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

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

Edited By Ramraj Saini | Updated on Dec 03, 2023 03:56 PM IST | #CBSE Class 12th

NCERT Solutions For Class 12 Maths Chapter 4 Exercise 4.1

NCERT Solutions for Exercise 4.1 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 exercise 4.1 consists of questions related to finding determinants of matrices of a different order. Determinant of the matrix is the factor by which its volume blows up. There are a lot of applications of determinants like calculating the value of square matrices, solving a set of linear equations, check whether a matrix has an inverse, evaluate the inverse of the matrix using cofactors, etc.

In exercise 4.1 Class 12 Maths, there are 8 questions related to finding determinants of matrices of a different order. Class 12th Maths chapter 4 exercise 4.1 questions are solved with necessary steps. Only knowing the answer is not enough, you need to know how to write solutions in board exams to get good marks. 12th class Maths exercise 4.1 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.

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Assess NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.1

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Determinants Exercise:4.1

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

Answer:

The determinant is evaluated as follows

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

Question:2(i) Evaluate the following determinant- \begin{vmatrix} \cos \theta & -\sin \theta \\ \sin \theta &\cos \theta \end{vmatrix}

Answer:

The given two by two determinant is calculated as follows

\dpi{100} \begin{vmatrix} \cos \theta & -\sin \theta \\ \sin \theta &\cos \theta \end{vmatrix} = cos \theta (\cos \theta) - (-\sin \theta)\sin \theta = \cos^2\theta + \sin ^2 \theta = 1

Question:2(ii) Evaluate the following determinant- \begin{vmatrix}x^2-x+1 & x-1\\x+1 &x+1 \end{vmatrix}

Answer:

We have determinant \begin{vmatrix}x^2-x+1 & x-1\\x+1 &x+1 \end{vmatrix}

So, \dpi{100} \begin{vmatrix}x^2-x+1 & x-1\\x+1 &x+1 \end{vmatrix} = (x^2-x+1)(x+1) - (x-1)(x+1)

= (x+1)(x^2-x+1-x+1) = (x+1)(x^2-2x+2)

=x^3-2x^2+2x +x^2-2x+2

= x^3-x^2+2

Question:3 If A = \begin{bmatrix} 1 & 2\\ 4 &2 \end{bmatrix} , then show that | 2 A |=4|A|

Answer:

Given determinant A = \begin{bmatrix} 1 & 2\\ 4 &2 \end{bmatrix} then we have to show that | 2 A |=4|A|,

So, A = \begin{bmatrix} 1 & 2\\ 4 &2 \end{bmatrix} then, 2A =2 \begin{bmatrix} 1 & 2\\ 4 &2 \end{bmatrix} = \begin{bmatrix} 2 & 4\\ 8 &4 \end{bmatrix}

Hence we have \left | 2A \right | = \begin{vmatrix} 2 &4 \\ 8& 4 \end{vmatrix} = 2(4) - 4(8) = -24

So, L.H.S. = |2A| = -24

then calculating R.H.S. 4\left | A \right |

We have,

\left | A \right | = \begin{vmatrix} 1 &2 \\ 4& 2 \end{vmatrix} = 1(2) - 2(4) = -6

hence R.H.S becomes 4\left | A \right | = 4\times(-6) = -24

Therefore L.H.S. =R.H.S.

Hence proved.

Question:4 If A =\begin{bmatrix} 1 &0 &1 \\ 0& 1& 2\\ 0& 0 &4 \end{bmatrix} then show that |3A|=27|A|

Answer:

Given MatrixA =\begin{bmatrix} 1 &0 &1 \\ 0& 1& 2\\ 0& 0 &4 \end{bmatrix}

Calculating 3A =3\begin{bmatrix} 1 &0 &1 \\ 0& 1& 2\\ 0& 0 &4 \end{bmatrix} = \begin{bmatrix} 3 &0 &3 \\ 0& 3& 6\\ 0& 0 &12 \end{bmatrix}

So, \left | 3A \right | = 3(3(12) - 6(0) ) - 0(0(12)-0(6)) + 3(0-0) = 3(36) = 108

calculating 27|A|,

|A| = \begin{vmatrix} 1 & 0 &1 \\ 0 & 1 & 2\\ 0& 0 &4 \end{vmatrix} = 1\begin{vmatrix} 1 &2 \\ 0 & 4 \end{vmatrix} - 0\begin{vmatrix} 0 &2 \\ 0& 4 \end{vmatrix} + 1\begin{vmatrix} 0 &1 \\ 0& 0 \end{vmatrix} = 4 -0 + 0 = 4

So, 27|A| = 27(4) = 108

Therefore |3A|=27|A|.

Hence proved.

Question:5(i) Evaluate the determinants.

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

Answer:

Given the determinant \begin{vmatrix}3 &-1 &-2 \\0 &0 &-1 \\3 &-5 & 0 \end{vmatrix};

now, calculating its determinant value,

\begin{vmatrix}3 &-1 &-2 \\0 &0 &-1 \\3 &-5 & 0 \end{vmatrix} = 3\begin{vmatrix} 0 &-1 \\ -5& 0 \end{vmatrix} -(-1)\begin{vmatrix} 0 &-1 \\ 3& 0 \end{vmatrix} +(-2)\begin{vmatrix} 0 &0 \\ 3& -5 \end{vmatrix}

= 3(0-5)+1(0+3) -2(0-0) = -15+3-0 = -12.

Question:5(ii) Evaluate the determinants.

\begin{vmatrix}3 &-4 &5 \\1 &1 &-2 \\2 &3 &1 \end{vmatrix}

Answer:

Given determinant \begin{vmatrix}3 &-4 &5 \\1 &1 &-2 \\2 &3 &1 \end{vmatrix};

Now calculating the determinant value;

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

= 3(1+6) +4(1+4) +5(3-2) = 21+20+5 = 46.

Question:5(iii) Evaluate the determinants.

\begin{vmatrix}0 & 1 & 2\\-1 &0 &-3 \\ -2 &3 &0 \end{vmatrix}

Answer:

Given determinant \begin{vmatrix}0 & 1 & 2\\-1 &0 &-3 \\ -2 &3 &0 \end{vmatrix};

Now calculating the determinant value;

\begin{vmatrix}0 & 1 & 2\\-1 &0 &-3 \\ -2 &3 &0 \end{vmatrix} = 0\begin{vmatrix} 0 &-1 \\ 3& 0 \end{vmatrix} -1\begin{vmatrix} -1 &-3 \\ -2& 0 \end{vmatrix}+2\begin{vmatrix} -1 &0 \\ -2& 3 \end{vmatrix}

= 0 - 1(0-6)+2(-3-0) = 6 -6 =0

Question:5(iv) Evaluate the determinants.

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

Answer:

Given determinant: \begin{vmatrix}2 &-1 &-2 \\0 &2 &-1 \\3 &-5 &0 \end{vmatrix},

We now calculate determinant value:

\begin{vmatrix}2 &-1 &-2 \\0 &2 &-1 \\3 &-5 &0 \end{vmatrix} =2\begin{vmatrix} 2 &-1 \\ -5 & 0 \end{vmatrix} -(-1)\begin{vmatrix} 0 &-1 \\ 3 & 0 \end{vmatrix}+(-2)\begin{vmatrix} 0 &2 \\ 3&-5 \end{vmatrix}

=2(0-5)+1(0+3)-2(0-6) = -10+3+12 = 5

Question:6 If A=\begin{bmatrix}1 & 1 & -2\\ 2& 1 &-3 \\5 &4 &-9 \end{bmatrix} , then find |A|.

Answer:

Given the matrix A=\begin{bmatrix}1 & 1 & -2\\ 2& 1 &-3 \\5 &4 &-9 \end{bmatrix} then,

Finding the determinant value of A;

|A| = 1\begin{vmatrix} 1 &-3 \\ 4& -9 \end{vmatrix} -1\begin{vmatrix} 2 &-3 \\ 5& -9 \end{vmatrix}-2\begin{vmatrix} 2 &1 \\ 5& 4 \end{vmatrix}

= 1(-9+12)-1(-18+15)-2(8-5) =3+3-6 =0

Question:7(i) Find values of x, if

\begin{vmatrix}2 &4 \\5 &1 \end{vmatrix} =\begin{vmatrix}2x &4 \\6 &x \end{vmatrix}

Answer:

Given that \begin{vmatrix}2 &4 \\5 &1 \end{vmatrix} =\begin{vmatrix}2x &4 \\6 &x \end{vmatrix}

First, we solve the determinant value of L.H.S. and equate it to the determinant value of R.H.S.,

\dpi{100} \begin{vmatrix}2 &4 \\5 &1 \end{vmatrix} =2-20 = -18 and \begin{vmatrix}2x &4 \\6 &x \end{vmatrix} = 2x(x)-24 = 2x^2-24

So, we have then,

-18= 2x^2-24 or 3= x^2 or x= \pm \sqrt{3}

Question:7(ii) Find values of x, if

\begin{vmatrix}2 &3 \\ 4 &5 \end{vmatrix}=\begin{vmatrix}x &3 \\2x &5 \end{vmatrix}

Answer:

Given \begin{vmatrix}2 &3 \\ 4 &5 \end{vmatrix}=\begin{vmatrix}x &3 \\2x &5 \end{vmatrix};

So, we here equate both sides after calculating each side's determinant values.

L.H.S. determinant value;

\dpi{100} \begin{vmatrix}2 &3 \\ 4 &5 \end{vmatrix}= 10 - 12 = -2

Similarly R.H.S. determinant value;

\begin{vmatrix}x &3 \\2x &5 \end{vmatrix} = 5(x) - 3(2x) = 5x - 6x =-x

So, we have then;

-2 = -x or x =2.

Question:8 If \begin{vmatrix}x &2 \\18 &x \end{vmatrix}=\begin{vmatrix} 6 &2 \\ 18 &6 \end{vmatrix} , then x is equal to

(A) 6 (B) \pm 6 (C) -6 (D) 0

Answer:

Solving the L.H.S. determinant ;

\dpi{100} \begin{vmatrix}x &2 \\18 &x \end{vmatrix}= x^2 - 36

and solving R.H.S determinant;

\begin{vmatrix} 6 &2 \\ 18 &6 \end{vmatrix} = 36-36 = 0

So equating both sides;

x^2 - 36 =0 or x^2 = 36 or x = \pm 6

Hence answer is (B).

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

The first exercise of this chapter of NCERT book is based on the basics of determinants and calculating determinants of a different order of the square matrices. There are 5 solved examples given before the Class 12 Maths ch 4 ex 4.1 that you can solve to get clarity. There are 7 long-answer types of questions and 1 multiple-type question in Class 12th Maths chapter 4 exercise 4.1. There are few questions in this exercise that could be solved using the properties of determinants.

Also Read| Determinants Class 12 Chapter 4 Notes

Benefits of NCERT Solutions for Class 12 Maths Chapter Exercise 4.1:-

  • NCERT Solutions for Class 12 Maths Chapter Exercise 4.1 are very helpful for the students to get conceptual clarity.
  • Exercise 4.1 Class 12 Maths solutions are provided in a very detailed manner, so you can understand them very easily.
  • Class 12 Maths chapter 4 exercise 4.1 solutions can be used for quick revision before the final exam.
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Key Features Of NCERT Solutions for Exercise 4.1 Class 12 Maths Chapter 4

  • Comprehensive Coverage: The solutions encompass all the topics covered in ex 4.1 class 12, ensuring a thorough understanding of the concepts.
  • Step-by-Step Solutions: In this class 12 maths ex 4.1, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
  • Accuracy and Clarity: Solutions for class 12 ex 4.1 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
  • Conceptual Clarity: In this 12th class maths exercise 4.1 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 4.1 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 4.1 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.
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Frequently Asked Questions (FAQs)

1. what is determinant ?

The determinant is a scalar value of a square matrix which characterize some properties of the matrix.

2. If square matrix A is order of 2 then | 2A | = ?

|2A| = 2^2 = 4 | A |

3. How many exercises are there in Class 12 Maths chapter 4 ?

There are 6 exercises and one miscellaneous exercise given in the NCERT textbook Class 12 Maths chapter 4.

4. Does CBSE provides NCERT solutions for Class 12 Maths chapter 4?

No, CBSE doesn't provide NCERT solutions, you can get NCERT solutions for chapter 4 Class 12 Maths.

5. What is the determinant of a singular matrix ?

The determinant of a singular matrix is always zero.

6. What is non-singular matrix?

A matrix that is not singular is called a non-singular matrix. The determinant of a non-singular matrix is non-zero.  

7. What is syllabus for CBSE Class 12 Maths ?
8. how many chapters are there in NCERT Class 12 Maths syllabus?

There are 13 chapters in the NCERT Class 12 Maths book.

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i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

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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.

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Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

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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.

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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:

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