NCERT Solutions Class 12 Maths Chapter 4 Exercise 4.1

NCERT Solutions Class 12 Maths Chapter 4 Exercise 4.1 - Determinants

Edited By Komal Miglani | Updated on May 09, 2025 11:48 AM IST | #CBSE Class 12th

Suppose your friend told you that the following $2 \times 2$ matrix is given: $[\begin{array}{ll} 2 & 3 \\ 4 & 5 \end{array}]$ and asked you to find its determinant. How would you approach this problem? This is where the concept of determinants comes in. Determinants provide a systematic way to find a scalar value from a square matrix, which is crucial in solving systems of equations, finding the area, and understanding matrix properties. The first exercise of this chapter of the NCERT book is based on the basics of determinants and calculating determinants of different orders of square matrices. NCERT Class 12 Maths Chapter 4 - Determinants, Exercise 4.1 introduces us to the basic idea of determinants and how to evaluate them. This article on the NCERT Solutions for Exercise 4.1 Class 12 Maths Chapter 4 offers clear and step-by-step solutions for the exercise problems to help the students understand the method and logic behind it. For syllabus, notes, and PDF, refer to this link: NCERT.

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Class 12 Maths Chapter 4 Exercise 4.1 Solutions: Download PDF

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Determinants Exercise:4.1

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

Answer:

The determinant is evaluated as follows

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

Question:2(i) Evaluate the following determinant- $| \begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix} |$

Answer:

The given two by two determinant is calculated as follows

$| \begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix} | = \cos θ (\cos θ) - (- \sin θ) \sin θ = \cos^{2} θ + \sin^{2} θ = 1$

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

Answer:

We have determinant $| \begin{matrix} x^{2} - x + 1 & x - 1 \\ x + 1 & x + 1 \end{matrix} |$

$| \begin{matrix} x^{2} - x + 1 & x - 1 \\ x + 1 & x + 1 \end{matrix} | = (x^{2} - x + 1) (x + 1) - (x - 1) (x + 1)$

$= (x + 1) (x^{2} - x + 1 - x + 1) = (x + 1) (x^{2} - 2 x + 2)$

$= x^{3} - 2 x^{2} + 2 x + x^{2} - 2 x + 2$

$= x^{3} - x^{2} + 2$

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

Answer:

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

So, $A = [\begin{matrix} 1 & 2 \\ 4 & 2 \end{matrix}]$ then, $2 A = 2 [\begin{matrix} 1 & 2 \\ 4 & 2 \end{matrix}] = [\begin{matrix} 2 & 4 \\ 8 & 4 \end{matrix}]$

Hence we have $| 2 A | = | \begin{matrix} 2 & 4 \\ 8 & 4 \end{matrix} | = 2 (4) - 4 (8) = - 24$

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

then calculating R.H.S. $4 | A |$

We have,

$| A | = | \begin{matrix} 1 & 2 \\ 4 & 2 \end{matrix} | = 1 (2) - 2 (4) = - 6$

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

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

Hence proved.

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

Answer:

Given Matrix $A = [\begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{matrix}]$

Calculating $3 A = 3 [\begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{matrix}] = [\begin{matrix} 3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12 \end{matrix}]$

So, $| 3 A | = 3 (3 (12) - 6 (0)) - 0 (0 (12) - 0 (6)) + 3 (0 - 0) = 3 (36) = 108$

calculating $27 | A |$ ,

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

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

Therefore $| 3 A | = 27 | A |$ .

Hence proved.

Question:5(i) Evaluate the determinants.

$| \begin{matrix} 3 & - 1 & - 2 \\ 0 & 0 & - 1 \\ 3 & - 5 & 0 \end{matrix} |$

Answer:

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

now, calculating its determinant value,

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

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

Question:5(ii) Evaluate the determinants.

$| \begin{matrix} 3 & - 4 & 5 \\ 1 & 1 & - 2 \\ 2 & 3 & 1 \end{matrix} |$

Answer:

Given determinant $| \begin{matrix} 3 & - 4 & 5 \\ 1 & 1 & - 2 \\ 2 & 3 & 1 \end{matrix} |$ ;

Now calculating the determinant value;

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

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

Question:5(iii) Evaluate the determinants.

$| \begin{matrix} 0 & 1 & 2 \\ - 1 & 0 & - 3 \\ - 2 & 3 & 0 \end{matrix} |$

Answer:

Given determinant $| \begin{matrix} 0 & 1 & 2 \\ - 1 & 0 & - 3 \\ - 2 & 3 & 0 \end{matrix} |$ ;

Now calculating the determinant value;

$| \begin{matrix} 0 & 1 & 2 \\ - 1 & 0 & - 3 \\ - 2 & 3 & 0 \end{matrix} | = 0 | \begin{matrix} 0 & - 1 \\ 3 & 0 \end{matrix} | - 1 | \begin{matrix} - 1 & - 3 \\ - 2 & 0 \end{matrix} | + 2 | \begin{matrix} - 1 & 0 \\ - 2 & 3 \end{matrix} |$

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

Question:5(iv) Evaluate the determinants.

$| \begin{matrix} 2 & - 1 & 2 \\ 0 & 2 & - 1 \\ 3 & - 5 & 0 \end{matrix} |$

Answer:

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

We now calculate determinant value:

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

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

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

Answer:

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

Finding the determinant value of A;

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

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

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

$| \begin{matrix} 2 & 4 \\ 5 & 1 \end{matrix} | = | \begin{matrix} 2 x & 4 \\ 6 & x \end{matrix} |$

Answer:

Given that $| \begin{matrix} 2 & 4 \\ 5 & 1 \end{matrix} | = | \begin{matrix} 2 x & 4 \\ 6 & x \end{matrix} |$

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

$| \begin{matrix} 2 & 4 \\ 5 & 1 \end{matrix} | = 2 (1) - 4 (5) = 2 - 20 = - 18$ and $| \begin{matrix} 2 x & 4 \\ 6 & x \end{matrix} | = 2 x (x) - 4 (6) = 2 x^{2} - 24$

So, we have then,

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

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

$| \begin{matrix} 2 & 3 \\ 4 & 5 \end{matrix} | = | \begin{matrix} x & 3 \\ 2 x & 5 \end{matrix} |$

Answer:

Given $| \begin{matrix} 2 & 3 \\ 4 & 5 \end{matrix} | = | \begin{matrix} x & 3 \\ 2 x & 5 \end{matrix} |$ ;

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

L.H.S. determinant value;

$| \begin{matrix} 2 & 3 \\ 4 & 5 \end{matrix} | = 2 (5) - 3 (4) = 10 - 12 = - 2$

Similarly R.H.S. determinant value;

$| \begin{matrix} x & 3 \\ 2 x & 5 \end{matrix} | = 5 (x) - 3 (2 x) = 5 x - 6 x = - x$

So, we have then;

$- 2 = - x$ or $x = 2$ .

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

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

Answer:

Solving the L.H.S. determinant ;

$| \begin{matrix} x & 2 \\ 18 & x \end{matrix} | = x (x) - 2 (18) = x^{2} - 36$

and solving R.H.S determinant;

$| \begin{matrix} 6 & 2 \\ 18 & 6 \end{matrix} | = 36 - 36 = 0$

So equating both sides;

$x^{2} - 36 = 0$ or $x^{2} = 36$ or $x = \pm 6$

Hence answer is (B).

Topics covered in Chapter 4, Determinants: Exercise 4.1

Here are the main topics covered in NCERT Class 12 Chapter 4, Determinants: Exercise 4.1.

1. Definition:

We know that multiplication and addition are basic operations in matrices. In determinants, we evaluate a scalar value from a square matrix using specific rules. For example, for a $2 \times 2$ matrix

$[\begin{array}{ll} a & b \\ c & d \end{array}], the determinant is given by a d - b c$

2. Determinant of a $2 \times 2$ Matrix: Let the matrix $A$ be defined as:

$A = [\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}]$

To evaluate the determinant of a $2 \times 2$ matrix, use the formula:

$\det (A) = a_{11} a_{22} - a_{12} a_{21}$

3. Determinant of a $3 \times 3$ Matrix:

For $3 \times 3$ matrices, the determinant is calculated using expansion by minors.

$| \begin{array}{lll} a & b & c \\ d & e & f \\ g & h & i \end{array} | = a (e i - f h) - b (d i - f g) + c (d h - e g)$

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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 ?

Here you will get 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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Changing from the CBSE board to the Odisha CHSE in Class 12 is generally difficult and often not ideal due to differences in syllabi and examination structures. Most boards, including Odisha CHSE , do not recommend switching in the final year of schooling. It is crucial to consult both CBSE and Odisha CHSE authorities for specific policies, but making such a change earlier is advisable to prevent academic complications.

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Manasvini Gupta 30 January,2025

quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

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Jefferson 25 September,2024

i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

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Seek Support:
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Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.

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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

Hi,

Qualifications:
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Hope you find this useful!

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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

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.

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Ashvadha 12 July,2024

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

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) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

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 .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

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.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

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

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

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NCERT Solutions Class 12 Maths Chapter 4 Exercise 4.1 - Determinants

Class 12 Maths Chapter 4 Exercise 4.1 Solutions: Download PDF

Determinants Exercise:4.1

Topics covered in Chapter 4, Determinants: Exercise 4.1

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