CBSE Class 12 Maths Chapter 4: Determinants NCERT Solutions

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NCERT Solutions PDF Class 12 PDF

CBSE Class 12 Mathematics Chapter 4, Determinants, introduces students to the fundamental concepts and applications of determinants. This chapter delves into calculating determinants for 2x2 and 3x3 matrices, exploring their essential properties, and applying these to solve systems of linear equations. The NCERT Solutions provide detailed, step-by-step explanations for various problems, including finding determinant values, verifying properties such as the relationship between the determinant of a scalar multiple of a matrix and the determinant of the original matrix (|kA| = k^n|A|), and solving for unknown variables within determinant expressions. These solutions are crafted to simplify complex topics, ensuring students gain a solid understanding and can confidently tackle exam questions. They serve as a crucial study aid for effective preparation and mastery of determinants.

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BoardCBSE
ClassClass 12
Subjectगणित
Session2026
LanguageEnglish
TypeNCERT Solutions
Chapter4. सारणिक

Chapter summary

Chapter 4, Determinants, for Class 12 Maths NCERT Solutions focuses on understanding and calculating determinants for matrices. It covers evaluating determinants of various orders, including 2x2 and 3x3 matrices, and applying properties to simplify calculations. The exercises involve finding determinant values, proving identities related to scalar multiplication of matrices and their determinants, and solving equations where determinants are equated. This chapter builds a foundational understanding of determinants, crucial for further topics in linear algebra.

Learning outcomes

  • Understand the concept of determinants for square matrices.
  • Calculate the determinant of 2x2 and 3x3 matrices.
  • Apply properties of determinants to simplify calculations.
  • Verify relationships between determinants and scalar multiples of matrices.
  • Solve equations involving determinants for unknown variables.

Topics covered

Paper topics

  • Determinants of a matrix
  • Value of determinants (2x2, 3x3)
  • Expansion of determinants
  • Properties of determinants
  • Scalar multiplication of matrices and determinants
  • Solving equations using determinants

Important topics

  • Calculating determinants of 2x2 and 3x3 matrices
  • Properties of determinants
  • Relationship between |kA| and |A|
  • Solving determinant equations

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Questions and Solutions

Question 1

Find the value of the determinant: \begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix}
Solution:

To find the value of the determinant

\begin{vmatrix}

2 & 4 \\

-5 & -1

\end{vmatrix}

, we use the formula for a 2x2 determinant:

\begin{vmatrix}

a & b \\

c & d

\end{vmatrix} = ad - bc

Applying this formula:

(2 \times -1) - (4 \times -5)

= -2 - (-20)

= -2 + 20

= 18

Thus, the value of the determinant is 18.

Question 2

Find the value of the following determinants:
  1. \begin{vmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{vmatrix}
  2. \begin{vmatrix} x^2 - x + 1 & x - 1 \\ x + 1 & x + 1 \end{vmatrix}
Solution:

Part (i):

We need to find the value of the determinant

\begin{vmatrix}

\cos \theta & -\sin \theta \\

\sin \theta & \cos \theta

\end{vmatrix} . Using the formula $ad - bc$:

(\cos \theta \times \cos \theta) - (-\sin \theta \times \sin \theta)

= \cos^2 \theta - (-\sin^2 \theta)

= \cos^2 \theta + \sin^2 \theta

Using the trigonometric identity $\cos^2 \theta + \sin^2 \theta = 1$, the value of the determinant is 1.

Part (ii):

We need to find the value of the determinant

\begin{vmatrix}

x^2 - x + 1 & x - 1 \\

x + 1 & x + 1

\end{vmatrix} . Using the formula $ad - bc$:

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

First, expand $(x^2 - x + 1)(x + 1)$. This is a sum of cubes formula: $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$. Here, $a=x$ and $b=1$, so $(x+1)(x^2 - x + 1) = x^3 + 1^3 = x^3 + 1$.

Next, expand $(x - 1)(x + 1)$. This is a difference of squares formula: $a^2 - b^2 = (a-b)(a+b)$. So, $(x - 1)(x + 1) = x^2 - 1^2 = x^2 - 1$.

Now substitute these back into the determinant calculation:

(x^3 + 1) - (x^2 - 1)

= x^3 + 1 - x^2 + 1

= x^3 - x^2 + 2

Thus, the value of the determinant is $x^3 - x^2 + 2$.

Question 3

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

Given the matrix

A = \begin{bmatrix}

1 & 2 \\

4 & 2

\end{bmatrix} . First, let's find the determinant of A, denoted as |A|:

|A| = \begin{vmatrix}

1 & 2 \\

4 & 2

\end{vmatrix}

= (1 \times 2) - (2 \times 4)

= 2 - 8

= -6

Now, let's find the matrix 2A. We multiply each element of A by 2:

2A = 2 \times \begin{bmatrix}

1 & 2 \\

4 & 2

\end{bmatrix} = \begin{bmatrix}

2 \times 1 & 2 \times 2 \\

2 \times 4 & 2 \times 2

\end{bmatrix} = \begin{bmatrix}

2 & 4 \\

8 & 4

\end{bmatrix}

Next, let's find the determinant of 2A, denoted as |2A|:

|2A| = \begin{vmatrix}

2 & 4 \\

8 & 4

\end{vmatrix}

= (2 \times 4) - (4 \times 8)

= 8 - 32

= -24

Now, let's calculate 4|A|:

4|A| = 4 \times (-6)

= -24

Comparing the values, we see that $|2A| = -24$ and $4|A| = -24$. Therefore, we have shown that $|2A| = 4|A|$. This is consistent with the property $|kA| = k^n|A|$ for an n x n matrix, where here n=2, so $|2A| = 2^2|A| = 4|A|$.

Question 4

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

Given the matrix

A = \begin{bmatrix}

1 & 0 & 1 \\

0 & 1 & 2 \\

0 & 0 & 4

\end{bmatrix} . First, let's find the determinant of A, denoted as |A|. We can expand along the first row:

|A| = \begin{vmatrix}

1 & 0 & 1 \\

0 & 1 & 2 \\

0 & 0 & 4

\end{vmatrix}

= 1 \times \begin{vmatrix}

1 & 2 \\

0 & 4

\end{vmatrix} - 0 \times \begin{vmatrix}

0 & 2 \\

0 & 4

\end{vmatrix} + 1 \times \begin{vmatrix}

0 & 1 \\

0 & 0

\end{vmatrix}

= 1 \times ((1 \times 4) - (2 \times 0)) - 0 + 1 \times ((0 \times 0) - (1 \times 0))

= 1 \times (4 - 0) - 0 + 1 \times (0 - 0)

= 1 \times 4 = 4

So, $|A| = 4$.

Now, let's find the matrix 3A. We multiply each element of A by 3:

3A = 3 \times \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}

Next, let's find the determinant of 3A, denoted as |3A|. Expanding along the first row:

|3A| = \begin{vmatrix}

3 & 0 & 3 \\

0 & 3 & 6 \\

0 & 0 & 12

\end{vmatrix}

= 3 \times \begin{vmatrix}

3 & 6 \\

0 & 12

\end{vmatrix} - 0 \times \begin{vmatrix}

0 & 6 \\

0 & 12

\end{vmatrix} + 3 \times \begin{vmatrix}

0 & 3 \\

0 & 0

\end{vmatrix}

= 3 \times ((3 \times 12) - (6 \times 0)) - 0 + 3 \times ((0 \times 0) - (3 \times 0))

= 3 \times (36 - 0) - 0 + 3 \times (0 - 0)

= 3 \times 36 = 108

So, $|3A| = 108$.

Now, let's calculate $27|A|$:

27|A| = 27 \times 4

= 108

Comparing the values, we see that $|3A| = 108$ and $27|A| = 108$. Therefore, we have shown that $|3A| = 27|A|$. This aligns with the property $|kA| = k^n|A|$ for an n x n matrix, where here n=3, so $|3A| = 3^3|A| = 27|A|$.

Question 5

Find the value of the following determinants:
  1. \begin{vmatrix} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{vmatrix}
  2. \begin{vmatrix} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{vmatrix}
  3. \begin{vmatrix} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{vmatrix}
  4. \begin{vmatrix} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{vmatrix}
Solution:

Part (i):

We need to find the value of the determinant

\begin{vmatrix}

3 & -1 & -2 \\

0 & 0 & -1 \\

3 & -5 & 0

\end{vmatrix}

. Let's expand along the first row:

= 3 \times \begin{vmatrix}

0 & -1 \\

-5 & 0

\end{vmatrix} - (-1) \times \begin{vmatrix}

0 & -1 \\

3 & 0

\end{vmatrix} + (-2) \times \begin{vmatrix}

0 & 0 \\

3 & -5

\end{vmatrix}

= 3 \times ((0 \times 0) - (-1 \times -5)) + 1 \times ((0 \times 0) - (-1 \times 3)) - 2 \times ((0 \times -5) - (0 \times 3))

= 3 \times (0 - 5) + 1 \times (0 - (-3)) - 2 \times (0 - 0)

= 3 \times (-5) + 1 \times 3 - 2 \times 0

= -15 + 3 - 0

= -12

The value of the determinant is -12.

Part (ii):

We need to find the value of the determinant

\begin{vmatrix}

3 & -4 & 5 \\

1 & 1 & -2 \\

2 & 3 & 1

\end{vmatrix}

. Let's expand along the first row:

= 3 \times \begin{vmatrix}

1 & -2 \\

3 & 1

\end{vmatrix} - (-4) \times \begin{vmatrix}

1 & -2 \\

2 & 1

\end{vmatrix} + 5 \times \begin{vmatrix}

1 & 1 \\

2 & 3

\end{vmatrix}

= 3 \times ((1 \times 1) - (-2 \times 3)) + 4 \times ((1 \times 1) - (-2 \times 2)) + 5 \times ((1 \times 3) - (1 \times 2))

= 3 \times (1 - (-6)) + 4 \times (1 - (-4)) + 5 \times (3 - 2)

= 3 \times (1 + 6) + 4 \times (1 + 4) + 5 \times 1

= 3 \times 7 + 4 \times 5 + 5

= 21 + 20 + 5

= 46

The value of the determinant is 46.

Part (iii):

We need to find the value of the determinant

\begin{vmatrix}

0 & 1 & 2 \\

-1 & 0 & -3 \\

-2 & 3 & 0

\end{vmatrix}

. Let's expand along the first row:

= 0 \times \begin{vmatrix}

0 & -3 \\

3 & 0

\end{vmatrix} - 1 \times \begin{vmatrix}

-1 & -3 \\

-2 & 0

\end{vmatrix} + 2 \times \begin{vmatrix}

-1 & 0 \\

-2 & 3

\end{vmatrix}

= 0 \times ((0 \times 0) - (-3 \times 3)) - 1 \times ((-1 \times 0) - (-3 \times -2)) + 2 \times ((-1 \times 3) - (0 \times -2))

= 0 \times (0 - (-9)) - 1 \times (0 - 6) + 2 \times (-3 - 0)

= 0 \times 9 - 1 \times (-6) + 2 \times (-3)

= 0 + 6 - 6 = 0 The value of the determinant is 0.

Part (iv):

We need to find the value of the determinant

\begin{vmatrix}

2 & -1 & -2 \\

0 & 2 & -1 \\

3 & -5 & 0

\end{vmatrix}

. Let's expand along the first row:

= 2 \times \begin{vmatrix}

2 & -1 \\

-5 & 0

\end{vmatrix} - (-1) \times \begin{vmatrix}

0 & -1 \\

3 & 0

\end{vmatrix} + (-2) \times \begin{vmatrix}

0 & 2 \\

3 & -5

\end{vmatrix}

= 2 \times ((2 \times 0) - (-1 \times -5)) + 1 \times ((0 \times 0) - (-1 \times 3)) - 2 \times ((0 \times -5) - (2 \times 3))

= 2 \times (0 - 5) + 1 \times (0 - (-3)) - 2 \times (0 - 6)

= 2 \times (-5) + 1 \times 3 - 2 \times (-6)

= -10 + 3 + 12 = 5 The value of the determinant is 5.

Question 6

If A = \begin{bmatrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{bmatrix} , then find the value of |A| .
Solution:

We need to find the determinant of the matrix

A = \begin{bmatrix}

1 & 1 & -2 \\

2 & 1 & -3 \\

5 & 4 & -9

\end{bmatrix} . |A| = \begin{vmatrix}

1 & 1 & -2 \\

2 & 1 & -3 \\

5 & 4 & -9

\end{vmatrix}

Let's expand the determinant along the first row:

= 1 \times \begin{vmatrix}

1 & -3 \\

4 & -9

\end{vmatrix} - 1 \times \begin{vmatrix}

2 & -3 \\

5 & -9

\end{vmatrix} + (-2) \times \begin{vmatrix}

2 & 1 \\

5 & 4

\end{vmatrix}

= 1 \times ((1 \times -9) - (-3 \times 4)) - 1 \times ((2 \times -9) - (-3 \times 5)) - 2 \times ((2 \times 4) - (1 \times 5))

= 1 \times (-9 - (-12)) - 1 \times (-18 - (-15)) - 2 \times (8 - 5)

= 1 \times (-9 + 12) - 1 \times (-18 + 15) - 2 \times 3

= 1 \times 3 - 1 \times (-3) - 6

= 3 + 3 - 6 = 0 The value of the determinant |A| is 0.

Question 7

Find the value(s) of x if:
  1. \begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix} = \begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}
  2. \begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} = \begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix}
Solution:

Part (i):

We are given the equation:

\begin{vmatrix}

2 & 4 \\

5 & 1

\end{vmatrix} = \begin{vmatrix}

2x & 4 \\

6 & x

\end{vmatrix}

First, calculate the determinant on the left side:

(2 \times 1) - (4 \times 5) = 2 - 20 = -18

Next, calculate the determinant on the right side:

(2x \times x) - (4 \times 6) = 2x^2 - 24

Now, equate the two values:

-18 = 2x^2 - 24

Solve for x:

2x^2 = -18 + 24

2x^2 = 6

x^2 = 3

x = \pm \sqrt{3}

The values of x are $\sqrt{3}$ and $-\sqrt{3}$.

Part (ii):

We are given the equation:

\begin{vmatrix}

2 & 3 \\

4 & 5

\end{vmatrix} = \begin{vmatrix}

x & 3 \\

2x & 5

\end{vmatrix}

First, calculate the determinant on the left side:

(2 \times 5) - (3 \times 4) = 10 - 12 = -2

Next, calculate the determinant on the right side:

(x \times 5) - (3 \times 2x) = 5x - 6x

Now, equate the two values:

-2 = 5x - 6x

-2 = -x

Solve for x:

x = 2

The value of x is 2.

Common mistakes

  • Errors in applying the cofactor expansion formula.
  • Sign errors during determinant calculation.
  • Incorrectly applying scalar multiplication properties to determinants.
  • Algebraic mistakes when solving equations involving determinants.

Revision tips

  • Practice calculating determinants for various matrices until the process is automatic.
  • Memorize the cofactor expansion formula and practice its application.
  • Focus on understanding the properties of determinants, as they significantly simplify problems.
  • Work through the examples and exercises systematically to build confidence.

Practice MCQs

Q1. What is the value of the determinant \begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix}?

Q2. For a 2x2 matrix A, if |kA| = m|A|, what is the value of m?

Q3. What is the value of the determinant \begin{vmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{vmatrix}?

Q4. If A is a 3x3 matrix, what is the relationship between |3A| and |A|?

Q5. What is the value of x if \begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix} = \begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}?

Frequently asked questions

What is a determinant in Class 12 Maths?

A determinant is a scalar value that can be computed from the elements of a square matrix. It provides important information about the matrix and is used in solving systems of linear equations.

How do you calculate the determinant of a 2x2 matrix?

For a matrix [a b; c d], the determinant is calculated as (a*d) - (b*c).

What is the key property related to scalar multiplication of a matrix and its determinant?

For an n x n matrix A and a scalar k, the determinant of kA is given by |kA| = k^n|A|.

Are these NCERT Solutions for Chapter 4 helpful for exam preparation?

Yes, these solutions provide clear, step-by-step explanations for all exercises in Chapter 4: Determinants, which is crucial for understanding the concepts and performing well in exams.

What kind of problems are covered in this chapter's solutions?

The solutions cover finding the values of determinants, verifying identities involving determinants and scalar multiples of matrices, and solving equations where determinants are equated.

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