Chapter 4: Determinants

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,

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

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

\[\therefore {{\text{A}}_{12}}\text{=}-b\]

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

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

\[\therefore {{\text{A}}_{21}}\text{=}-\text{c}\]

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

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

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

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

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

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

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

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

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

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

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

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

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

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, \[\Delta \text{=21}\].

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

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.