Question

Find the ad-joint of the matrix, {\left( {\dfrac{{{{\text{7}}^{ - 4}}}}{{{4^{ - 2}}}}} \right)^{\dfrac{1}{4}}}\left[ {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\\ 3&0&{ - 2} \\\ 1&0&3 \end{array}} \right].

Answer 1

In order to find the adjoint of the given matrix, we follow the procedure to find the adjoint of a 3 × 3 matrix. We do it in a stepwise manner. The Cofactor of each of the elements from each of the given matrix is found, all the respective cofactors are written in their relative positions in form of a matrix. The transpose of this matrix gives us the adjoint of the given matrix.

Complete step-by-step answer:
Given Data,
{\left( {\dfrac{{{{\text{7}}^{ - 4}}}}{{{4^{ - 2}}}}} \right)^{\dfrac{1}{4}}}\left[ {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\\ 3&0&{ - 2} \\\ 1&0&3 \end{array}} \right]
Let us start by simplifying the coefficient of the given matrix, ${\left( {\dfrac{{{{\text{7}}^{ - 4}}}}{{{4^{ - 2}}}}} \right)^{\dfrac{1}{4}}}$
$\Rightarrow {\left( {\dfrac{{{{\text{7}}^{ - 4}}}}{{{4^{ - 2}}}}} \right)^{\dfrac{1}{4}}} \\\ \Rightarrow {\left( {\dfrac{{\left( {\dfrac{1}{{{7^4}}}} \right)}}{{\left( {\dfrac{1}{{{4^2}}}} \right)}}} \right)^{\dfrac{1}{4}}} \\\ \Rightarrow {\left( {\dfrac{{{{\left( {{2^2}} \right)}^2}}}{{{7^4}}}} \right)^{\dfrac{1}{4}}} \\\ \Rightarrow {\left( {\dfrac{2}{7}} \right)^{4 \times \dfrac{1}{4}}} \\\ \Rightarrow \dfrac{2}{7} \\\$
So the given matrix can be expressed as:
\Rightarrow \left( {\dfrac{2}{7}} \right)\left[ {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\\ 3&0&{ - 2} \\\ 1&0&3 \end{array}} \right]
\Rightarrow \left[ {\begin{array}{*{20}{c}} {\dfrac{2}{7}}&{ - \dfrac{2}{7}}&{\dfrac{4}{7}} \\\ {\dfrac{6}{7}}&0&{ - \dfrac{4}{7}} \\\ {\dfrac{2}{7}}&0&{\dfrac{6}{7}} \end{array}} \right]
Now we find the adjoint of the given matrix as follows:
\Rightarrow \left[ {\begin{array}{*{20}{c}} {\dfrac{2}{7}}&{ - \dfrac{2}{7}}&{\dfrac{4}{7}} \\\ {\dfrac{6}{7}}&0&{ - \dfrac{4}{7}} \\\ {\dfrac{2}{7}}&0&{\dfrac{6}{7}} \end{array}} \right]
To find the adjoint of this matrix, we write the cofactor matrix for the respective elements in the relative positions and attribute them to the respective sign in each position.
The respective signs of each position in the third order matrix while finding its adjoint is given by\left[ {\begin{array}{*{20}{c}} \+ & \- & \+ \\\ \- & \+ & \- \\\ \+ & \- & \+ \end{array}} \right]
And the cofactor of an element of a third order matrix of the type \left[ {\begin{array}{*{20}{c}} {\text{a}}&{\text{b}}&{\text{c}} \\\ {\text{d}}&{\text{e}}&{\text{f}} \\\ {\text{g}}&{\text{h}}&{\text{i}} \end{array}} \right]is given as follows:
The cofactor of the element ‘e’ is given by, (ai – cg) multiplied by its respective sign. The respective sign in the position of element ‘e’ is ‘+’ from the above.
Therefore the cofactor of ‘e’ is + (ai – cg).
In this manner we find all the cofactors of the elements of the given matrix and make a cofactor matrix.
Each of the elements of the matrix is represented by${{\text{M}}_{{\text{xy}}}}$, where x represents the row and y represents the column.
\Rightarrow \left[ {\begin{array}{*{20}{c}} {\dfrac{2}{7}}&{ - \dfrac{2}{7}}&{\dfrac{4}{7}} \\\ {\dfrac{6}{7}}&0&{ - \dfrac{4}{7}} \\\ {\dfrac{2}{7}}&0&{\dfrac{6}{7}} \end{array}} \right]
Cofactor of ${{\text{M}}_{11}}$ is ${\text{ + }}\left( {\left( {0 \times \dfrac{6}{7}} \right) - \left( {0 \times - \dfrac{4}{7}} \right)} \right) = {\text{ }}\left( {0{\text{ - 0}}} \right) = 0$
Cofactor of ${{\text{M}}_{12}}$ is ${\text{ - }}\left( {\left( {\dfrac{6}{7} \times \dfrac{6}{7}} \right) - \left( {\dfrac{2}{7} \times - \dfrac{4}{7}} \right)} \right) = {\text{ - }}\left( {\dfrac{{36}}{{49}}{\text{ + }}\dfrac{8}{{49}}} \right) = - \dfrac{{44}}{{49}}$
Cofactor of ${{\text{M}}_{13}}$ is ${\text{ + }}\left( {\left( {\dfrac{6}{7} \times 0} \right) - \left( {\dfrac{2}{7} \times 0} \right)} \right) = {\text{ + }}\left( {0 + 0} \right) = 0$
Cofactor of ${{\text{M}}_{21}}$ is ${\text{ - }}\left( {\left( { - \dfrac{2}{7} \times \dfrac{6}{7}} \right) - \left( {0 \times \dfrac{4}{7}} \right)} \right) = {\text{ - }}\left( { - \dfrac{{12}}{{49}}{\text{ - 0}}} \right) = \dfrac{{12}}{{49}}$
Cofactor of ${{\text{M}}_{22}}$ is ${\text{ + }}\left( {\left( {\dfrac{2}{7} \times \dfrac{6}{7}} \right) - \left( {\dfrac{2}{7} \times \dfrac{4}{7}} \right)} \right) = {\text{ + }}\left( {\dfrac{{12}}{{49}}{\text{ - }}\dfrac{8}{{49}}} \right) = \dfrac{4}{{49}}$
Cofactor of ${{\text{M}}_{23}}$ is ${\text{ - }}\left( {\left( {\dfrac{2}{7} \times 0} \right) - \left( {\dfrac{2}{7} \times \dfrac{-2}{7}} \right)} \right) = {\text{ - }}\left( {{\text{0 + }}\dfrac{4}{{49}}} \right) = - \dfrac{4}{{49}}$
Cofactor of ${{\text{M}}_{31}}$ is ${\text{ + }}\left( {\left( { - \dfrac{2}{7} \times - \dfrac{4}{7}} \right) - \left( {0 \times \dfrac{4}{7}} \right)} \right) = {\text{ }}\left( {\dfrac{8}{{49}}{\text{ - 0}}} \right) = \dfrac{8}{{49}}$
Cofactor of ${{\text{M}}_{32}}$ is ${\text{ - }}\left( {\left( {\dfrac{2}{7} \times - \dfrac{4}{7}} \right) - \left( {\dfrac{6}{7} \times \dfrac{4}{7}} \right)} \right) = {\text{ - }}\left( { - \dfrac{8}{{49}}{\text{ - }}\dfrac{{24}}{{49}}} \right) = \dfrac{{32}}{{49}}$
Cofactor of ${{\text{M}}_{33}}$ is ${\text{ + }}\left( {\left( {\dfrac{2}{7} \times 0} \right) - \left( { - \dfrac{2}{7} \times \dfrac{6}{7}} \right)} \right) = {\text{ }}\left( {{\text{0 + }}\dfrac{{12}}{{49}}} \right) = \dfrac{{12}}{{49}}$
Therefore the cofactor matrix of the given matrix is given by:
\left[ {\begin{array}{*{20}{c}} 0&{\dfrac{{44}}{{49}}}&0 \\\ {\dfrac{{12}}{{49}}}&{\dfrac{4}{{49}}}&{-\dfrac{4}{{49}}} \\\ {\dfrac{8}{{49}}}&{\dfrac{{32}}{{49}}}&{\dfrac{{12}}{{49}}} \end{array}} \right]
Now we know the adjoint matrix of the given matrix is the transpose of the cofactor matrix.
The transpose of the cofactor matrix is given by interchanging its rows and columns.

0&{\dfrac{{44}}{{49}}}&0 \\\ {\dfrac{{12}}{{49}}}&{\dfrac{4}{{49}}}&{-\dfrac{4}{{49}}} \\\ {\dfrac{8}{{49}}}&{\dfrac{{32}}{{49}}}&{\dfrac{{12}}{{49}}} \end{array}} \right]^{\text{T}}} = \left[ {\begin{array}{*{20}{c}} 0&{\dfrac{{12}}{{49}}}&{\dfrac{8}{{49}}} \\\ {\dfrac{{44}}{{49}}}&{\dfrac{4}{{49}}}&{\dfrac{{32}}{{49}}} \\\ 0&{-\dfrac{4}{{49}}}&{\dfrac{{12}}{{49}}} \end{array}} \right]$$ Therefore the adjoint matrix of the given, ${\left( {\dfrac{{{{\text{7}}^{ - 4}}}}{{{4^{ - 2}}}}} \right)^{\dfrac{1}{4}}}\left[ {\begin{array}{*{20}{c}} 1&{ - 1}&2 \\\ 3&0&{ - 2} \\\ 1&0&3 \end{array}} \right]$ is given by $$\left[ {\begin{array}{*{20}{c}} 0&{\dfrac{{12}}{{49}}}&{\dfrac{8}{{49}}} \\\ {\dfrac{{44}}{{49}}}&{\dfrac{4}{{49}}}&{\dfrac{{32}}{{49}}} \\\ 0&{-\dfrac{4}{{49}}}&{\dfrac{{12}}{{49}}} \end{array}} \right]$$. **Note:** In order to solve this type of questions the key is to know how to write a cofactor of a respective term of a matrix and the signs each element of the cofactor matrix takes respectively. This process has to be done after multiplying the given matrix with the coefficients it has, if any out the matrix, this is an important step to look out for. The transpose of the cofactor matrix of the given matrix gives us the adjoint matrix. While transposing a matrix, all the rows of the given matrix become columns and all the columns of the given matrix become its rows.