Anand Classes brings you the most accurate and easy-to-understand NCERT Solutions for Determinants Exercise 4.3 Class 12 Math Chapter-4, prepared strictly according to the latest CBSE guidelines. These solutions help students master essential determinant properties, cofactor expansion, adjoint methods, and problem-solving strategies important for board examinations and competitive entrance tests. Designed for clarity and step-by-step learning, these notes ensure strong conceptual understanding and error-free practice for Class 12 Maths. Click the print button to download study material and notes.
NCERT Question.1 : Write Minors and Cofactors of the elements of following determinants:
$$(i)\quad \begin{vmatrix}3 & -4\\0 & 3\end{vmatrix}\qquad(ii) \quad \begin{vmatrix}a & c\\b & d\end{vmatrix}$$
Solution
(i) For $$\begin{vmatrix}2 & -4\\0 & 3\end{vmatrix}$$
Minors of the elements (denote $M_{ij}$ as the minor of $a_{ij}$):
$$M_{11}=3\quad M_{12}=0\quad M_{21}=-4\quad M_{22}=2$$
Cofactors (denote $A_{ij}=(-1)^{i+j}M_{ij}$):
$$A_{11}=(-1)^{2}M_{11}=3$$
$$A_{12}=(-1)^{3}M_{12}=0$$
$$A_{21}=(-1)^{3}M_{21}=4$$
$$A_{22}=(-1)^{4}M_{22}=2$$
(ii) For $$\begin{vmatrix}a & c\\b & d\end{vmatrix}$$
Minors:
$$M_{11}=d\quad M_{12}=b\quad M_{21}=c\quad M_{22}=a$$
Cofactors:
$$A_{11}=(-1)^{2}M_{11}=d$$
$$A_{12}=(-1)^{3}M_{12}=-b$$
$$A_{21}=(-1)^{3}M_{21}=-c$$
$$A_{22}=(-1)^{4}M_{22}=a$$
Published by Anand Classes, Written by Neeraj Anand — offering high-quality NCERT Class 12 Maths solutions, perfect for JEE, CUET, and CBSE exam preparation. These expertly crafted notes make Determinants, Matrices, and all advanced topics easy to understand for every student.
NCERT Question.2 : Write Minors and Cofactors of the elements of following determinants:
$$(i) \quad\begin{vmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{vmatrix} \qquad(ii) \quad \begin{vmatrix}1 & 0 & 4\\3 & 5 & -1\\0 & 1 & 2\end{vmatrix}$$
Solution
(i)
We find the minors $M_{ij}$ and cofactors $A_{ij}$ for each element of
$$\begin{vmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{vmatrix}$$
Minors $M_{ij}$
$M_{11}=\begin{vmatrix}1 & 0\\0 & 1\end{vmatrix}=1-0=1$
$M_{12}=\begin{vmatrix}0 & 0\\0 & 1\end{vmatrix}=0-0=0$
$M_{13}=\begin{vmatrix}0 & 1\\0 & 0\end{vmatrix}=0-0=0$
$M_{21}=\begin{vmatrix}0 & 0\\0 & 1\end{vmatrix}=0$
$M_{22}=\begin{vmatrix}1 & 0\\0 & 1\end{vmatrix}=1$
$M_{23}=\begin{vmatrix}1 & 0\\0 & 0\end{vmatrix}=0$
$M_{31}=\begin{vmatrix}0 & 0\\1 & 0\end{vmatrix}=0$
$M_{32}=\begin{vmatrix}1 & 0\\0 & 0\end{vmatrix}=0$
$M_{33}=\begin{vmatrix}1 & 0\\0 & 1\end{vmatrix}=1$
Cofactors $A_{ij}=(-1)^{i+j}M_{ij}$
- $A_{11}=1,\quad A_{12}=0,\quad A_{13}=0$
- $A_{21}=0,\quad A_{22}=1,\quad A_{23}=0$
- $A_{31}=0,\quad A_{32}=0,\quad A_{33}=1$
(ii)
Now for
$$\begin{vmatrix}1 & 0 & 4\\3 & 5 & -1\\0 & 1 & 2\end{vmatrix}$$
Minors $M_{ij}$
$M_{11}=\begin{vmatrix}5 & -1\\1 & 2\end{vmatrix}=5(2)-(-1)(1)=10+1=11$
$M_{12}=\begin{vmatrix}3 & -1\\0 & 2\end{vmatrix}=3(2)-(-1)(0)=6$
$M_{13}=\begin{vmatrix}3 & 5\\0 & 1\end{vmatrix}=3(1)-5(0)=3$
$M_{21}=\begin{vmatrix}0 & 4\\1 & 2\end{vmatrix}=0(2)-4(1)=−4$
$M_{22}=\begin{vmatrix}1 & 4\\0 & 2\end{vmatrix}=1(2)-4(0)=2$
$M_{23}=\begin{vmatrix}1 & 0\\0 & 1\end{vmatrix}=1(1)-0=1$
$M_{31}=\begin{vmatrix}0 & 4\\5 & -1\end{vmatrix}=0(−1)-4(5)=−20$
$M_{32}=\begin{vmatrix}1 & 4\\3 & -1\end{vmatrix}=1(−1)-4(3)=−1−12=−13$
$M_{33}=\begin{vmatrix}1 & 0\\3 & 5\end{vmatrix}=1(5)-0=5$
Cofactors $A_{ij}$
Using $A_{ij}=(-1)^{i+j}M_{ij}$:
- $A_{11}=11,\quad A_{12}=(-1)^{1+2}(6)= -6,\quad A_{13}=3$
- $A_{21}=(-1)^{2+1}(−4)=4,\quad A_{22}=2,\quad A_{23}=(-1)^{2+3}(1)= -1$
- $A_{31}=(-1)^{3+1}(−20)= -20,\quad A_{32}=(-1)^{3+2}(−13)=13,\quad A_{33}=5$
📌 Published by Anand Classes, Written by Neeraj Anand — offering high-quality NCERT Class 12 Maths solutions, perfect for JEE, CUET, and CBSE exam preparation. These expertly crafted notes make Determinants, Minors, and Cofactors easy to master for every student.
NCERT Question.3 : Using Cofactors of elements of second row, evaluate
$$\bigtriangleup=\begin{vmatrix}5 & 3 & 8\\2 & 0 & 1\\1 & 2 & 3\end{vmatrix}$$
Solution
Using cofactors of the elements of the second row.
Compute the cofactors $A_{2j}=(-1)^{2+j}M_{2j}$ where $M_{2j}$ is the minor obtained by deleting row 2 and column $j$.
$$M_{21}=\begin{vmatrix}3 & 8\\2 & 3\end{vmatrix}=3\cdot 3-8\cdot 2=9-16=-7$$
$$A_{21}=(-1)^{3}M_{21}=-1\cdot(-7)=7$$
$$M_{22}=\begin{vmatrix}5 & 8\\1 & 3\end{vmatrix}=5\cdot 3-8\cdot 1=15-8=7$$
$$A_{22}=(-1)^{4}M_{22}=1\cdot 7=7$$
$$M_{23}=\begin{vmatrix}5 & 3\\1 & 2\end{vmatrix}=5\cdot 2-3\cdot 1=10-3=7$$
$$A_{23}=(-1)^{5}M_{23}=-1\cdot 7=-7$$
Now expand along the second row:
$$\bigtriangleup=a_{21}A_{21}+a_{22}A_{22}+a_{23}A_{23}$$
$$\bigtriangleup=2\cdot 7+0\cdot 7+1\cdot(-7)=14+0-7=7$$
Final Result
$$\boxed{\bigtriangleup=7}$$
NCERT Question.4 : Using Cofactors of elements of third column, evaluate
$$\bigtriangleup=\begin{vmatrix}1 & x & yz\\1 & y & zx\\1 & z & xy\end{vmatrix}$$
Solution
We expand the determinant along the third column using cofactors:
If $A_{ij}=(-1)^{i+j}M_{ij}$ where $M_{ij}$ is the minor of element $a_{ij}$, then
Step 1: Compute Cofactors for the third column
$$M_{13}=\begin{vmatrix}1 & y\\1 & z\end{vmatrix}=1\cdot z-y\cdot 1=z-y$$
$$A_{13}=(-1)^{1+3}(z-y)=(-1)^{4}(z-y)=z-y$$
$$M_{23}=\begin{vmatrix}1 & x\\1 & z\end{vmatrix}=1\cdot z-x\cdot 1=z-x$$
$$A_{23}=(-1)^{2+3}(z-x)=(-1)^{5}(z-x)=x-z$$
$$M_{33}=\begin{vmatrix}1 & x\\1 & y\end{vmatrix}=1\cdot y-x\cdot 1=y-x$$
$$A_{33}=(-1)^{3+3}(y-x)=(-1)^{6}(y-x)=y-x$$
Step 2: Expand the determinant
Using expansion along the third column:
$$\bigtriangleup = a_{13}A_{13}+a_{23}A_{23}+a_{33}A_{33}$$
Substitute values:
$$\bigtriangleup=yz(z-y)+zx(x-z)+xy(y-x)$$
Expand each product:
$$\bigtriangleup=yz^2-y^2z+zx^2-zx^2+xy^2-x^2y$$
Combine like terms:
$$\bigtriangleup=yz^2-y^2z+zx^2-z^2x+xy^2-x^2y$$
Factor the result to simplify:
$$\bigtriangleup=(x-y)(y-z)(z-x)$$
Final Answer
$$\boxed{(x-y)(y-z)(z-x)}$$
Published by Anand Classes, Director by Neeraj Anand — comprehensive step-by-step NCERT solutions for Class 12 Maths Determinants exercise, ideal for CBSE boards, CUET and JEE prep with clear cofactor expansion methods.
NCERT Question.5 : If
$$\Delta=\begin{vmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{vmatrix}$$
and $A_{ij}$ is the cofactor of $a_{ij}$ then the value of $\Delta$ is given by:
(A) $a_{11}A_{31}+a_{12}A_{32}+a_{13}A_{33}$
(B) $a_{11}A_{11}+a_{12}A_{21}+a_{13}A_{31}$
(C) $a_{21}A_{11}+a_{22}A_{12}+a_{23}A_{13}$
(D) $a_{11}A_{11}+a_{21}A_{21}+a_{31}A_{31}$
Solution
Expand the determinant $\Delta$ by the first column using cofactors. By definition of expansion along column 1 we have:
$$\Delta=a_{11}A_{11}+a_{21}A_{21}+a_{31}A_{31}$$
This expression matches option (D). To check the other options:
- Option (A) is expansion along the third row but with entries from the first row incorrectly paired with cofactors of the third row — not a valid cofactor expansion formula for $\Delta$.
- Option (B) lists $a_{11}A_{11}+a_{12}A_{21}+a_{13}A_{31}$ which mixes column and row indices incorrectly (not a standard expansion).
- Option (C) is $a_{21}A_{11}+a_{22}A_{12}+a_{23}A_{13}$ which is expansion along the first row but with cofactors transposed; it is not the correct expansion.
Therefore the correct choice is (D).
Final Result
$$\boxed{\Delta=a_{11}A_{11}+a_{21}A_{21}+a_{31}A_{31}}$$
Published by Anand Classes, Main Head Faculty by Neeraj Anand — precise and exam-ready NCERT Class 12 Maths solutions with clear cofactor expansion explanations to strengthen your preparation for CBSE, JEE and CUET.
✅ FAQ Section
Q1. What is covered in NCERT Class 12 Determinants Exercise 4.3?
Exercise 4.3 includes questions based on the application of determinants, evaluation using properties, and problem-solving involving cofactor expansion.
Q2. Are these NCERT Solutions for Determinants Exercise 4.3 helpful for CBSE Board Exams?
Yes. The solutions follow the exact CBSE marking scheme and cover every important step needed to score full marks.
Q3. Can I download the Determinants Exercise 4.3 Class 12 solutions in PDF format?
Absolutely. Anand Classes provides a free downloadable PDF for quick revision and offline study.
Q4. Who has prepared these Determinants Class 12 solutions?
All solutions are prepared by the expert Maths faculty at Anand Classes, ensuring accuracy and conceptual clarity.
Q5. Are these solutions useful for competitive exams like JEE?
Yes. Determinants are a key topic in competitive exams, and these solutions strengthen your basics for higher-level questions.
Summary
Exercise 4.4 of Chapter 4 on Determinants presents a comprehensive set of advanced problems that challenge students to apply their knowledge of determinants in diverse mathematical contexts. This exercise emphasizes the interconnections between determinants, matrices, geometry, and algebraic systems. Students are required to prove complex identities, solve non-linear systems of equations, and explore the geometric implications of determinant properties. The problems in this set are designed to deepen students’ understanding of determinants as a versatile mathematical tool, applicable in various areas of mathematics. By working through these questions, students develop advanced problem-solving skills and gain a more profound appreciation for the elegance and power of determinants in higher mathematics.
