1}

Hence proved.

Question:13 If \small A=\begin{bmatrix} 3 &1 \\ -1 &2 \end{bmatrix}? , show that A^2-5A+7I=O. Hence find \small A^-^1

Answer:

Given \small A=\begin{bmatrix} 3 &1 \\ -1 &2 \end{bmatrix} then we have to show the relation A^2-5A+7I=0

So, calculating each term;

A^2 = \begin{bmatrix} 3& 1\\ -1& 2 \end{bmatrix}\begin{bmatrix} 3&1 \\ -1& 2 \end{bmatrix} = \begin{bmatrix} 9-1 &3+2 \\ -3-2&-1+4 \end{bmatrix} = \begin{bmatrix} 8 &5 \\ -5& 3 \end{bmatrix}

therefore A^2-5A+7I;

=\begin{bmatrix} 8 &5 \\ -5& 3 \end{bmatrix} - 5\begin{bmatrix} 3 &1 \\ -1& 2 \end{bmatrix} + 7 \begin{bmatrix} 1 &0 \\ 0 & 1 \end{bmatrix}

=\begin{bmatrix} 8 &5 \\ -5& 3 \end{bmatrix} - \begin{bmatrix} 15 &5 \\ -5& 10 \end{bmatrix} + \begin{bmatrix} 7 &0 \\ 0 & 7 \end{bmatrix}

\begin{bmatrix} 8-15+7 &&5-5+0 \\ -5+5+0 && 3-10+7 \end{bmatrix} = \begin{bmatrix} 0 &&0 \\ 0 && 0 \end{bmatrix}

Hence A^2-5A+7I = 0.

\therefore A.A -5A = -7I

\Rightarrow A.A(A^{-1}) - 5AA^{-1} = -7IA^{-1}

[Post multiplying by A^{-1}, also |A| \neq 0]

\Rightarrow A(AA^{-1}) - 5I = -7A^{-1}

\Rightarrow AI - 5I = -7A^{-1}

\Rightarrow -\frac{1}{7}(AI - 5I)= \frac{1}{7}(5I-A)

\therefore A^{-1} = \frac{1}{7}(5\begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}-\begin{bmatrix} 3 &1 \\ -1& 2 \end{bmatrix}) = \frac{1}{7}\begin{bmatrix} 2 &-1 \\ 1& 3 \end{bmatrix}

Question:14 For the matrix \small A=\begin{bmatrix} 3 &2 \\ 1 & 1 \end{bmatrix} , find the numbers \small a and \small b such that A^2+aA+bI=0.

Answer:

Given \small A=\begin{bmatrix} 3 &2 \\ 1 & 1 \end{bmatrix} then we have the relation A^2+aA+bI=O

So, calculating each term;

A^2 = \begin{bmatrix} 3& 2\\ 1& 1 \end{bmatrix}\begin{bmatrix} 3&2 \\ 1& 1 \end{bmatrix} = \begin{bmatrix} 9+2 &6+2 \\ 3+1&2+1 \end{bmatrix} = \begin{bmatrix} 11 &8 \\ 4& 3 \end{bmatrix}

therefore A^2+aA+bI=O;

=\begin{bmatrix}11 &8 \\ 4& 3 \end{bmatrix} + a\begin{bmatrix} 3 &2 \\ 1& 1 \end{bmatrix} + b \begin{bmatrix} 1 &0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 &0 \\ 0& 0 \end{bmatrix}

\begin{bmatrix} 11+3a+b & 8+2a \\ 4+a & 3+a+b \end{bmatrix} = \begin{bmatrix} 0 &0 \\ 0 & 0 \end{bmatrix}

So, we have equations;

11+3a+b = 0,\ 8+2a = 0 and 4+a = 0,and\ \ 3+a+b = 0

We get a = -4\ and\ b= 1.

Question:15 For the matrix \small A=\begin{bmatrix} 1 &1 &1 \\ 1 &2 &-3 \\ 2 &-1 &3 \end{bmatrix} Show that \small A^3-6A^2+5A+11I=O Hence, find \small A^-^1.

Answer:

Given matrix: \small A=\begin{bmatrix} 1 &1 &1 \\ 1 &2 &-3 \\ 2 &-1 &3 \end{bmatrix};

To show: \small A^3-6A^2+5A+11I=O

Finding each term:

A^{2} = \begin{bmatrix} 1 & 1& 1\\ 1 & 2& -3\\ 2& -1 & 3 \end{bmatrix} \begin{bmatrix} 1 & 1& 1\\ 1 & 2& -3\\ 2& -1 & 3 \end{bmatrix}

= \begin{bmatrix} 1+1+2 &&1+2-1 &&1-3+3 \\ 1+2-6 &&1+4+3 &&1-6-9 \\ 2-1+6 &&2-2-3 && 2+3+9 \end{bmatrix}

= \begin{bmatrix} 4 &&2 &&1 \\ -3 &&8 &&-14 \\ 7 &&-3 && 14 \end{bmatrix}

A^{3} = \begin{bmatrix} 4 &&2 &&1 \\ -3 &&8 &&-14 \\ 7 &&-3 && 14 \end{bmatrix}\begin{bmatrix} 1 &1 &1 \\ 1 &2 &-3 \\ 2 &-1 &3 \end{bmatrix}

= \begin{bmatrix} 4+2+2 &4+4-1 &4-6+3 \\ -3+8-28 &-3+16+14 & -3-24-42 \\ 7-3+28&7-6-14 &7+9+42 \end{bmatrix}

= \begin{bmatrix} 8 &7 &1 \\ -23 &27 & -69 \\ 32&-13 &58 \end{bmatrix}

So now we have, \small A^3-6A^2+5A+11I

= \begin{bmatrix} 8 &7 &1 \\ -23 &27 & -69 \\ 32&-13 &58 \end{bmatrix}-6\begin{bmatrix} 4 &&2 &&1 \\ -3 &&8 &&-14 \\ 7 &&-3 && 14 \end{bmatrix}+5\begin{bmatrix} 1 &1 &1 \\ 1 &2 &-3 \\ 2 &-1 &3 \end{bmatrix}+11\begin{bmatrix} 1 &0 &0 \\ 0 &1 & 0\\ 0& 0& 1 \end{bmatrix}

= \begin{bmatrix} 8 &7 &1 \\ -23 &27 & -69 \\ 32&-13 &58 \end{bmatrix}-\begin{bmatrix} 24 &&12 &&6 \\ -18 &&48 &&-84 \\ 42 &&-18 && 84 \end{bmatrix}+\begin{bmatrix} 5 &5 &5 \\ 5 &10 &-15 \\ 10 &-5 &15 \end{bmatrix}+\begin{bmatrix} 11 &0 &0 \\ 0 &11 & 0\\ 0& 0& 11 \end{bmatrix}

= \begin{bmatrix} 8-24+5+11 &7-12+5 &1-6+5 \\ -23+18+5&27-48+10+11 &-69+84-15 \\ 32-42+10&-13+18-5 & 58-84+15+11 \end{bmatrix}

= \begin{bmatrix} 0 &0 &0 \\ 0&0 &0 \\ 0&0 & 0 \end{bmatrix} = 0

Now finding the inverse of A;

Post-multiplying by A^{-1} as, |A| \neq 0

\Rightarrow (AAA)A^{-1}-6(AA)A^{-1} +5AA^{-1}+11IA^{-1} = 0

\Rightarrow AA(AA^{-1})-6A(AA^{-1}) +5(AA^{-1})=- 11IA^{-1}

\Rightarrow A^{2}-6A +5I=- 11A^{-1}

A^{-1} = \frac{-1}{11}(A^{2}-6A+5I) ...................(1)

Now,

From equation (1) we get;

A^{-1} = \frac{-1}{11}( \begin{bmatrix} 4 &&2 &&1 \\ -3 &&8 &&-14 \\ 7 &&-3 && 14 \end{bmatrix}-6\begin{bmatrix} 1 &1 &1 \\ 1 &2 &-3 \\ 2 &-1 &3 \end{bmatrix}+5\begin{bmatrix} 1 & 0& 0\\ 0&1 &0 \\ 0& 0&1 \end{bmatrix})


A^{-1} = \frac{-1}{11}( \begin{bmatrix} 4-6+5 &&2-6 &&1-6 \\ -3-6 &&8-12+5 &&-14+18 \\ 7-12 &&-3+6 && 14-18+5 \end{bmatrix}


A^{-1} = \frac{-1}{11}( \begin{bmatrix} 3 &&-4 &&-5 \\ -9 &&1 &&4 \\ -5 &&3 && 1 \end{bmatrix}

Question:16 If \small A=\begin{bmatrix} 2 &-1 &1 \\ -1 &2 &-1 \\ 1 &-1 &2 \end{bmatrix} , verify that \small A^3-6A^2+9A-4I=O. Hence find \small A^-^1.

Answer:

Given matrix: \small A=\begin{bmatrix} 2 &-1 &1 \\ -1 &2 &-1 \\ 1 &-1 &2 \end{bmatrix};

To show: \small A^3-6A^2+9A-4I

Finding each term:

A^{2} = \begin{bmatrix} 2 &-1 &1 \\ -1 &2 &-1 \\ 1 &-1 &2 \end{bmatrix}\begin{bmatrix} 2 &-1 &1 \\ -1 &2 &-1 \\ 1 &-1 &2 \end{bmatrix}

= \begin{bmatrix} 4+1+1 &&-2-2-1 &&2+1+2 \\ -2-2-1 &&1+4+1 &&-1-2-2 \\ 2+1+2 &&-1-2-2 && 1+1+4 \end{bmatrix}

= \begin{bmatrix} 6 &&-5 &&5 \\ -5 &&6 &&-5 \\ 5 &&-5 && 6 \end{bmatrix}

A^{3} =\begin{bmatrix} 6 &&-5 &&5 \\ -5 &&6 &&-5 \\ 5 &&-5 && 6 \end{bmatrix}\begin{bmatrix} 2 &-1 &1 \\ -1 &2 &-1 \\ 1 &-1 &2 \end{bmatrix}

= \begin{bmatrix} 12+5+5 &-6-10-5 &6+5+10 \\ -10-6-5 &5+12+5 & -5-6-10 \\ 10+5+6&-5-10-6 &5+5+12 \end{bmatrix}

= \begin{bmatrix} 22 &-21 &21 \\ -21 &22 & -21 \\ 21&-21 &22 \end{bmatrix}

So now we have, \small A^3-6A^2+9A-4I

=\begin{bmatrix} 22 &-21 &21 \\ -21 &22 & -21 \\ 21&-21 &22 \end{bmatrix}-6 \begin{bmatrix} 6 &&-5 &&5 \\ -5 &&6 &&-5 \\ 5 &&-5 && 6 \end{bmatrix}+9\begin{bmatrix} 2 &-1 &1 \\ -1 &2 &-1 \\ 1 &-1 &2 \end{bmatrix}-4\begin{bmatrix} 1 &0 &0 \\ 0 &1 & 0\\ 0& 0& 1 \end{bmatrix}

=\begin{bmatrix} 22 &-21 &21 \\ -21 &22 & -21 \\ 21&-21 &22 \end{bmatrix}- \begin{bmatrix} 36 &&-30 &&30 \\ -30 &&36 &&-30 \\30 &&-30 && 36 \end{bmatrix}+\begin{bmatrix} 18 &-9 &9 \\ -9 &18 &-9 \\ 9 &-9 &18 \end{bmatrix}-\begin{bmatrix} 4 &0 &0 \\ 0 &4 & 0\\ 0& 0& 4 \end{bmatrix}

= \begin{bmatrix} 22-36+18-4 &-21+30-9 &21-30+9 \\ -21+30-9&22-36+18-4 &-21+30-9 \\ 21-30+9&-21+30-9 & 22-36+18-4 \end{bmatrix}

= \begin{bmatrix} 0 &0 &0 \\ 0&0 &0 \\ 0&0 & 0 \end{bmatrix} = O

Now finding the inverse of A;

Post-multiplying by A^{-1} as, |A| \neq 0

\Rightarrow (AAA)A^{-1}-6(AA)A^{-1} +9AA^{-1}-4IA^{-1} = 0

\Rightarrow AA(AA^{-1})-6A(AA^{-1}) +9(AA^{-1})=4IA^{-1}

\Rightarrow A^{2}-6A +9I=4A^{-1}

A^{-1} = \frac{1}{4}(A^{2}-6A+9I) ...................(1)

Now,

From equation (1) we get;

A^{-1} = \frac{1}{4}(\begin{bmatrix} 6 &&-5 &&5 \\ -5 &&6 &&-5 \\ 5 &&-5 && 6 \end{bmatrix}-6\begin{bmatrix} 2 &-1 &1 \\ -1 &2 &-1 \\ 1 &-1 &2 \end{bmatrix}+9\begin{bmatrix} 1 & 0& 0\\ 0&1 &0 \\ 0& 0&1 \end{bmatrix})

A^{-1} = \frac{1}{4} \begin{bmatrix} 6-12+9 &&-5+6 &&5-6 \\ -5+6 &&6-12+9 &&-5+6 \\ 5-6 &&-5+6 && 6-12+9 \end{bmatrix}

Hence inverse of A is :

A^{-1} = \frac{1}{4} \begin{bmatrix} 3 &&1 &&-1 \\ 1 &&3 &&1 \\ -1 &&1 && 3 \end{bmatrix}

Question:17 Let A be a nonsingular square matrix of order \small 3\times 3. Then \small |adjA| is equal to

(A) \small |A| (B) \small |A|^2 (C) \small |A|^3 (D) \small 3|A|

Answer:

We know the identity (adjA)A = |A| I

Hence we can determine the value of |(adjA)|.

Taking both sides determinant value we get,

|(adjA)A| = ||A| I| or |(adjA)||A| = ||A||| I|

or taking R.H.S.,

||A||| I| = \begin{vmatrix} |A| & 0&0 \\ 0&|A| &0 \\ 0&0 &|A| \end{vmatrix}

= |A| (|A|^2) = |A|^3

or, we have then |(adjA)||A| = |A|^3

Therefore |(adjA)| = |A|^2

Hence the correct answer is B.

Question:18 If A is an invertible matrix of order 2, then det \small (A^-^1) is equal to

(A) \small det(A) (B) \small \frac{1}{det (A)} (C) \small 1 (D) \small 0

Answer:

Given that the matrix is invertible hence A^{-1} exists and A^{-1} = \frac{1}{|A|}adjA

Let us assume a matrix of the order of 2;

A = \begin{bmatrix} a &b \\ c &d \end{bmatrix}.

Then |A| = ad-bc.

adjA = \begin{bmatrix} d &-b \\ -c & a \end{bmatrix} and |adjA| = ad-bc

Now,

A^{-1} = \frac{1}{|A|}adjA

Taking determinant both sides;

|A^{-1}| = |\frac{1}{|A|}adjA| = \begin{bmatrix} \frac{d}{|A|} &\frac{-b}{|A|} \\ \\ \frac{-c}{|A|} & \frac{a}{|A|} \end{bmatrix}

\therefore|A^{-1}| = \begin{vmatrix} \frac{d}{|A|} &\frac{-b}{|A|} \\ \\ \frac{-c}{|A|} & \frac{a}{|A|} \end{vmatrix} = \frac{1}{|A|^2}\begin{vmatrix} d &-b \\ -c& a \end{vmatrix} = \frac{1}{|A|^2}(ad-bc) =\frac{1}{|A|^2}.|A| = \frac{1}{|A|}

Therefore we get;

|A^{-1}| = \frac{1}{|A|}

Hence the correct answer is B.

More About NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5:-

In the NCERT Solutions for Class 12 Maths chapter 4 Exercise 4.5 there are 18 questions including two multiple choice type questions. These questions are related to the most important concept of determinant i.e finding adjoint and inverse of a matrix. There are four examples given in the NCERT book before the exercise 4.5 that you can solve. Solving these examples will help you get conceptual clarity and you will be able to solve exercise questions of NCERT syllabus very easily.

Also Read| Determinants Class 12 Chapter 4 Notes

Benefits of NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5:-

  • Class 12 Maths chapter 4 exercise 4.5 solutions are helpful for the students to get conceptual clarity as these solutions are designed in a detailed manner.
  • Only knowing the answer is not enough to get good marks in the board exams, one must know how to write in the board exams in order to perform well.
  • NCERT solutions for Class 12 Maths chapter 4 exercise 4.4 are prepared by experts who know how to write in board exams, so it will help you to perform better.
  • You can take Class 12 Maths ch 4 ex 4.5 solutions for reference.
JEE Main Highest Scoring Chapters & Topics
Just Study 40% Syllabus and Score upto 100%
Download EBook

Key Features Of NCERT Solutions for Exercise 4.5 Class 12 Maths Chapter 4

  • Comprehensive Coverage: The solutions encompass all the topics covered in ex 4.5 class 12, ensuring a thorough understanding of the concepts.
  • Step-by-Step Solutions: In this class 12 maths ex 4.5, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
  • Accuracy and Clarity: Solutions for class 12 ex 4.5 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
  • Conceptual Clarity: In this 12th class maths exercise 4.5 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
  • Inclusive Approach: Solutions for ex 4.5 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
  • Relevance to Curriculum: The solutions for class 12 maths ex 4.5 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.
JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

Apply

Also see-

NCERT Solutions of Class 12 Subject Wise

Subject Wise NCERT Exampler Solutions

Happy learning!!!

Frequently Asked Questions (FAQs)

1. Does singular matrices are invertible ?

No, singular matrices are not invertible.

2. Does non-singular matrices are invertible ?

Yes, non-singular matrices are invertible.

3. what is an invertible matrix ?

If an inverse of a square matrix exists then it is called an invertible matrix.

4. If |A| = 5 and order of A is 2 then |3A| = ?​

|3A| = 3^2|A| = 45

5. If A is a symmetric matrix then the transpose of A is?

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

6. If A is a skew-symmetric matrix then the transpose of A is?

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

7. If A is a matrix and A' is the transpose of matrix A then what is |A|?

If A is a matrix and A' is the transpose of matrix A then |A| = |A'|.

8. Does square diagonal marix is a symmetric matrix ?

Yes, every square diagonal matrix is a symmetric matrix.

Also Read

Articles

Upcoming School Exams

National Means Cum-Merit Scholarship

Application Date:05 August,2024 - 20 September,2024

National Rural Talent Scholarship Examination

Application Date:05 September,2024 - 20 September,2024

National Institute of Open Schooling 10th examination

Admit Card Date:13 September,2024 - 07 October,2024

National Institute of Open Schooling 12th Examination

Admit Card Date:13 September,2024 - 07 October,2024

Nagaland Board of Secondary Education 12th Examination

Application Date:17 September,2024 - 30 September,2024

View All School Exams

Certifications By Top Providers

Edx
 1113 courses
Coursera
 804 courses
Udemy
 394 courses
Futurelearn
 222 courses
Harvard University, Cambridge
 98 courses
IBM
 85 courses

Explore Top Universities Across Globe

University of Essex, Colchester
  Wivenhoe Park Colchester CO4 3SQ
University College London, London
  Gower Street, London, WC1E 6BT
The University of Edinburgh, Edinburgh
  Old College, South Bridge, Edinburgh, Post Code EH8 9YL
University of Bristol, Bristol
  Beacon House, Queens Road, Bristol, BS8 1QU
University of Delaware, Newark
  Newark, DE 19716
University of Nottingham, Nottingham
  University Park, Nottingham NG7 2RD

Related E-books & Sample Papers

Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?
i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Read Complete Answer
Ashvadha 12 July,2024
Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Read Complete Answer
kumardurgesh1802 10 July,2024
I have completed my 12th with arts cbse. I want to repeat my 12th with pcmb. How to become a private candidate and what are deadline to be able to appear in 2025 exam?

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

  • No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
  • You have to appear for the 2025 12th board exams.
  • Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
  • Aim to register before late October to avoid extra fees.
  • Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

Good luck with your studies!

Read Complete Answer
kumardurgesh1802 10 July,2024
I passed my 12th cbse boards in 2023 , and first time appear for jee mains and advance in 2024 so , am i eligible for 2025 jee mains and advance ?

Hello Aspirant , Hope your doing great . As per your query , your eligible for JEE mains in the year of 2025 , Every candidate can appear for the JEE Main exam 6 times over three consecutive years . The JEE Main exam is held two times every year, in January and April.

Read Complete Answer
satyamercedes0626 23 July,2024
I've passed my class 12th board exam from cbse in PCB in 2024. Now I'll give mathematics exam from cbse as a additional subject in 2025. However, they'll provide separate marksheet for maths. Will i be eligible for jee?? And how I'll fill the form??

Hi there,

Hope you are doing fine

Yes you are certainly eligible for giving the jee exam in the year 2025. You must pass the maths exam with at least 75% criteria as required by jee and provide the marksheet and the passing certificate while registering for the exam.


Pursuing maths as an additional subject while taking biology as your main subject does not offer any hindrance in you appearing for the jee examination. It is indeed an privilege to pursue both maths and biology as the subjects and prepare for the same.

There will be no issue in filling the form while registering for the exam as it will only require your basic details and marksheet which you can provide by attaching the marksheet of maths also. Also, a detailed roadmap is also available on the official websites on how to fill the registration form. So you can fill the form easily.


Hope this resolves your query.

Read Complete Answer
Abhishek 12 July,2024

Popular CBSE Class 12th Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

 Option 2)

0.16\; J
Option 3)

1.00\; J

 Option 4)

0.67\; J
Read More

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

 Option 2)

 6.45×10−3 kg
Option 3)

 9.89×10−3 kg

 Option 4)

12.89×10−3 kg

 
Read More

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

 Option 2)

200 \, \, J - 500 \, \, J
Option 3)

2\times 10^{5}J-3\times 10^{5}J

 Option 4)

20,000 \, \, J - 50,000 \, \, J
Read More

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
Read More

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
Read More

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

Colleges After 12th

Applications for Admissions are open.

VMC VIQ Scholarship Test
VMC VIQ Scholarship Test
Apply

Register for Vidyamandir Intellect Quest. Get Scholarship and Cash Rewards.

Tallentex 2025 - ALLEN's Talent Encouragement Exam
Tallentex 2025 - ALLEN's Talent Encouragement Exam
Apply

Register for Tallentex '25 - One of The Biggest Talent Encouragement Exam

JEE Main Important Physics formulas
JEE Main Important Physics formulas
Apply

As per latest 2024 syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters

JEE Main Important Chemistry formulas
JEE Main Important Chemistry formulas
Apply

As per latest 2024 syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters

TOEFL ® Registrations 2024
TOEFL ® Registrations 2024
Apply

Accepted by more than 11,000 universities in over 150 countries worldwide

Pearson | PTE
Pearson | PTE
Apply

Register now for PTE & Unlock 10% OFF : Use promo code: 'C360SPL10'. Limited Period Offer! Trusted by 3,500+ universities globally

View All Application Forms
News and Notifications
Submit list of candidates for CBSE board exams 2025 correctly, on time: Schools reminded
September 12, 2024 - 01:16 PM IST
CBSE sample paper 2024-25, marking scheme out for skill education subjects
September 02, 2024 - 12:13 PM IST
CBSE to overhaul internal assessment system, develop framework with British Council
August 28, 2024 - 04:58 PM IST
65 lakh students didn't clear Class 10, 12 board exams last year: Education ministry
August 21, 2024 - 10:24 PM IST
CBSE Compartment Result 2024: Marks verification, answer book photocopy, revaluation dates out
August 06, 2024 - 12:37 PM IST
CBSE Class 12 compartment results 2024 declared at cbseresults.nic.in; 29.78% pass
August 02, 2024 - 04:46 PM IST
CBSE best, Kerala board worst in curriculum, infrastructure: NCERT PARAKH report
July 30, 2024 - 04:48 PM IST

Stay up-to date with CBSE Class 12th News

Back to top