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# NCERT Solutions for Class 12 Maths Chapter 4 - Determinants Exercise 4.3

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## NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.3 Determinants - FREE PDF Download

NCERT for Ex 4.3 Class 12, "Determinants," delves into important concepts like minors and cofactors. Exercise is specifically focused on these topics, which are crucial for understanding how determinants work. Grasping minors and cofactors is essential as they form the basis for more advanced topics in determinants.

Table of Content
1. NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.3 Determinants - FREE PDF Download
2. Glance on NCERT Solutions Class 12 Maths Chapter 4 Exercise 4.3 | Vedantu
3. Access NCERT Solutions for Maths Class 12 Chapter 4 - Determinants Exercise 4.3
4. Class 12 Maths Chapter 4: Exercises Breakdown
5. CBSE Class 12 Maths Chapter 4 Other Study Materials
6. Chapter-Specific NCERT Solutions for Class 12 Maths
FAQs

In Exercise 4.3 Class 12, you will learn how to calculate minors and cofactors, which are key elements in matrix algebra. Pay close attention to the methods of finding these values and practice regularly to master the concepts. Vedantu's comprehensive solutions will help you understand each step clearly, ensuring you build a strong foundation in this chapter.  To boost your exam preparations, you can download the FREE PDF for NCERT Solutions for Class 12 Maths from Vedantu’s website.

## Glance on NCERT Solutions Class 12 Maths Chapter 4 Exercise 4.3 | Vedantu

• The NCERT of ex 4.3 class 12 maths solutions of chapter Determinants, deals with the topics minors, cofactors and using cofactors evaluating the determinants.

• A minor is a specific type of sub-determinant. For any element in a matrix, the minor is the determinant of the smaller matrix that is obtained by removing the row and column that contain that element.

• A cofactor is also known as the signed minor. For an element say ‘mij', the cofactor shall be denoted as ‘Mij’ where the said cofactor shall be defined with the help of the formula M = (-1)i+j B, where B is minor of the particular element 'mij'.

• The calculation of minor and cofactor in this chapter has been covered in a very simple and thorough manner in the entire chapter.

• There are 5 fully solved Questions and Solutions in Class 12th Maths Chapter 4 Exercise 4.3 Determinants.

Competitive Exams after 12th Science

## Access NCERT Solutions for Maths Class 12 Chapter 4 - Determinants Exercise 4.3

1. Write Minors and Cofactors of the elements of following determinants:

1. $\mathbf{\left| \begin{matrix} \text{2} & \text{-4} \\ \text{0} & \text{3} \\ \end{matrix} \right|}$

Ans: Given, $\left| \begin{matrix} \text{2} & \text{-4} \\ \text{0} & \text{3} \\ \end{matrix} \right|$

Minor of an element is termed as the determinant obtained by removing the row and the column in which that element is present.

Minor of element ${{\text{a}}_{ij}}$ is denoted by ${{M}_{ij}}$,where

$i$ and $j$ denotes the row and the column of the determinant respectively.

$\therefore {{\text{M}}_{11}}\text{=3}$

${{\text{M}}_{12}}\text{=0}$

${{\text{M}}_{21}}\text{=-4}$

${{\text{M}}_{22}}\text{=2}$

Cofactor of an element is termed as the determinant obtained by removing the row and the column in which that element is present preceded by a negative or a positive sign based on the position of the element.

Thus,

Cofactor of ${{\text{a}}_{ij}}$ is ${{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}$

$\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{\text{1+1}}}{{\text{M}}_{11}}$

$\Rightarrow {{\text{A}}_{11}}={{\left( \text{-1} \right)}^{\text{2}}}\left( \text{3} \right)$

$\therefore {{\text{A}}_{11}}\text{=3}$

Similarly,

$\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{\text{1+2}}}{{\text{M}}_{12}}$

$\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{\text{3}}}\left( \text{0} \right)$

$\therefore {{\text{A}}_{12}}\text{=0}$

$\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{\text{2+1}}}{{M}_{21}}$

$\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{\text{3}}}\left( \text{-4} \right)$

$\therefore {{\text{A}}_{21}}\text{=4}$

$\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{\text{2+2}}}{{M}_{22}}$

$\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{\text{4}}}\left( \text{2} \right)$

$\therefore {{\text{A}}_{22}}\text{=2}$

1. $\mathbf{\left| \begin{matrix} \text{a} & \text{c} \\ \text{b} & \text{d} \\ \end{matrix} \right|}$

Ans: Given, $\left| \begin{matrix} \text{a} & \text{c} \\ \text{b} & \text{d} \\ \end{matrix} \right|$

Minor of element ${{\text{a}}_{ij}}$ is denoted by ${{M}_{ij}}$.

$\therefore {{\text{M}}_{11}}\text{=d}$

${{\text{M}}_{12}}\text{=b}$

${{\text{M}}_{21}}\text{=c}$

${{\text{M}}_{22}}\text{=a}$

Cofactor of ${{\text{a}}_{ij}}$ is ${{\text{A}}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{\text{M}}_{ij}}$

$\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{\text{1+1}}}{{\text{M}}_{11}}$

$\Rightarrow {{\text{A}}_{11}}={{\left( \text{-1} \right)}^{\text{2}}}\left( d \right)$

$\therefore {{\text{A}}_{11}}\text{=d}$

Similarly,

$\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{\text{1+2}}}{{\text{M}}_{12}}$

$\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{\text{3}}}\left( b \right)$

$\therefore {{\text{A}}_{12}}\text{=}-b$

$\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{\text{2+1}}}{{M}_{21}}$

$\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{\text{3}}}\left( c \right)$

$\therefore {{\text{A}}_{21}}\text{=}-\text{c}$

$\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{\text{2+2}}}{{M}_{22}}$

$\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{\text{4}}}\left( a \right)$

$\therefore {{\text{A}}_{22}}\text{=a}$

2. Write Minors and Cofactors of the elements of following determinants:

1. $\mathbf{\left| \begin{matrix} \text{1} & \text{0} & \text{0} \\ \text{0} & \text{1} & \text{0} \\ \text{0} & \text{0} & \text{1} \\ \end{matrix} \right|}$

Ans: Given determinant, $\left| \begin{matrix} \text{1} & \text{0} & \text{0} \\ \text{0} & \text{1} & \text{0} \\ \text{0} & \text{0} & \text{1} \\ \end{matrix} \right|$.

Minor of element ${{\text{a}}_{ij}}$ is denoted by ${{M}_{ij}}$.

$\therefore {{\text{M}}_{11}}\text{=}\left| \begin{matrix} \text{1} & \text{0} \\ \text{0} & \text{1} \\ \end{matrix} \right|\text{=1}$

$\Rightarrow {{\text{M}}_{12}}\text{=}\left| \begin{matrix} \text{0} & \text{0} \\ \text{0} & \text{1} \\ \end{matrix} \right|\text{=0}$

$\Rightarrow {{\text{M}}_{13}}\text{=}\left| \begin{matrix} \text{0} & \text{1} \\ \text{0} & \text{0} \\ \end{matrix} \right|\text{=0}$

$\Rightarrow {{\text{M}}_{21}}\text{=}\left| \begin{matrix} \text{0} & \text{0} \\ \text{0} & \text{1} \\ \end{matrix} \right|\text{=0}$

$\Rightarrow {{\text{M}}_{22}}\text{=}\left| \begin{matrix} \text{1} & \text{0} \\ \text{0} & \text{1} \\ \end{matrix} \right|\text{=1}$

$\Rightarrow {{\text{M}}_{23}}\text{=}\left| \begin{matrix} \text{1} & \text{0} \\ \text{0} & \text{0} \\ \end{matrix} \right|\text{=0}$

$\Rightarrow {{\text{M}}_{31}}\text{=}\left| \begin{matrix} \text{0} & \text{0} \\ \text{1} & \text{0} \\ \end{matrix} \right|\text{=0}$

$\Rightarrow {{\text{M}}_{32}}\text{=}\left| \begin{matrix} \text{1} & \text{0} \\ \text{0} & \text{0} \\ \end{matrix} \right|\text{=0}$

$\Rightarrow {{\text{M}}_{33}}\text{=}\left| \begin{matrix} \text{1} & \text{0} \\ \text{0} & \text{1} \\ \end{matrix} \right|\text{=1}$

Cofactor of ${{\text{a}}_{ij}}$ is ${{A}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{M}_{ij}}$.

$\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}=1$

$\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}=0$

$\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}=0$

$\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}=0$

$\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}=1$

$\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}=0$

$\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}=0$

$\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}=0$

$\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}=1$

1. $\mathbf{\left| \begin{matrix} \text{1} & \text{0} & \text{4} \\ \text{3} & \text{5} & \text{-1} \\ \text{0} & \text{1} & \text{2} \\ \end{matrix} \right|}$

Ans: Given determinant, $\left| \begin{matrix} \text{1} & \text{0} & \text{4} \\ \text{3} & \text{5} & \text{-1} \\ \text{0} & \text{1} & \text{2} \\ \end{matrix} \right|$

Minor of element ${{\text{a}}_{ij}}$ is denoted by ${{M}_{ij}}$.

$\Rightarrow {{\text{M}}_{11}}\text{=}\left| \begin{matrix} \text{5} & \text{-1} \\ \text{1} & \text{2} \\ \end{matrix} \right|\text{=10+1=11}$

$\Rightarrow {{\text{M}}_{12}}\text{=}\left| \begin{matrix} \text{3} & \text{-1} \\ \text{0} & \text{2} \\ \end{matrix} \right|\text{=6-0=6}$

$\Rightarrow {{\text{M}}_{13}}\text{=}\left| \begin{matrix} \text{3} & \text{5} \\ \text{0} & \text{1} \\ \end{matrix} \right|\text{=3-0=3}$

$\Rightarrow {{\text{M}}_{21}}\text{=}\left| \begin{matrix} \text{0} & \text{4} \\ \text{1} & \text{2} \\ \end{matrix} \right|\text{=0-4=-4}$

$\Rightarrow {{\text{M}}_{22}}\text{=}\left| \begin{matrix} \text{1} & \text{4} \\ \text{0} & \text{2} \\ \end{matrix} \right|\text{=2-0=2}$

$\Rightarrow {{\text{M}}_{23}}\text{=}\left| \begin{matrix} \text{1} & \text{0} \\ \text{0} & \text{1} \\ \end{matrix} \right|\text{=1-0=1}$

$\Rightarrow {{\text{M}}_{31}}\text{=}\left| \begin{matrix} \text{0} & \text{4} \\ \text{5} & \text{-1} \\ \end{matrix} \right|\text{=0-20=-20}$

$\Rightarrow {{\text{M}}_{32}}\text{=}\left| \begin{matrix} \text{1} & \text{4} \\ \text{3} & \text{-1} \\ \end{matrix} \right|\text{=-1-12=-13}$

$\Rightarrow {{\text{M}}_{33}}\text{=}\left| \begin{matrix} \text{1} & \text{0} \\ \text{3} & \text{5} \\ \end{matrix} \right|\text{=5-0=5}$

Cofactor of ${{\text{a}}_{ij}}$ is ${{A}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{M}_{ij}}$.

$\Rightarrow {{\text{A}}_{11}}\text{=}{{\left( \text{-1} \right)}^{1+1}}{{\text{M}}_{11}}=11$

$\Rightarrow {{\text{A}}_{12}}\text{=}{{\left( \text{-1} \right)}^{1+2}}{{\text{M}}_{12}}=6$

$\Rightarrow {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{1+3}}{{\text{M}}_{13}}=3$

$\Rightarrow {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{2+1}}{{\text{M}}_{21}}=-4$

$\Rightarrow {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{2+2}}{{\text{M}}_{22}}=2$

$\Rightarrow {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{2+3}}{{\text{M}}_{23}}=1$

$\Rightarrow {{\text{A}}_{31}}\text{=}{{\left( \text{-1} \right)}^{3+1}}{{\text{M}}_{31}}=-20$

$\Rightarrow {{\text{A}}_{32}}\text{=}{{\left( \text{-1} \right)}^{3+2}}{{\text{M}}_{32}}=-13$

$\Rightarrow {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{3+3}}{{\text{M}}_{33}}=5$

3. Using Cofactors of elements of second row, evaluate $\mathbf{\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{5} & \text{3} & \text{8} \\ \text{2} & \text{0} & \text{1} \\ \text{1} & \text{2} & \text{3} \\ \end{matrix} \right|}$.

Ans: Given determinant, $\left| \begin{matrix} \text{5} & \text{3} & \text{8} \\ \text{2} & \text{0} & \text{1} \\ \text{1} & \text{2} & \text{3} \\ \end{matrix} \right|$

Determining the minors and cofactors, we get:

$\Rightarrow {{\text{M}}_{21}}\text{=}\left| \begin{matrix} \text{3} & \text{8} \\ \text{2} & \text{3} \\ \end{matrix} \right|=9-16=-7$

$\therefore {{\text{A}}_{21}}\text{=}{{\left( \text{-1} \right)}^{\text{2+1}}}{{M}_{21}}\text{=7}$

$\Rightarrow {{\text{M}}_{22}}\text{=}\left| \begin{matrix} \text{5} & \text{8} \\ \text{1} & \text{3} \\ \end{matrix} \right|=15-8=7$

$\therefore {{\text{A}}_{22}}\text{=}{{\left( \text{-1} \right)}^{\text{2+2}}}{{M}_{22}}\text{=7}$

$\Rightarrow {{\text{M}}_{23}}\text{=}\left| \begin{matrix} \text{5} & \text{3} \\ \text{1} & \text{2} \\ \end{matrix} \right|=10-3=7$

$\therefore {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{\text{2+1}}}{{M}_{23}}=-7$

Since, $\text{ }\!\!\Delta\!\!\text{ }$ is equal to the sum of the product of the elements of the second row with their corresponding cofactors.

$\therefore \Delta ={{a}_{21}}{{A}_{21}}+{{a}_{22}}{{A}_{22}}+{{a}_{23}}{{A}_{23}}$

$\Rightarrow \Delta \text{=2}\left( \text{7} \right)\text{+0}\left( \text{7} \right)\text{+1}\left( \text{7} \right)$

Hence, $\Delta \text{=21}$.

4. Using Cofactors of elements of third column, evaluate  $\mathbf{\text{ }\!\!\Delta\!\!\text{ =}\left| \begin{matrix} \text{1} & \text{x} & \text{yz} \\ \text{1} & \text{y} & \text{zx} \\ \text{1} & \text{z} & \text{xy} \\ \end{matrix} \right|}$.

Ans: Given determinant, $\left| \begin{matrix} \text{1} & \text{x} & \text{yz} \\ \text{1} & \text{y} & \text{zx} \\ \text{1} & \text{z} & \text{xy} \\ \end{matrix} \right|$

Determining the minors and cofactors, we get:

$\Rightarrow {{\text{M}}_{13}}\text{=}\left| \begin{matrix} 1 & y \\ 1 & z \\ \end{matrix} \right|=z-y$

$\therefore {{\text{A}}_{13}}\text{=}{{\left( \text{-1} \right)}^{\text{1+3}}}{{M}_{13}}\text{=}\left( z-y \right)$

$\Rightarrow {{\text{M}}_{23}}\text{=}\left| \begin{matrix} 1 & x \\ \text{1} & z \\ \end{matrix} \right|=z-x$

$\therefore {{\text{A}}_{23}}\text{=}{{\left( \text{-1} \right)}^{\text{2+3}}}{{M}_{23}}\text{=}\left( x-z \right)$

$\Rightarrow {{\text{M}}_{33}}\text{=}\left| \begin{matrix} 1 & x \\ \text{1} & y \\ \end{matrix} \right|=y-x$

$\therefore {{\text{A}}_{33}}\text{=}{{\left( \text{-1} \right)}^{\text{3+3}}}{{M}_{33}}=\left( y-x \right)$

Since, $\text{ }\!\!\Delta\!\!\text{ }$ is equal to the sum of the product of the elements of the first row with their corresponding cofactors.

$\therefore \Delta ={{a}_{13}}{{A}_{13}}+{{a}_{23}}{{A}_{23}}+{{a}_{33}}{{A}_{33}}$

$\Rightarrow \Delta =yz\left( z-y \right)+zx\left( x-z \right)+xy\left( y-x \right)$

$\Rightarrow \Delta =y{{z}^{2}}-{{y}^{2}}z+{{x}^{2}}z-x{{z}^{2}}+x{{y}^{2}}-{{x}^{2}}y$

$\Rightarrow \Delta =\left( {{x}^{2}}z-{{y}^{2}}z \right)+\left( y{{z}^{2}}-x{{z}^{2}} \right)+\left( x{{y}^{2}}-{{x}^{2}}y \right)$

$\Rightarrow \Delta =\left( x-y \right)\left[ zx+zy-{{z}^{2}}-xy \right]$

$\Rightarrow \Delta =\left( x-y \right)\left[ z\left( x-z \right)+y\left( z-x \right) \right]$

Thus, $\text{ }\!\!\Delta\!\!\text{ =}\left( \text{x-y} \right)\left( \text{y-z} \right)\left( \text{z-x} \right)$.

5. For the matrices $\mathbf{\text{A}}$ and $\mathbf{\text{B}}$ , verify that $\mathbf{\left( \text{AB} \right)\text{ }\!\!'\!\!\text{ =B }\!\!'\!\!\text{ A }\!\!'\!\!\text{ }}$ where

If $\mathbf{\Delta =\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right|}$ and ${{A}_{ij}}$ is Cofactors of ${{a}_{ij}}$ , then value of $\Delta$ is given by

1. $\mathbf{{{a}_{11}}{{A}_{31}}+{{a}_{12}}{{A}_{32}}+{{a}_{13}}{{A}_{33}}}$

2. $\mathbf{{{a}_{11}}{{A}_{11}}+{{a}_{12}}{{A}_{21}}+{{a}_{13}}{{A}_{31}}}$

3. $\mathbf{{{a}_{21}}{{A}_{11}}+{{a}_{22}}{{A}_{12}}+{{a}_{23}}{{A}_{13}}}$

4. $\mathbf{{{a}_{11}}{{A}_{11}}+{{a}_{21}}{{A}_{21}}+{{a}_{31}}{{A}_{31}}}$

Ans: It is given that $\Delta =\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\ \end{matrix} \right|$.

The value of $\Delta$ by expanding along first column is obtained as,

${{a}_{11}}\left( {{a}_{22}}.{{a}_{33}}-{{a}_{23}}.{{a}_{32}} \right)-{{a}_{21}}\left( {{a}_{12}}.{{a}_{33}}-{{a}_{13}}.{{a}_{32}} \right)+{{a}_{31}}\left( {{a}_{12}}.{{a}_{23}}-{{a}_{13}}.{{a}_{22}} \right)$ ….. 1

Now, the cofactor ${{A}_{ij}}$ of element ${{a}_{ij}}$ is given by ${{\left( -1 \right)}^{i+j}}{{M}_{ij}}$, where ${{M}_{ij}}$ is the minor. Minor is the determinant obtained by cancelling the ith row and jth column of the original matrix.

Now for element ${{a}_{11}}$, the minor is ${{M}_{11}}=\left| \begin{matrix} {{a}_{22}} & {{a}_{23}} \\ {{a}_{32}} & {{a}_{33}} \\ \end{matrix}\right|={{a}_{22}}.{{a}_{33}}-{{a}_{23}}.{{a}_{32}}$and the cofactor is ${{A}_{11}}={{\left( -1 \right)}^{1+1}}\left( {{a}_{22}}.{{a}_{33}}-{{a}_{23}}.{{a}_{32}} \right)\Rightarrow {{A}_{11}}=\left( {{a}_{22}}.{{a}_{33}}-{{a}_{23}}.{{a}_{32}} \right)$.

Next for element ${{a}_{21}}$, the minor is ${{M}_{21}}=\left| \begin{matrix} {{a}_{12}} & {{a}_{13}} \\ {{a}_{32}} & {{a}_{33}} \\ \end{matrix}\right|={{a}_{12}}.{{a}_{33}}-{{a}_{13}}.{{a}_{32}}$ and the cofactor is ${{A}_{21}}={{\left( -1 \right)}^{2+1}}\left( {{a}_{12}}.{{a}_{33}}-{{a}_{13}}.{{a}_{32}} \right)\Rightarrow {{A}_{21}}=-\left( {{a}_{12}}.{{a}_{33}}-{{a}_{13}}.{{a}_{32}} \right)$.

Next for element ${{a}_{31}}$, the minor is ${{M}_{31}}=\left| \begin{matrix} {{a}_{12}} & {{a}_{13}} \\ {{a}_{22}} & {{a}_{23}} \\ \end{matrix}\right|={{a}_{12}}.{{a}_{23}}-{{a}_{13}}.{{a}_{22}}$ and the cofactor is ${{A}_{31}}={{\left( -1 \right)}^{3+1}}\left( {{a}_{12}}.{{a}_{23}}-{{a}_{13}}.{{a}_{22}} \right)\Rightarrow {{A}_{31}}=\left( {{a}_{12}}.{{a}_{23}}-{{a}_{13}}.{{a}_{22}} \right)$.

Now substituting the terms as obtained from above computation in equation 1,

${{a}_{11}}{{A}_{11}}-{{a}_{21}}\left( -{{A}_{21}} \right)+{{a}_{31}}{{A}_{31}}$

${{a}_{11}}{{A}_{11}}+{{a}_{21}}{{A}_{21}}+{{a}_{31}}{{A}_{31}}$

This matches with option d.

## Conclusion

The Exercise 4.3 Class 12 Chapter 4 Maths is essential for understanding the concepts of minors and cofactors. These topics are the building blocks for understanding more complex operations involving determinants. It's important to focus on the methods of calculating minors and cofactors accurately, as these skills are frequently tested in exams.Ensuring you can solve these problems with confidence will greatly benefit your overall understanding of determinants.

Vedantu's Exercise 4.3 Class 12 Maths solutions provide detailed, step-by-step explanations that make these concepts easier to understand and apply. Practice consistently, and review each solution carefully to solidify your knowledge and prepare effectively for your exams.

## Class 12 Maths Chapter 4: Exercises Breakdown

 Exercises Number of Questions Exercise 4.1 8 Questions and Solutions Exercise 4.2 5 Questions and Solutions Exercise 4.4 18  Questions and Solutions Exercise 4.5 16 Questions and Solutions

## Chapter-Specific NCERT Solutions for Class 12 Maths

Given below are the chapter-wise NCERT Solutions for Class 12 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 12 Maths Chapter 4 - Determinants Exercise 4.3

1. Can I download NCERT Solutions Class 12 Maths Chapter 4 from Vedantu?

Yes, you can download the Class 12 Maths NCERT Solutions Chapter 4 by signing into the Vedantu’s website. These solutions are solved by our expert teachers in a very easy and step by step method. We have also added class 12 Maths Chapter 4 exercise 4.3 pdf on this site. Regular solving of these solutions will help the students to secure better marks.

2. Are the PDFs provided by Vedantu Free of Cost?

Yes, the pdfs provided by Vedantu for various NCERT Solutions and notes are free of cost. Vedantu is an online learning platform which provides free PDF downloads of solved NCERT Solutions and sample papers along with previous years solutions for various state boards and CBSE exams. These solutions are solved by expert teachers on Vedantu.com.

Also, register yourself to the Vedantu app to get the latest information, study materials and PDFs CBSE and various state board examinations.

3. Which books shall I refer along with NCERT Class 12 Maths Textbook?

Although NCERT textbooks provide detailed information and solutions for every chapter, it is advisable to the students to refer to R.S. Agarwal and R.D. Sharma textbooks for better understanding. There are proper chapter-wise solutions provided in the textbook which are categorised as very short, short, long, multiple-type and high order thinking type questions. Therefore, CBSE students can refer to both R.S. Agarwal and R. D. Sharma books for gaining better concepts regarding any topic. The students can also download free PDFs of solved R.S.Agarwal's questions from the Vedantu’s website. These solutions are solved by our expert teachers who are having years of experience in this subject.

4. How can I score well in the Class 12 Maths Final Exam?

Class 12 final examination holds an important part in every student's life. Therefore, they need to solve the NCERT solutions and practice papers regularly. Candidates aiming for JEE should have a good grasp over the maths subject. They are advised to revise the important mathematical formulas and attempt maximum mock tests for better understanding and marks.

Before starting their preparation, students must have a deep understanding of the Class 12 maths syllabus and should focus on his/her weak areas. He/She must also follow a regular time table and segregate the syllabus accordingly.

6. What is ex 4.3 class 12 in Chapter 4 about?

Exercise 4.3 focuses on calculating minors and cofactors in determinants. This exercise is part of Chapter 4, which covers determinants in the NCERT Class 12 Maths syllabus. Understanding minors and cofactors is crucial for mastering the concepts of determinants. This exercise provides foundational knowledge necessary for more complex matrix operations. It helps in solving various mathematical problems related to determinants.

7. Why are minors and cofactors important in class 12 maths ex 4.3?

Minors and cofactors are essential for understanding how to calculate determinants and solve matrix equations. They play a critical role in finding the inverse of a matrix. These concepts are fundamental in advanced mathematics and various applications in science and engineering. Mastering minors and cofactors enables students to handle complex problems efficiently. They form the basis for many operations in linear algebra.

8. How do I find the minor of an element in class 12 ex 4.3?

To find the minor of an element, remove the row and column of that element. Then, find the determinant of the resulting submatrix. This submatrix is obtained by excluding the row and column where the element resides. The minor is the determinant of this smaller matrix. Practice with different elements to become proficient in this method.

9. What should I focus on in 4.3 maths class 12?

Focus on accurately calculating minors and understanding the cofactor expansion process. Ensure you are comfortable with the steps involved in finding submatrices. Pay attention to the sign conventions used in cofactor calculations. Practice problems regularly to build confidence and accuracy. Understanding these concepts deeply will help in solving more complex problems later.

10. Are there any tips for solving Exercise 4.3 maths class 12 problems?

Practice regularly and ensure you understand each step in finding minors and cofactors. Double-check your work for accuracy, especially the signs in cofactor calculations. Use systematic approaches to avoid confusion and errors. Study solved examples to grasp the methodology thoroughly. Seeking help from teachers or peers when stuck can also be beneficial.