NCERT Solutions for Class 12 Maths Chapter 4 Determinants Exercise 4.3 – 2025-26

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,

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

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

\[\therefore {{\text{A}}_{12}}\text{=0}\]

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

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

\[\therefore {{\text{A}}_{21}}\text{=4}\]

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

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

\[\therefore {{\text{A}}_{22}}\text{=2}\]

2. \[\mathbf{\left| \begin{matrix} \text{a} & \text{c}  \\ \text{b} & \text{d}  \\ \end{matrix} \right|}\]

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 \[{{A}_{ij}}\text{=}{{\left( \text{-1} \right)}^{\text{i+j}}}{{M}_{ij}}\]. 

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

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

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

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

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

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

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

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

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

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

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.

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