CBSE Class 12 Mathematics NCERT Solutions: Determinants

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NCERT Solutions PDF Class 12 PDF

This section provides detailed NCERT Solutions for Class 12 Mathematics, focusing on Chapter 4: Determinants. It covers the fundamental concepts of evaluating determinants for 2x2 and 3x3 matrices, along with understanding and applying properties of determinants. The solutions guide students through step-by-step calculations for problems involving determinant evaluation and verification of properties like |kA| = k^n|A|. These solutions are designed to clarify complex concepts and provide a strong foundation for students preparing for their board examinations, ensuring a thorough understanding of determinant calculations and their applications.

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BoardCBSE
ClassClass 12
SubjectMathematics
Session2026
LanguageEnglish
TypeNCERT Solutions
ChapterDeterminants

Chapter summary

Chapter 4, Determinants, for Class 12 Mathematics introduces the concept of determinants as a scalar value associated with a square matrix. This chapter's NCERT Solutions focus on evaluating determinants of order 2 and 3, and exploring key properties such as the relationship between a matrix and its scalar multiple's determinant. The exercises provide practice in applying these rules to solve problems and verify given identities.

Learning outcomes

  • Understand the definition and calculation of a determinant for 2x2 and 3x3 matrices.
  • Evaluate determinants using cofactor expansion.
  • Apply properties of determinants to simplify calculations.
  • Verify identities involving determinants and scalar multiples of matrices.
  • Solve problems related to the evaluation of determinants.

Topics covered

Paper topics

  • Determinants of order 2
  • Determinants of order 3
  • Evaluation of determinants
  • Cofactors and minors
  • Properties of determinants
  • Scalar multiplication of matrices and determinants
  • Relationship between |kA| and |A|

Important topics

  • Evaluation of 2x2 and 3x3 determinants
  • Cofactor expansion method
  • Property: |kA| = k^n|A|
  • Determinant of diagonal and triangular matrices

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Questions and Solutions

Question 1

Evaluate the determinant:

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

Solution:

To evaluate the determinant of a 2x2 matrix \begin{vmatrix} a & b \\ c & d \end{vmatrix}, we use the formula $ad - bc$.

For the given determinant:

\begin{vmatrix}

2 & -5 \\

4 & -1

\end{vmatrix} = (2)(-1) - (-5)(4)

Calculating the products:

= -2 - (-20)

Simplifying the expression:

= -2 + 20 = 18

Thus, the value of the determinant is 18.

Question 2 (i)

Evaluate the determinant:

\begin{vmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{vmatrix}

Solution:

We use the formula for a 2x2 determinant, $ad - bc$.

For the given determinant:

\begin{vmatrix}

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

\sin\theta & \cos\theta

\end{vmatrix} = (\cos\theta)(\cos\theta) - (-\sin\theta)(\sin\theta)

This simplifies to:

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

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

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

Question 2 (ii)

Evaluate the determinant:

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

Solution:

We apply the determinant formula $ad - bc$ to the given 2x2 matrix.

Here, $a = x^2 - x + 1$, $b = x - 1$, $c = x + 1$, and $d = x + 1$.

The determinant is:

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

We can factor out $(x + 1)$ from both terms:

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

Simplify the expression inside the brackets:

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

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

Alternatively, we can expand directly:

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

Expand the first product: $(x^2 - x + 1)(x + 1) = x^3 + x^2 - x^2 - x + x + 1 = x^3 + 1$.

Expand the second product: $(x - 1)(x + 1) = x^2 - 1$.

Now substitute these back:

= (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, we find the matrix $2A$ by multiplying each element of $A$ by 2:

2A = 2 \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}

Now, we calculate the determinant of $2A$ (L.H.S.):

|2A| = \begin{vmatrix} 2 & 4 \\ 8 & 4 \end{vmatrix} = (2)(4) - (4)(8)

= 8 - 32 = -24

Next, we calculate the determinant of $A$:

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

= 2 - 8 = -6

Now, we calculate $4|A|$ (R.H.S.):

4|A| = 4 \times (-6) = -24

Comparing the L.H.S. and R.H.S., we see that $|2A| = -24$ and $4|A| = -24$.

\therefore |2A| = 4|A|

This verifies the property for the given matrix.

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, we calculate the determinant of $A$. Since the first column has two zeros, we expand along the first column ($C_1$) for easier calculation.

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

Evaluating the remaining 2x2 determinant:

|A| = 1 [(1)(4) - (2)(0)] - 0 + 0 = 1(4 - 0) = 4

Now, we calculate $27|A|$:

27|A| = 27 \times 4 = 108

Next, we find the matrix $3A$ by multiplying each element of $A$ by 3:

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

Now, we calculate the determinant of $3A$. We can expand along the first column ($C_1$) again, as it contains two zeros.

|3A| = 3 \begin{vmatrix} 3 & 6 \\ 0 & 12 \end{vmatrix} - 0 \begin{vmatrix} 0 & 3 \\ 0 & 12 \end{vmatrix} + 0 \begin{vmatrix} 0 & 3 \\ 3 & 6 \end{vmatrix}

Evaluating the remaining 2x2 determinant:

|3A| = 3 [(3)(12) - (6)(0)] - 0 + 0 = 3(36 - 0) = 3(36) = 108

Comparing the results:

|3A| = 108 \quad \text{and} \quad 27|A| = 108

\therefore |3A| = 27|A|

This confirms the property for the given matrix.

Common mistakes

  • Errors in arithmetic calculations during determinant evaluation.
  • Incorrectly applying the formula for cofactor expansion.
  • Misunderstanding the property |kA| = k^n|A| for matrices of different orders.
  • Sign errors when calculating cofactors or minors.

Revision tips

  • Practice evaluating determinants of various orders using both direct formula and cofactor expansion.
  • Memorize the key properties of determinants and practice applying them to simplify problems.
  • Work through the solved examples carefully to understand the step-by-step approach.
  • Focus on verifying the identities like |kA| = k^n|A| to solidify understanding of scalar multiplication effects.

Practice MCQs

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

Q2. If \begin{vmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{vmatrix} = 1, what is the value of the determinant?

Q3. For a 2x2 matrix A, if |A| = -6, what is |2A|?

Q4. What is the determinant of the matrix A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{bmatrix}?

Q5. For a 3x3 matrix A, if |A| = 4, what is |3A|?

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, such as whether it is invertible.

How do you evaluate a 2x2 determinant?

For a matrix \begin{bmatrix} a & b \\ c & d \end{bmatrix}, the determinant is calculated as ad - bc.

How do you evaluate a 3x3 determinant?

A 3x3 determinant can be evaluated using cofactor expansion along any row or column. The formula for expansion along the first row is a(ei − fh) − b(di − fg) + c(dh − eg) for matrix \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}.

What is the property relating |kA| and |A| for a square matrix A of order n?

For a square matrix A of order n, the property is |kA| = k^n|A|, where k is a scalar.

Are these NCERT Solutions for Class 12 Maths Determinants helpful for exams?

Yes, these solutions provide clear, step-by-step explanations for all exercises, helping students understand concepts and practice problem-solving for their board exams.

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