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NCERT Solutions for Class 12 Maths Chapter 4: Determinants - Exercise 4.1

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NCERT Solutions for Class 12 Maths Chapter 4 (Ex 4.1)

Determinants are a very important concept of mathematics. It has a vast range of applications in the fields of  Engineering, Science, Economics, Social Science, etc. The solution to exercise 4.1 of NCERT Class 12th Mathematics is provided below. 


There are five examples and eight questions covered in NCERT Class 12th Maths Chapter 4 Exercise 4.1. Students are required to solve all sums given in the exercises to understand the concepts of this chapter. Solving these exercises will help them to increase their problem-solving speed, efficiency, capability, and many other factors.


Class:

NCERT Solutions for Class 12

Subject:

Class 12 Maths

Chapter Name:

Chapter 4 - Determinants

Exercise:

Exercise - 4.1

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2023-24

Medium:

English and Hindi

Available Materials:

  • Chapter Wise

  • Exercise Wise

Other Materials

  • Important Questions

  • Revision Notes



NCERT Class 12 Maths Exercise 4.1 Solutions offered by Vedantu are compiled by learned teaching professionals according to the latest CBSE syllabus. These solutions for Class 12 Maths Exercise 4.1 are available in the PDF format to help students gain access to them from anywhere and at any time. Our Class 12 Maths NCERT Solutions consists of shortcut techniques as well as elaborate explanations. You can download NCERT Solutions PDF from our portal for all subjects and fetch higher grades in your boards.

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Access NCERT Solutions for Class 12 Maths Chapter 4 - Determinants

Exercise (4.1)

1. Evaluate the determinant: $\begin{vmatrix}2 & 4 \\ -5& -1 \\\end{vmatrix}$

Ans: Solving the determinant  $\begin{vmatrix}2 & 4 \\ -5& -1 \\\end{vmatrix}$ we have:

$\Rightarrow \begin{vmatrix} 2 & 4 \\ -5& -1 \\ \end{vmatrix}=2(-1)-4(-5)$ $\Rightarrow \begin{vmatrix} 2 & 4 \\ -5& -1 \\ \end{vmatrix}=-2+20$ $\Rightarrow \begin{vmatrix} 2 & 4 \\ -5& -1 \\ \end{vmatrix}=18$

2. Evaluate the determinants.

 i. \[\left| \begin{matrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \\ \end{matrix} \right|\] Ans: Solving the determinant \[\left| \begin{matrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \\ \end{matrix} \right|\], we have: \[\Rightarrow \left| \begin{matrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \\ \end{matrix} \right|=\left( \cos \theta \right)\left( \cos \theta \right)-\left( -\sin \theta \right)\left( \sin \theta \right)\] \[\Rightarrow \left| \begin{matrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \\ \end{matrix} \right|=\text{ }{{\cos }^{2}}\theta +{{\sin }^{2}}\theta \] We know, \[{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\] \[\therefore \left| \begin{matrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \\ \end{matrix} \right|=1\] ii. \[\left| \begin{matrix} {{\text{x}}^{\text{2}}}\text{-x+1} & \text{x-1} \\ \text{x+1} & \text{x+1} \\ \end{matrix} \right|\] Ans: Solving the determinant \[\left| \begin{matrix} {{\text{x}}^{\text{2}}}\text{-x+1} & \text{x-1} \\ \text{x+1} & \text{x+1} \\ \end{matrix} \right|\], we have: \[\Rightarrow \left| \begin{matrix} {{\text{x}}^{\text{2}}}\text{-x+1} & \text{x-1} \\ \text{x+1} & \text{x+1} \\ \end{matrix} \right|\text{=}\left( {{\text{x}}^{\text{2}}}\text{-x+1} \right)\left( \text{x+1} \right)\text{-}\left( \text{x-1} \right)\left( \text{x+1} \right)\] \[\Rightarrow \left| \begin{matrix} {{\text{x}}^{\text{2}}}\text{-x+1} & \text{x-1} \\ \text{x+1} & \text{x+1} \\ \end{matrix} \right|\text{= }{{\text{x}}^{\text{3}}}\text{-}{{\text{x}}^{\text{2}}}\text{+x+}{{\text{x}}^{\text{2}}}\text{-x+1-}\left( {{\text{x}}^{\text{2}}}\text{-1} \right)\] So, \[\left| \begin{matrix} {{\text{x}}^{\text{2}}}\text{-x+1} & \text{x-1} \\ \text{x+1} & \text{x+1} \\ \end{matrix} \right|\text{= }{{\text{x}}^{\text{3}}}\text{+1-}{{\text{x}}^{\text{2}}}\text{+1}\] \[\therefore \left| \begin{matrix} {{\text{x}}^{\text{2}}}\text{-x+1} & \text{x-1} \\ \text{x+1} & \text{x+1} \\ \end{matrix} \right|\text{= }{{\text{x}}^{\text{3}}}\text{-}{{\text{x}}^{\text{2}}}\text{+2}\].

3. If $\text{A=}$\[\left| \begin{matrix}1 & 2  \\   4 & 2  \\ \end{matrix} \right|\], then show that \[\left| \text{2A} \right|\text{=4}\left| \text{A} \right|\].

Ans: Given that,\[\text{A=}\left[ \begin{matrix} \text{1} & \text{2} \\ \text{4} & \text{2} \\ \end{matrix} \right]\] Multiplying $\text{A}$ by $2$, we have: \[\Rightarrow \text{2A= 2}\left[ \begin{matrix} \text{1} & \text{2} \\ \text{4} & \text{2} \\ \end{matrix} \right]\text{=}\left[ \begin{matrix} \text{2} & \text{4} \\ \text{8} & \text{4} \\ \end{matrix} \right]\] \[\Rightarrow \text{2A=}\left[ \begin{matrix} \text{2} & \text{4} \\ \text{8} & \text{4} \\ \end{matrix} \right]\] \[\therefore \] L.H.S \[\text{=}\left| \text{2A} \right|\text{=}\left| \begin{matrix} \text{2} & \text{4} \\ \text{8} & \text{4} \\ \end{matrix} \right|\] \[\Rightarrow \left| \text{2A} \right|\text{=}2\times 4-4\times 8\] \[\Rightarrow \left| \text{2A} \right|\text{=}8-32\] \[\therefore \left| \text{2A} \right|\text{=}-24\] The value of determinant $\text{A}$ is \[\Rightarrow \left| \text{A} \right|\text{=}\left| \begin{matrix} \text{1} & \text{2} \\ \text{4} & \text{2} \\ \end{matrix} \right|\] \[\Rightarrow \left| \text{A} \right|\text{=}2-8\] \[\therefore \left| \text{A} \right|\text{=}-6\] R.H.S is given as $\text{4}\left| \text{A} \right|$. \[\therefore \text{4}\left| \text{A} \right|\text{=4 }\!\!\times\!\!\text{ }\left( \text{-6} \right)\text{=-24}\] Hence, we have L.H.S $=$ R.H.S \[\therefore \left| \text{2A} \right|\text{=4}\left| \text{A} \right|\].

4. If \[\text{A=}\left[ \begin{matrix} 1 & 0 & 1  \\  0 & 1 & 2  \\ 0 & 0 & 4  \\  \end{matrix} \right]\], then show that \[\left| {\text{3A}} \right|\text{=27}\left| \text{A} \right|\].


Ans: Given, 

\[\text{A=}\left[ \begin{matrix} \text{1} & \text{0} & \text{1} \\ \text{0} & \text{1} & \text{2} \\ \text{0} & \text{0} & \text{4} \\ \end{matrix} \right]\] Determining the value of determinant $\text{A}$, by expanding along the first column, i.e., \[\text{C1}\], we get: \[\Rightarrow \left| \text{A} \right|\text{=1}\left| \begin{matrix} \text{1} & \text{2} \\ \text{0} & \text{4} \\ \end{matrix} \right|\text{-0}\left| \begin{matrix} \text{0} & \text{1} \\ \text{0} & \text{4} \\ \end{matrix} \right|\text{+0}\left| \begin{matrix} \text{0} & \text{1} \\ \text{1} & \text{2} \\ \end{matrix} \right|\] \[\Rightarrow \left| \text{A} \right|\text{=}1\left( 4-0 \right)-0+0\] \[\Rightarrow \left| \text{A} \right|\text{=4}\] \[\therefore 27\left| \text{A} \right|\text{=27}\times \text{4}=\text{108}\] ……(1) The value of $\left| 3\text{A} \right|$ is obtained as: \[\begin{align} & \Rightarrow \text{3A=3}\left[ \begin{matrix} \text{1} & \text{0} & \text{1} \\ \text{0} & \text{1} & \text{2} \\ \text{0} & \text{0} & \text{4} \\ \end{matrix} \right] \\ & \Rightarrow \text{3A=}\left[ \begin{matrix} \text{3} & \text{0} & \text{3} \\ \text{0} & \text{3} & \text{6} \\ \text{0} & \text{0} & \text{12} \\ \end{matrix} \right] \\ \end{align}\] \[\therefore \left| \text{3A} \right|\text{=3}\left| \begin{matrix} \text{3} & \text{6} \\ \text{0} & \text{12} \\ \end{matrix} \right|\text{-0}\left| \begin{matrix} \text{0} & \text{3} \\ \text{0} & \text{12} \\ \end{matrix} \right|\text{+0}\left| \begin{matrix} \text{0} & \text{3} \\ \text{3} & \text{6} \\ \end{matrix} \right|\] \[\Rightarrow \text{3}\left( \text{36-0} \right)\text{+0+0}\] \[\Rightarrow \left| \text{3A} \right|\text{=3}\times \text{36}\] Thus, \[\left| \text{3A} \right|\text{=}108\] ……(2) From equations (1) and (2), we have: \[\left| \text{3A} \right|\text{=27}\left| \text{A} \right|\]

Hence proved.

5. Evaluate the determinants

i. \[\left| \begin{matrix} \text{3} & \text{-1} & \text{-2} \\ \text{0} & \text{0} & \text{-1} \\ \text{3} & \text{-5} & \text{0} \\ \end{matrix} \right|\]

Ans:Let \[\text{A=}\left| \begin{matrix} \text{3} & \text{-1} & \text{-2} \\ \text{0} & \text{0} & \text{-1} \\ \text{3} & \text{-5} & \text{0} \\ \end{matrix} \right|\] Determining the value of $\text{A}$ by expanding along the second row, we have: \[\Rightarrow \left| \text{A} \right|\text{=}-0\left| \begin{matrix} -1 & -2 \\ -5 & 0 \\ \end{matrix} \right|+0\left| \begin{matrix} 3 & -2 \\ 3 & 0 \\ \end{matrix} \right|-\left( -1 \right)\left| \begin{matrix} 3 & -1 \\ 3 & -5 \\ \end{matrix} \right|\] \[\Rightarrow \left| \text{A} \right|\text{=}\left( \text{-15+3} \right)\] \[\therefore \left| \text{A} \right|\text{=-12}\]


ii.\[\left| \begin{matrix} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \\ \end{matrix} \right|\]

Ans: Let \[\text{A=}\left| \begin{matrix} \text{3} & \text{-4} & \text{5} \\ \text{1} & \text{1} & \text{-2} \\ \text{2} & \text{3} & \text{1} \\ \end{matrix} \right|\] Determining the value of $\text{A}$ by expanding along the first row, we have: \[\Rightarrow \left| \text{A} \right|\text{=3}\left| \begin{matrix} \text{1} & \text{-2} \\ \text{3} & \text{1} \\ \end{matrix} \right|\text{+4}\left| \begin{matrix} \text{1} & \text{-2} \\ \text{2} & \text{1} \\ \end{matrix} \right|\text{+5}\left| \begin{matrix} \text{1} & \text{1} \\ \text{2} & \text{3} \\ \end{matrix} \right|\] \[\Rightarrow \left| \text{A} \right|\text{=3}\left( \text{1+6} \right)\text{+4}\left( \text{1+4} \right)\text{+5}\left( \text{3-2} \right)\] \[\Rightarrow \left| \text{A} \right|\text{=21+20+5}\] \[\therefore \left| \text{A} \right|\text{=}46\]

iii. \[\left| \begin{matrix} \text{0} & \text{1} & \text{2} \\ \text{-1} & \text{0} & \text{-3} \\ \text{-2} & \text{3} & \text{0} \\ \end{matrix} \right|\]

Ans: Let \[\text{A=}\left| \begin{matrix} \text{0} & \text{1} & \text{2} \\ \text{-1} & \text{0} & \text{-3} \\ \text{-2} & \text{3} & \text{0} \\ \end{matrix} \right|\] Determining the value of $\text{A}$ by expanding along the first row, we have: \[\Rightarrow \left| \text{A} \right|\text{=0}\left| \begin{matrix} \text{0} & \text{-3} \\ \text{3} & \text{0} \\ \end{matrix} \right|\text{-1}\left| \begin{matrix} \text{-1} & \text{-3} \\ \text{-2} & \text{0} \\ \end{matrix} \right|\text{+2}\left| \begin{matrix} \text{-1} & \text{0} \\ \text{-2} & \text{3} \\ \end{matrix} \right|\] \[\Rightarrow \left| \text{A} \right|\text{=0}\left( 9 \right)-\left( -6 \right)+2\left( -3 \right)\] \[\therefore \left| \text{A} \right|\text{=0}\]

iv. $\left| \begin{matrix} \text{2} & \text{-1} & \text{-2} \\ \text{0} & \text{2} & \text{-1} \\ \text{3} & \text{-5} & \text{0} \\ \end{matrix} \right|$

Ans: Let $\text{A=}\left| \begin{matrix} \text{2} & \text{-1} & \text{-2} \\ \text{0} & \text{2} & \text{-1} \\ \text{3} & \text{-5} & \text{0} \\ \end{matrix} \right|$ Determining the value of $\text{A}$ by expanding along the first column, we have: \[\Rightarrow \left| \text{A} \right|\text{=2}\left| \begin{matrix} \text{2} & \text{-1} \\ \text{-5} & \text{0} \\ \end{matrix} \right|\text{-0}\left| \begin{matrix} \text{-1} & \text{-2} \\ \text{-5} & \text{0} \\ \end{matrix} \right|\text{+3}\left| \begin{matrix} \text{-1} & \text{-2} \\ \text{2} & \text{-1} \\ \end{matrix} \right|\] \[\Rightarrow \left| \text{A} \right|\text{=}2\left( -5 \right)-0+3\left( 5 \right)\] \[\Rightarrow \left| \text{A} \right|\text{=}-\text{10+15}\] \[\therefore \left| \text{A} \right|\text{=5}\]

6. If \[\text{A=}\left[ \begin{matrix} 1 & 1 & -2  \\   2 & 1 & -3  \\ 5 & 4 & -9  \\ \end{matrix} \right]\], find \[\left| \text{A} \right|\].

Ans: Given,\[\text{A=}\left[ \begin{matrix} \text{1} & \text{1} & \text{-2} \\ \text{2} & \text{1} & \text{-3} \\ \text{5} & \text{4} & \text{-9} \\ \end{matrix} \right]\] Determining the value of $\text{A}$ by expanding along the first row, we have: \[\Rightarrow \left| \text{A} \right|\text{=1}\left| \begin{matrix} \text{1} & \text{-3} \\ \text{4} & \text{-9} \\ \end{matrix} \right|\text{-1}\left| \begin{matrix} \text{2} & \text{-3} \\ \text{5} & \text{-9} \\ \end{matrix} \right|\text{-2}\left| \begin{matrix} \text{2} & \text{1} \\ \text{5} & \text{4} \\ \end{matrix} \right|\] \[\Rightarrow \left| \text{A} \right|\text{=}1\left( -9+12 \right)-1\left( -18+15 \right)-2\left( 8-5 \right)\] \[\Rightarrow \left| \text{A} \right|\text{=}3+3-6\] \[\therefore \left| \text{A} \right|\text{=}0\]

7. Find values of \[x\] , if

i. \[\left| \begin{matrix} \text{2} & \text{4} \\ \text{5} & \text{1} \\ \end{matrix} \right|\text{=}\left| \begin{matrix} \text{2x} & \text{4} \\ \text{6} & \text{x} \\ \end{matrix} \right|\] Ans: Given, \[\left| \begin{matrix} \text{2} & \text{4} \\ \text{5} & \text{1} \\ \end{matrix} \right|\text{=}\left| \begin{matrix} \text{2x} & \text{4} \\ \text{6} & \text{x} \\ \end{matrix} \right|\] Solving it, we have: \[\Rightarrow \left( 2\times 1 \right)-\left( 5\times 4 \right)=\left( 2x\times x \right)-\left( 6\times 4 \right)\] \[\Rightarrow 2-20=2{{x}^{2}}-24\] \[\Rightarrow -18+24=2{{x}^{2}}\] \[\Rightarrow 3={{x}^{2}}\] Applying square root on both the sides, we obtain: \[\Rightarrow \text{x = }\!\!\pm\!\!\text{ }\sqrt{\text{3}}\] ii. \[\left| \begin{matrix} \text{2} & \text{3} \\ \text{4} & \text{5} \\ \end{matrix} \right|\text{=}\left| \begin{matrix} \text{x} & \text{3} \\ \text{2x} & \text{5} \\ \end{matrix} \right|\] Ans: Given, \[\left| \begin{matrix} \text{2} & \text{3} \\ \text{4} & \text{5} \\ \end{matrix} \right|\text{=}\left| \begin{matrix} \text{x} & \text{3} \\ \text{2x} & \text{5} \\ \end{matrix} \right|\] Solving it, we have: \[\Rightarrow \left( \text{2 }\!\!\times\!\!\text{ 5} \right)-\left( \text{3 }\!\!\times\!\!\text{ 4} \right)\text{=}\left( \text{x }\!\!\times\!\!\text{ 5} \right)-\left( \text{3 }\!\!\times\!\!\text{ 2x} \right)\] \[\Rightarrow \text{10}-\text{12=5x}-\text{6x}\] \[\Rightarrow -\text{2=}-\text{x}\] Multiplying by $\left( -1 \right)$ on both the sides, we obtain: \[\Rightarrow \text{x = 2}\]

8. If \[\left| \begin{matrix} x & 2  \\ 18 & x  \\ \end{matrix} \right|\text{=}\left| \begin{matrix} 6 & 2 \\  18 & 6  \\ \end{matrix} \right|\], then \[x\] is equal to

  1. \[\text{6}\]

  2. \[\text{ }\!\!\pm\!\!\text{ 6}\]

  3. \[\text{-6}\]

  4. \[\text{0}\]

Ans: Given,\[\left| \begin{matrix} \text{x} & \text{2} \\ \text{18} & \text{x} \\ \end{matrix} \right|\text{=}\left| \begin{matrix} \text{6} & \text{2} \\ \text{18} & \text{6} \\ \end{matrix} \right|\] Solving it, we have: \[\Rightarrow {{\text{x}}^{\text{2}}}-\text{36 = 36}-\text{36}\] \[\Rightarrow {{\text{x}}^{\text{2}}}-\text{36=0}\] \[\Rightarrow {{\text{x}}^{\text{2}}}\text{=36}\] Applying square root on both the sides, we obtain: \[\Rightarrow \text{x= }\!\!\pm\!\!\text{ 6}\] Hence, B. \[\text{ }\!\!\pm\!\!\text{ 6}\] is the correct answer.


NCERT Solutions for Class 12 Maths PDF Download

 

NCERT Solution Class 12 Maths of Chapter 4 All Exercises

Chapter 4 - Determinants Exercises in PDF Format

Exercise 4.1

8 Questions & Solutions (3 Short Answers, 5 Long Answers)

Exercise 4.2

10 Questions & Solutions (4 Short Answers, 10 Long Answers)

Exercise 4.3

5 Questions & Solutions (2 Short Answers, 3 Long Answers)

Exercise 4.4

5 Questions & Solutions (2 Short Answers, 3 Long Answers)

Exercise 4.5

18 Questions & Solutions (4 Short Answers, 14 Long Answers)

Exercise 4.6

16 Questions & Solutions (3 Short Answers, 13 Long Answers)


NCERT Solutions for Class 12 Maths Chapter 4 - Determinants

Class 12 Maths Exercise 4.1 Solutions

‘Determinants’ is an essential chapter for students taking their Class 12 board exams. A determinant is calculated on a square matrix, and the obtained value is primarily a scalar quantity. The final value is to be computed based on particular properties and parameters of linear algebra.


Initially, it might seem confusing to students; however, with our Class 12 Maths Exercise 4.1 Solutions, you can avoid the mayhem. Consequently, you can also get deeper into the concept of determinants starting right from the basics. It will not only strengthen your fundamentals but also make ready for the twisted questions too.


Our solutions by Vedantu will help you to understand the topics clearly and then solve the numerical without any error. It will aid you in fetching higher grades in the exam too. Here is an overview of the exercise and the questions covered under it.


  • Exercise 4.1 Class 12 Maths – Question 1

A 2x2 matrix is given in the first question, and you are asked to find the determinant of it. The elements present in the matrix are integers in this case; however, from our Class 12 Maths Exercise 4.1 Solutions, students will know that they need to consider that these elements could be anything – natural numbers, fractions, decimals or trigonometric identities, etc.


The method of finding the determinant although remains same in all the cases as mentioned above. However, students need to be careful with the process of solving these. Identifying the rows and columns is of utmost importance, and it is the first when delving into determinants.


Therefore, to know the process in a step by step manner, you can refer to our study material for Determinants and Matrices Exercise 4.1. Our standard study materials will guide you throughout the solution of this question and gaining an in-depth knowledge of the same.


  • Exercise 4.1 Maths Class 12 – Question 2

The elements present in this 2x2 matrix are trigonometric identities and algebraic expressions. Students should be well versed with the process of evaluating the determinant to follow the same rule even here.


In our Class 12 Maths Exercise 4.1 Solutions, you will know how to replace the numbers with the provided elements irrespective of their type. However, basic trigonometric rules should also be kept in mind to find an absolute determinant.


The process involves simple expansion of the given matrix and similarly finding the value, as it is done in case of natural numbers. With regular practice, you can get a firm grasp on the chapter and also on the processes. It will help you avoid unnecessary confusion during the exam, and also gain an edge over others by scoring higher.


  • Exercise 4.1 Class 12 – Question 3

This question asks to prove the given equation. You should have prior understanding of multiplying a natural number with the matrix. On the other hand, understanding how to multiply a natural number to the obtained determinant is equally essential.


You should know that both of these are different from each other. However, both the resultants can be equal when a matrix is multiplied by different natural numbers. You can learn to find the results from our Exercise 4.1 Class 12 Maths NCERT solutions in the most natural way.


The best part of such questions is that you can check whether your answer is correct or not right from your solutions. If both the sides of that equation are same, then it is correct otherwise not.

Hence, after every calculation, make sure you equate both the sides and check whether they are equivalent or not. Keeping this small trick in mind will help you in gaining more marks too without worrying about committing any error.


  • Exercise 4.1 Class 12 Maths – Question 4

Here, a 3x3 matrix is given, and you are asked to prove that multiplying its determinant with 27 is equal to multiplying the matrix with 3. A clear understanding of the previous question will help you in determining the approach to these questions.


Besides, you have to know how to multiply a natural number with a 3x3 matrix. Students often confuse between rows and columns while multiplying. To avoid such confusion, you can always refer to our PDF of Class 12 Maths Exercise 4.1 Solutions for clarity.


Our solutions will help you easily determine the elements while multiplying and also keep track of the element that you have already used in your calculations. As a result, you will commit lesser mistakes, and even improve your scores in exam significantly.


  •  Exercise 4.1 Maths Class 12 – Question 5

Under question 5, there are four matrices for which you need to calculate the determinants. All these are 3x3 matrices, and they are evaluated similarly as you evaluate for a 2x2 matrix.


However, there is a catch here; if you find a particular row to have more than one element as zero, then you can reduce the determinant equation to a single term. The other two terms get cancelled as they are multiplied by zero.

From our Class 12 Maths Chapter 4 Exercise 4.1 solutions, you can learn these tricks easily. We have explained the concept as well as have provided short tricks to help you memorise these rules effortlessly.


  • Maths Chapter 4 Exercise 4.1 – Question 6

Question 6 involves a very straightforward approach as you are asked to determine the value of the determinant of this 3x3 matrix. You should be careful while expanding the rows for calculation. With every expansion, one of the elements in the first row gets omitted, and you have to consider the other elements.


Besides, make sure you are well acquainted with the process of determinant calculation to score high in your board exam. Also, you can cross-check with our Class 12 Maths Exercise 4.1 Solutions for strict evaluation of the answers that you have calculated.


It will give you an insight into the step by solutions of the same questions. Also, you may learn the simple tricks to calculate faster in the exam.


  •  Exercise 4.1 Class 12 – Question 7

Here you have two matrices given in each question, where you need to find the value of ‘x’ from one of the matrices by equating them. To compare them, you need to frame the equation at first.


Be careful when dealing with x2 terms, as you are likely to obtain two values for the same. One of the values is negative, and the other is positive; therefore, you need to consider both the values while writing your final answer.


To get a stronghold on such calculations, you can avail our Class 12 Maths Exercise 4.1 Solutions and know how to use the rules of finding determinants.


  • Exercise 4.1 Maths Class 12 – Question 8

It is the last question in exercise 4.1, and it is quite similar to the previous one. You are required to calculate the value of ‘x’ from the given two matrices which are given to be equal.


Hence, at first, you have to find the determinant of each matrix in terms of ‘x’ and thereafter equate them to solve for the value of ‘x’. Knowing it for the first time often seems a little complicated; however, with regular practice, you can gain a stronghold on the topic.


Here, you are provided with four options to select the answer out of them. The answer is in square root form, which shows that there are two values. However, do not get confused when you get such an answer, because a determinant can hold any value as an answer.


Simultaneously, you can also avail our Class 12 Maths Exercise 4.1 Solutions for an elaborate understanding. We have compiled the solutions to these questions in a compact way to guide you in solving them in the right way.


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With our Class 12 Maths Exercise 4.1 Solutions, you can significantly benefit as you will get the top quality study materials from Vedantu.


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FAQs on NCERT Solutions for Class 12 Maths Chapter 4: Determinants - Exercise 4.1

1. What is the best solution book for Exercise 4.1 of  Chapter 4 Class 12 Maths? 

Vedantu's NCERT Solutions are the finest study materials for Exercise 4.1 of  Chapter 4 Class 12 Maths. These solutions are designed to help students understand the exercise questions and make the exam preparation process more fun. The content thoroughly covers all of the solutions and practise problems from Chapter 4's first exercise. The answers are simple to comprehend and error-free. Look no further since this is the most up-to-date, accurate, and reliable online math study material available. The PDF of the Solutions may be obtained for free from Vedantu's official website or the Vedantu app.

2. Can you please provide a detailed Stepwise Study Plan to ace Class 12 Maths Chapter 4 Exercise 4.1? 

Before solving problems from Exercise 4.1 of  Chapter 4 Class 12 Maths, the first step is to memorise the formulae and comprehend how they are used. Then, complete Exercise 4.1 by solving all of the questions. To help you with this exercise, you may use Vedantu's NCERT Solutions for Exercise 4.1 by going to the page NCERT Solutions Class 12 Maths Exercise 4.1 or using the Vedantu app; these solutions are accessible for free. Last but not the least, previous year questions from this exercise should be practised in order to ace the CBSE Board Exam.

3. Do I need to practice all the questions provided in Exercise 4.1 of  Chapter 4 Class 12 Maths? 

From an examination standpoint, each question in NCERT of Exercise 4.1 of  Chapter 4 Class 12 Maths is essential. To do well in the Maths exam, you must practise all of the different types of questions included in Exercise 4.1. You will learn something new from each question. For the best results on the Maths exam, practise the problems you find difficult many times and focus more on your areas of weakness.

4. Is it sufficient to practice Exercise 4.1 of  Chapter 4 Class 12 Maths or do I need to practice it from other reference books too? 

To ace Class 12 Maths Board exams, you must, first, be clear with all the questions of Exercise 4.1 from the NCERT Maths standard textbook. It is always better to practice more, so after completing NCERT questions, you can practice the most important questions from other reference books such as RD Sharma and RS Aggarwal. The more you practice, the more you will strengthen your preparation of concepts from Exercise 4.1. Thus, to become a Maths topper in Class 12, you must also practice additional questions based on Exercise 4.1. 

5. What are the most important questions in Exercise 4.1 of  Chapter 4 Class 12 Maths? 

Exercise 4.1 of Class 12 Maths contains various different kinds of questions on Determinants. The most important types of questions given in this exercise are as follows- finding a determinant when a matrix (having elements such as natural numbers, fractions, decimals or trigonometric identities); proving an equation having matrices and determinants; finding the value of 'x' when an equation having two equal matrices are given; determining the value of a determinant. By solving each of these types of questions, you will be able to understand the concept of determinants better.