Determinants: Chapter 4 Links
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This exercise of Determinants 4.1 introduces the basic meaning and properties of determinants in Class 12 Maths. We will learn how determinants are defined for square matrices and how they help in simplifying matrix calculations and solving algebraic problems.
In this exercise, we will study determinants of matrices of order 1, 2 and 3 along with the method of expansion using rows and columns. Special attention should be given to sign conventions and choosing suitable rows or columns for easier calculations.
These concepts form the foundation for advanced applications of determinants such as finding area, inverse of matrices and solving systems of linear equations in the upcoming exercises.
Key Concepts
- Meaning of Determinant: For a square matrix A, |A| is read as the determinant of A and not modulus of A.
- Square Matrix Requirement: Determinants are defined only for square matrices.
- Determinant of Order 1: If A = [a] , then the determinant of A is defined as |A| = a .
- Determinant of Order 2: If A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} , then |A| = a_{11}a_{22} - a_{21}a_{12} .
- Determinant of Order 3: Determinants of order 3 are evaluated by expanding along a row or a column.
- Expansion Along Row or Column: For easier calculations, it is better to expand along the row or column containing the maximum number of zeros.
- Sign Convention: While expanding determinants, the signs alternate between positive and negative according to the position of elements.
- Position Rule: While expanding, the sign is positive if (i+j) is even and negative if (i+j) is odd.
Now, let’s begin this exercise of Determinants 4.1 with Question 1
Question 1: Determinants 4.1
1. Evaluate the determinant \begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix}
Solution
For a determinant of order 2, we use:
\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc
Therefore,
\begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix} = (2 \times -1) - (4 \times -5)
= -2 + 20 = 18
Hence, the value of the determinant is 18.
You may already be following Maths Better for important concepts and exam-oriented questions, like integration tricks and solving linear equations using the Matrix Method etc. Similarly, this NCERT Solutions series for Class 12 Maths will help strengthen your concepts and improve problem-solving skills.
Now, let’s move on to the next question of Determinants 4.1.
Question 2: Determinants Ex 4.1
2 (i). Evaluate the determinant \begin{vmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{vmatrix}
Solution
Using the determinant formula for order 2:
\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc
Therefore,
\begin{vmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{vmatrix} = (\cos \theta \times \cos \theta) - (-\sin \theta \times \sin \theta)
= \cos^2 \theta + \sin^2 \theta
Using the identity:
\sin^2 \theta + \cos^2 \theta = 1
Hence,
\begin{vmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{vmatrix} = 1
Therefore, the value of the determinant is 1.
2 (ii). Evaluate the determinant \begin{vmatrix} x^2 - x + 1 & x - 1 \\ x + 1 & x + 1 \end{vmatrix}
Solution
Using the determinant formula:
\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc
Therefore,
\begin{vmatrix} x^2 - x + 1 & x - 1 \\ x + 1 & x + 1 \end{vmatrix} = (x^2 - x + 1)(x + 1) - (x - 1)(x + 1)
Taking (x+1) common,
= (x+1)\left[(x^2 - x + 1) - (x - 1)\right]
= (x+1)(x^2 - 2x + 2)
Or it is
= x^3 - x^2 + 2
Hence, the value of the determinant is x^3 - x^2 + 2 .
Question 3: Determinants 4.1
3. If A = \begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix} , then show that |2A| = 4|A|
Solution
First, find |A| .
|A| = \begin{vmatrix} 1 & 2 \\ 4 & 2 \end{vmatrix}
= (1 \times 2) - (4 \times 2)
i.e.
= 2 - 8 = -6
Now,
2A = \begin{bmatrix} 2 & 4 \\ 8 & 4 \end{bmatrix}
Therefore,
|2A| = \begin{vmatrix} 2 & 4 \\ 8 & 4 \end{vmatrix}
= (2 \times 4) - (8 \times 4)
i.e.
= 8 - 32 = -24
Also,
4|A| = 4(-6) = -24
Hence,
|2A| = 4|A|
Question 4: Ex 4.1 Determinants
4. If A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{bmatrix} , then show that |3A| = 27|A|
Solution
First, find |A| .
|A| = \begin{vmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{vmatrix}
Expanding along the first column,
|A| = 1 \begin{vmatrix} 1 & 2 \\ 0 & 4 \end{vmatrix} - 0 + 0
= 1[(1 \times 4) - (0 \times 2)]
i.e.
= 4
Now, multiply each element of A by 3.
3A = \begin{bmatrix} 3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12 \end{bmatrix}
Therefore,
|3A| = \begin{vmatrix} 3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12 \end{vmatrix}
Expanding along the first column,
|3A| = 3 \begin{vmatrix} 3 & 6 \\ 0 & 12 \end{vmatrix} - 0 + 0
= 3[(3 \times 12) - (0 \times 6)]
i.e.
= 3(36) = 108
Also,
27|A| = 27 \times 4 = 108
Hence,
|3A| = 27|A|
Try solving the next question on your own first, and then compare your approach with the solution given below.
Question 5: Determinants 4.1
5 (i). Evaluate the determinant \begin{vmatrix} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{vmatrix}
Solution
Expanding along the second row, since it contains maximum zeros,
\begin{vmatrix} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{vmatrix} = - 0 + 0 - (-1)\begin{vmatrix} 3 & -1 \\ 3 & -5 \end{vmatrix}
(You can also try expanding along another row or column for practice)
= (+1)\left[(3 \times -5) - (3 \times -1)\right]
= (+1)(-15 + 3)
(All signs and steps are shown clearly for better understanding)
= (+1)(-12)
= -12
Hence, the value of the determinant is -12.
5 (ii). Evaluate the determinant \begin{vmatrix} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{vmatrix}
Solution
Expanding along the first row,
\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}
(Use the simple formula for determinants of order 2)
= 3[(1 \times 1) - (3 \times -2)] + 4[(1 \times 1) - (2 \times -2)] + 5[(1 \times 3) - (2 \times 1)]
= 3(1 + 6) + 4(1 + 4) + 5(3 - 2)
i.e.
= 3(7) + 4(5) + 5(1)
= 21 + 20 + 5
So, it is
= 46
Hence, the value of the determinant is 46.
5 (iii). Evaluate the determinant \begin{vmatrix} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{vmatrix}
Solution
Expanding along the first row,
\begin{vmatrix} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{vmatrix}
(Notice the alternating signs in expansion.)
= 0 - 1\begin{vmatrix} -1 & -3 \\ -2 & 0 \end{vmatrix} + 2\begin{vmatrix} -1 & 0 \\ -2 & 3 \end{vmatrix}
= -[(-1 \times 0) - (-2 \times -3)] + 2[(-1 \times 3) - (-2 \times 0)]
(Watch the negative signs carefully)
= -(0 - 6) + 2(-3 - 0)
= 6 - 6
So, it is
= 0
Hence, the value of the determinant is 0.
5 (iv). Evaluate the determinant \begin{vmatrix} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{vmatrix}
Solution
Expanding along the second row,
\begin{vmatrix} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{vmatrix}
= 0 + 2\begin{vmatrix} 2 & -2 \\ 3 & 0 \end{vmatrix} - (-1)\begin{vmatrix} 2 & -1 \\ 3 & -5 \end{vmatrix}
(After evaluating all the 2×2 determinants), we get
= 2[(2 \times 0) - (3 \times -2)] + [(2 \times -5) - (3 \times -1)]
= 2(0 + 6) + (-10 + 3)
(Just be careful with the signs)
= 12 - 7
= 5
Hence, the value of the determinant is 5.
Question 6: Determinants Ex 4.1
6. If A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & -1 & 1 \\ 1 & 3 & 4 \end{bmatrix} , find |A| .
Solution
We have,
|A| = \begin{vmatrix} 1 & 2 & 3 \\ 2 & -1 & 1 \\ 1 & 3 & 4 \end{vmatrix}
Expanding along the first row,
|A| = 1\begin{vmatrix} -1 & 1 \\ 3 & 4 \end{vmatrix} - 2\begin{vmatrix} 2 & 1 \\ 1 & 4 \end{vmatrix} + 3\begin{vmatrix} 2 & -1 \\ 1 & 3 \end{vmatrix}
= 1[(-1 \times 4) - (3 \times 1)] - 2[(2 \times 4) - (1 \times 1)] + 3[(2 \times 3) - (1 \times -1)]
i.e. |A| = 1(-4 - 3) - 2(8 - 1) + 3(6 + 1)
= -7 - 14 + 21
= 0
Hence, |A| = 0 .
Expanding determinants becomes much easier once you understand the correct row or column selection.
Question 7: Determinants 4.1
7 (i). Find the value of x , if
\begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix} = \begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}
Solution
Evaluating the determinants,
(2 \times 1) - (5 \times 4) = (2x \times x) - (6 \times 4)
2 - 20 = 2x^2 - 24
or -18 = 2x^2 - 24
or 2x^2 = 6
x^2 = 3
i.e. x = \pm \sqrt{3}
Hence, the values of x are \pm \sqrt{3} .
7 (ii). Find the value of x , if
\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} = \begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix}
Solution
Evaluating the determinants,
(2 \times 5) - (4 \times 3) = (x \times 5) - (2x \times 3)
10 - 12 = 5x - 6x
or -2 = -x
i.e. x = 2
Hence, the value of x is 2 .
NCERT Class 12 Maths has a total of 61 exercises across Part 1 and Part 2 including the miscellaneous ones. I’ll cover each exercise one by one with clear explanations and exam-oriented solutions. Many of these questions are also available in video format on my YouTube channel, @Mathsbetter.
Here’s the next and last question of Determinants 4.1.
Question 8: Determinants Ex 4.1
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
Solution
Evaluating the determinants,
(x \times x) - (18 \times 2) = (6 \times 6) - (18 \times 2)
x^2 - 36 = 36 - 36
or x^2 = 36
So, x = \pm 6
Hence, the correct option is (B) \pm 6 .
Common Mistakes to Avoid
- Using determinant for non-square matrices: Remember that determinants are defined only for square matrices.
- Confusing determinant with modulus: |A| represents determinant of matrix A and not modulus.
- Sign errors in expansion: While expanding determinants of order 3, students often forget the alternating sign pattern.
- Wrong multiplication in order 2 determinants: Carefully apply ad - bc and avoid changing the order.
- Ignoring simpler expansion choices: Expanding along a row or column containing maximum zeros makes calculations easier and reduces mistakes.
- Arithmetic mistakes: Be careful while simplifying negative signs and products during determinant expansion.
Continue Learning
With this, we have completed Determinants 4.1 and revised the basic concepts of determinants, expansion methods and determinant evaluation for matrices of different orders.
These concepts will now help us study properties of determinants, minors, cofactors and inverse of matrices in the upcoming exercises of Determinants.
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You may also explore more NCERT Solutions, important questions and concept-based explanations on Maths Better to strengthen your preparation step-by-step.
All the best and keep learning 👍
