Anand Classes provides the most accurate and exam-oriented NCERT Solutions for Determinants Exercise 4.1 of Class 12 Maths, designed to help students strengthen their understanding of key concepts from Chapter 4. These solutions follow the latest CBSE guidelines and are presented in a step-by-step format to make learning easier and more effective. Students preparing for board exams will find these notes highly useful for revision, practice, and building strong conceptual clarity in determinants. Click the print button to download study material and notes.
Access exam-oriented NCERT Solutions for Chapter 4 Determinants Exercise 4.1 of Class 12 Maths
NCERT Question 1 : Evaluate the determinant:
$$
\begin{vmatrix}2 & 4 \\ -5 & -1\end{vmatrix}
$$
Solution
For any 2×2 determinant
$$
\begin{vmatrix}a & b \\ c & d\end{vmatrix} = ad – bc.
$$
Here:
$$
a = 2,\quad b = 4,\quad c = -5,\quad d = -1.
$$
So,
$$\begin{vmatrix}2 & 4 \\ -5 & -1\end{vmatrix}
= (2)(-1) – (4)(-5) = -2 + 20 = 18$$
Answer:
$$
\begin{vmatrix}2 & 4 \\ -5 & -1\end{vmatrix} = 18.
$$
Understanding how to evaluate determinants is essential for mastering NCERT Class 12 Mathematics Chapter 4 – Determinants. This foundational problem reinforces how to apply the determinant formula for 2×2 matrices, a key concept that appears frequently in solving systems of linear equations, matrix inversion, and higher-order algebraic applications. Practicing such examples helps students build confidence and improve accuracy in board exams and competitive tests where matrix and determinant problems are commonly tested.
NCERT Question 2 : Evaluate the determinant
(i)
$$
\begin{vmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{vmatrix}
$$
(ii)
$$
\begin{vmatrix}x^2 – x + 1 & x – 1 \\ x + 1 & x + 1\end{vmatrix}
$$
Solution
For a $2\times 2$ determinant
$$
\begin{vmatrix}a & b \\ c & d\end{vmatrix} = ad – bc.
$$
Here $a=\cos\theta, b=-\sin\theta, c=\sin\theta, d=\cos\theta$. Thus
$$
\begin{vmatrix}\cos\theta & -\sin\theta \\sin\theta & \cos\theta\end{vmatrix}= (\cos\theta)(\cos\theta) – (-\sin\theta)(\sin\theta)
= \cos^2\theta + \sin^2\theta = 1$$
$\text{(using the identity }\cos^2\theta+\sin^2\theta=1\text{).}$
Answer:
$$
\begin{vmatrix}\cos\theta & -\sin\theta \\sin\theta & \cos\theta\end{vmatrix} = 1.
$$
(ii) Evaluate the determinant
$$
\begin{vmatrix}x^2 – x + 1 & x – 1 \\ x + 1 & x + 1\end{vmatrix}
$$
Using the $2\times 2$ formula,
$$
\begin{vmatrix}x^2 – x + 1 & x – 1 \\ x + 1 & x + 1\end{vmatrix}= (x^2 – x + 1)(x + 1) – (x – 1)(x + 1)= \bigl(x^3 + 1\bigr) – \bigl(x^2 – 1\bigr)$$
$$
\begin{vmatrix}x^2 – x + 1 & x – 1 \\ x + 1 & x + 1\end{vmatrix} = x^3 – x^2 + 2$$
Answer:
$$
\begin{vmatrix}x^2 – x + 1 & x – 1 \\ x + 1 & x + 1\end{vmatrix} = x^3 – x^2 + 2.
$$
Mastering determinants builds essential skills for NCERT Class 12 Chapter 4 on Determinants, helping students confidently solve problems on matrix properties, trigonometric identities, and polynomial expressions—perfect for exam preparation and strengthening linear algebra foundations.
NCERT Question 3 : If
$$A=\begin{vmatrix}1 & 2 \\ 4 & 2\end{vmatrix}$$
Show that $|2A| = 4|A|$.
Solution
For any $2\times 2$ matrix $M$ and scalar $k$ we have the property
$$|kM| = k^{2}|M|$$
Since multiplying every row (or column) by $k$ scales the determinant by $k$ for each of the 2 rows.
Compute $|A|$ directly:
$$|A| = \begin{vmatrix}1 & 2 \\ 4 & 2\end{vmatrix}= (1)(2) – (2)(4)= 2 – 8 =-6 $$
Form $2A$ by multiplying each entry of the matrix by $2$:
For any matrix :
$$A=\begin{bmatrix}1 & 2 \\ 4 & 2\end{bmatrix}$$
$$
2A = \begin{bmatrix}2\cdot 1 & 2\cdot 2 \\ 2\cdot 4 & 2\cdot 2\end{bmatrix}= \begin{bmatrix}2 & 4 \\ 8 & 4\end{bmatrix}.
$$
Now compute $|2A|$:
$$
|2A| = \begin{vmatrix}2 & 4 \\ 8 & 4\end{vmatrix}= (2)(4) – (4)(8)= 8 – 32 = -24$$
Compare $|2A|$ and $|A|$:
$$
4|A| = 4(-6) = -24 = |2A|.
$$
Final Result
$$\boxed{|2A| = 4|A|}$$
This problem verifies the determinant scaling property for $2times 2$ matrices and reinforces core NCERT Class 12 concepts on determinants—useful for mastering matrix operations, solving systems of linear equations, and board exam preparation. Practice similar examples to build speed and accuracy for competitive and school-level mathematics.
NCERT Question 4 : If
$$A=\begin{vmatrix}1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4\end{vmatrix}$$
then show that $|3A| = 27|A|$.
Solution
Given Matrix
$$A=\begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4\end{bmatrix}$$
Multiply every entry of $A$ by $3$ to form $3A$:
$$
3A=\begin{bmatrix}3\cdot 1 & 3\cdot 0 & 3\cdot 1 \\ 3\cdot 0 & 3\cdot 1 & 3\cdot 2 \\ 3\cdot 0 & 3\cdot 0 & 3\cdot 4\end{bmatrix}
=\begin{bmatrix}3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12\end{bmatrix}.
$$
Its determinant is
$$|3A|=\begin{vmatrix}3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12\end{vmatrix}.
$$
$$
|3A|=3 \begin{vmatrix}3 & 6 \\ 0 & 12\end{vmatrix}-0 \begin{vmatrix}0 & 6 \\ 0 & 12\end{vmatrix}+3 \begin{vmatrix}0 & 3 \\ 0 & 0\end{vmatrix}$$
$$|3A|=3(36)−0(0)+3(0)=108 $$
$|3A|$ is upper triangular, so its determinant equal the product of diagonal entries.
$$|3A|=3\cdot 3\cdot 12=108 $$
Compute $|A|$:
$$A=\begin{vmatrix}1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4\end{vmatrix}$$
$$
|A|=1 \begin{vmatrix}1 & 2 \\ 0 & 4\end{vmatrix}-0 \begin{vmatrix}0 & 2 \\ 0 & 4\end{vmatrix}+1 \begin{vmatrix}0 & 1 \\ 0 & 0\end{vmatrix}$$
$$|A|=1(4)−0(0)+1(0)=4$$
$|A|$ is upper triangular, so its determinant equal the product of diagonal entries
$$
|A|=1\cdot 1\cdot 4=4.
$$
Compute $|3A|$:
$$
|3A|=3\cdot 3\cdot 12=108.
$$
Since $3^{3}=27$, we have
$$
27|A|=27\times 4=108=|3A|.
$$
For any $3\times 3$ matrix $M$ and scalar $k$ we have the property
$$|kM| = k^{3}|M|$$
Final Result
$$\boxed{|3A| = 27|A|}$$
This exercise demonstrates the determinant scaling property for $3\times 3$ matrices: multiplying all entries by $k$ scales the determinant by $k^{3}$. Understanding such properties in NCERT Class 12 Chapter 4 on Determinants strengthens problem-solving skills for matrix operations, helps in quicker evaluation of triangular matrices, and is invaluable for exam preparation and applications in linear algebra.
NCERT Question 5 : Evaluate the determinants
$$(i) \quad\begin{vmatrix}3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0\end{vmatrix}\qquad (ii)\quad \begin{vmatrix}3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1\end{vmatrix}$$
$$(iii)\quad\begin{vmatrix}0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0\end{vmatrix}\qquad (iv)\quad\begin{vmatrix}2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0\end{vmatrix}$$
Solution (i)
We choose to expand along the second row because it has the most zeros (simplifies calculations).
The determinant expands as
$$
\begin{vmatrix}3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0\end{vmatrix}
= 0\cdot C_{21} – 0\cdot C_{22} + (-1)\cdot C_{23}
$$
where $C_{2j}$ are the cofactors of row 2. Note the sign pattern for cofactors in row 2 is $(-1)^{2+j}$, so the third term carries sign $(-1)^{2+3}= -1$; combining with the entry $-1$ yields $-(-1),(\text{minor})$.
Compute the minor corresponding to entry in position $(2,3)$ (delete row 2 and column 3):
$$
M_{23}=\begin{vmatrix}3 & -1 \\ 3 & -5\end{vmatrix} = (3)(-5) – (-1)(3) = -15 + 3 = -12.
$$
Putting the sign back:
$$
\text{term from }(2,3) = -(-1)\cdot M_{23} = +1\cdot(-12) = -12.
$$
Final Result
$$\boxed{\begin{vmatrix}3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0\end{vmatrix} = -12.}$$
(ii) $$\begin{vmatrix}3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1\end{vmatrix}$$
Expand along the first row (standard approach for a 3×3):
$$
\begin{vmatrix}3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1\end{vmatrix}
= 3\begin{vmatrix}1 & -2 \\ 3 & 1\end{vmatrix}-(-4)\begin{vmatrix}1 & -2 \\ 2 & 1\end{vmatrix}+5\begin{vmatrix}1 & 1 \\ 2 & 3\end{vmatrix}.
$$
Compute the 2×2 minors:
$$
\begin{vmatrix}1 & -2 \\ 3 & 1\end{vmatrix} = (1)(1) – (-2)(3) = 1 + 6 = 7,
$$
$$
\begin{vmatrix}1 & -2 \\ 2 & 1\end{vmatrix} = (1)(1) – (-2)(2) = 1 + 4 = 5,
$$
$$
\begin{vmatrix}1 & 1 \\ 2 & 3\end{vmatrix} = (1)(3) – (1)(2) = 3 – 2 = 1.
$$
Substitute back:
$$
3(7) + 4(5) + 5(1) = 21 + 20 + 5 = 46.
$$
Final Result
$$\boxed{\begin{vmatrix}3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1\end{vmatrix} = 46.}$$
(iii) $$\begin{vmatrix}0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0\end{vmatrix}$$
Expand along the first row (it has a zero) or proceed directly; we use the first row:
$$
\begin{vmatrix}0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0\end{vmatrix}
= 0\cdot\begin{vmatrix}0 & -3 \\ 3 & 0\end{vmatrix}-1\cdot\begin{vmatrix}-1 & -3 \\ -2 & 0\end{vmatrix}
+2\cdot\begin{vmatrix}-1 & 0 \\ -2 & 3\end{vmatrix}$$
Compute the minors:
$$
\begin{vmatrix}-1 & -3 \\ -2 & 0\end{vmatrix} = (-1)(0) – (-3)(-2) = 0 – 6 = -6,
$$
$$
\begin{vmatrix}-1 & 0 \\ -2 & 3\end{vmatrix} = (-1)(3) – (0)(-2) = -3 – 0 = -3.
$$
Substitute:
$$
\begin{vmatrix}0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0\end{vmatrix}
=0 – 1\cdot(-6) + 2\cdot(-3) = 0 + 6 – 6 = 0.
$$
Important remark: The matrix is skew-symmetric ($A^{T} = -A$). For every skew-symmetric matrix of odd order the determinant is zero, which agrees with the calculation above.
Final Result
$$\boxed{\begin{vmatrix}0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0\end{vmatrix} = 0.}$$
(iv) $$\begin{vmatrix}2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0\end{vmatrix}$$
Again, expand along the second row because it contains a zero and two relatively simple entries:
$$
\begin{vmatrix}2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0\end{vmatrix}= -0\cdot C_{21} + 2\cdot C_{22} -(-1)\cdot C_{23}$$
Compute the minors:
For $(2,2)$ minor (delete row 2 column 2):
$$
M_{22}=\begin{vmatrix}2 & -2 \\ 3 & 0\end{vmatrix} = (2)(0) – (-2)(3) = 0 + 6 = 6.
$$
For $(2,3)$ minor (delete row 2 column 3):
$$
M_{23}=\begin{vmatrix}2 & -1 \\ 3 & -5\end{vmatrix} = (2)(-5) – (-1)(3) = -10 + 3 = -7.
$$
Apply cofactors and signs: cofactor for $(2,2)$ is $(-1)^{2+2}=+1$ and for $(2,3)$ is $(-1)^{2+3}=-1$; the expansion therefore gives
$$
2\cdot(6) -(-1)\cdot\bigl((-1)\cdot M_{23}\bigr)
$$
but it is simpler to track directly:
$$
\text{Contribution from }(2,2)=2\cdot 6 = 12,
$$
$$
\text{Contribution from }(2,3) = -(-1)\cdot M_{23} = +1\cdot(-7) = -7.
$$
So total:
$$
12 – 7 = 5.
$$
Final Result
$$\boxed{\begin{vmatrix}2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0\end{vmatrix} = 5.}$$
These worked examples for NCERT Question 5 consolidate key determinant techniques—expansion by row/column, selecting rows with zeros to simplify computation, computing 2×2 minors, and recognizing special matrix types (for example, skew-symmetric matrices have zero determinant when of odd order). Mastering these steps strengthens problem solving for NCERT Class 12 Chapter 4 on Determinants and builds confidence for board exams and competitive tests; practice similar 3×3 determinants to improve speed and accuracy for linear algebra problems.
NCERT Question 6 : If
$$A=\begin{bmatrix}1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9\end{bmatrix}$$
find $|A|$.
Solution
Given
$$A=\begin{bmatrix}1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9\end{bmatrix}$$
Expand the determinant along the first row using the $3\times 3$ expansion formula:
$$
|A|= 1\cdot\begin{vmatrix}1 & -3 \\ 4 & -9\end{vmatrix}-1\cdot\begin{vmatrix}2 & -3 \\ 5 & -9\end{vmatrix}+(-2)\cdot\begin{vmatrix}2 & 1 \\ 5 & 4\end{vmatrix}$$
Compute the $2\times 2$ minors:
$$
\begin{vmatrix}1 & -3 \\ 4 & -9\end{vmatrix}= (1)(-9) – (-3)(4) = -9 + 12 = 3$$
$$\begin{vmatrix}2 & -3 \\ 5 & -9\end{vmatrix}= (2)(-9) – (-3)(5) =-18 + 15 = -3$$
$$\begin{vmatrix}2 & 1 \\ 5 & 4\end{vmatrix}= (2)(4) – (1)(5) = 8 – 5 = 3$$
Substitute these back into the expansion:
$$|A|= 1\cdot 3 – 1\cdot(-3) + (-2)\cdot 3 = 3 + 3 – 6 = 0$$
Final Result
$$\boxed{|A| = 0}$$
Practicing determinant calculations sharpens skills needed for solving systems of linear equations, understanding matrix rank, and preparing effectively for board and competitive exams.
NCERT Question 7 :Find the values of $x$ if
(i) $$\begin{vmatrix}2 & 4 \\ 5 & 1\end{vmatrix} = \begin{vmatrix}2x & 4 \\ 6 & x\end{vmatrix}$$
(ii) $$\begin{vmatrix}2 & 3 \\ 4 & 5\end{vmatrix} = \begin{vmatrix}x & 3 \\ 2x & 5\end{vmatrix}$$
Solution (i)
Compute the left determinant:
$$\begin{vmatrix}2 & 4 \\ 5 & 1\end{vmatrix} = (2)(1) – (4)(5) = 2 – 20 = -18.$$
Compute the right determinant:
$$\begin{vmatrix}2x & 4 \\ 6 & x\end{vmatrix}= (2x)(x) – (4)(6) = 2x^{2} – 24.$$
Equate and solve:
$$2x^{2} – 24 = -18$$
$$2x^{2} = 6$$
$$x^{2} = 3$$
$$\boxed{x = \pm\sqrt{3}}$$
(ii) $$\begin{vmatrix}2 & 3 \\ 4 & 5\end{vmatrix} = \begin{vmatrix}x & 3 \\ 2x & 5\end{vmatrix}$$
Compute the left determinant:
$$\begin{vmatrix}2 & 3 \\ 4 & 5\end{vmatrix} = (2)(5) – (3)(4) = 10 – 12 = -2.$$
Compute the right determinant:
$$\begin{vmatrix}x & 3 \\ 2x & 5\end{vmatrix}= x\cdot 5 – 3\cdot(2x) = 5x – 6x = -x$$
Equate and solve:
$$-x = -2$$
$$\boxed{x = 2}$$
Solving determinant equations reinforces key NCERT Class 12 skills in evaluating 2×2 determinants and using them to form algebraic equations for unknowns—practice that strengthens algebraic manipulation, matrix intuition, and exam readiness for board and competitive tests.
NCERT 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) $0$
Solution
Evaluate both determinants using the $2\times 2$ formula $$\begin{vmatrix}a & b \\ c & d\end{vmatrix} = ad – bc.$$
Left side:
$$
\begin{vmatrix}x & 2 \\ 18 & x\end{vmatrix} = x\cdot x – 2\cdot 18 = x^{2} – 36.
$$
Right side:
$$
\begin{vmatrix}6 & 2 \\ 18 & 6\end{vmatrix} = 6\cdot 6 – 2\cdot 18 = 36 – 36 = 0.
$$
Equate the two:
$$
x^{2} – 36 = 0
$$
$$
x^{2} = 36
$$
$$
\boxed{x = \pm 6}
$$
Final Result
$$\boxed{x = \pm 6\quad\text{(Option B)}}$$
Solving determinant equations reinforces essential NCERT Class 12 skills in evaluating 2×2 determinants and turning determinant equalities into algebraic equations for unknowns—practice that boosts accuracy and speed for board exams and competitive maths tests.
✅ FAQ Section
Q1. What is included in the Determinants Exercise 4.1 NCERT Solutions for Class 12?
The solutions include step-by-step explanations for each question, clear methods, and simplified calculations based on the NCERT Class 12 Maths textbook for Chapter 4.
Q2. Are these Determinants Exercise 4.1 solutions useful for CBSE board exams?
Yes. These solutions strictly follow the latest CBSE pattern and help students score higher by improving conceptual clarity and solving speed.
Q3. Can I download the Determinants Exercise 4.1 Class 12 Maths PDF for free?
Yes, Anand Classes provides the complete Exercise 4.1 NCERT Solutions PDF free of cost for all Class 12 students.
Q4. Are the solutions prepared by expert faculty?
Yes. All solutions are prepared by experienced Anand Classes faculty members who specialize in Class 11–12 Maths and competitive exam preparation.
Q5. Do these solutions include shortcut methods for determinants?
Where applicable, simplified approaches and smart methods are explained to make learning faster and more exam-oriented.
