Matrices and Determinants
5.0 Adjoint of a square Matrix
5.0 Adjoint of a square Matrix
Let $A = \left[ {{a_{ij}}} \right]$ be a square matrix of order $n$ and let ${{C_{ij}}}$ be cofactor of ${{C_{ij}}}$ in $A$.Then the transpose of the matrix of cofactors of elements of $A$ is called the adjoint of $A$ and is denoted by $adj A$
Thus ,$adj A$=${\left[ {{C_{ij}}} \right]^T}$.
If $A = \left( {\begin{array}{c} {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\ {{a_{31}}}&{{a_{32}}}&{{a_{33}}} \end{array}} \right)$,then, $adj A$ = ${\left( {\begin{array}{c} {{C_{11}}}&{{C_{12}}}&{{C_{13}}} \\ {{C_{21}}}&{{C_{22}}}&{{C_{23}}} \\ {{C_{31}}}&{{C_{32}}}&{{C_{33}}} \end{array}} \right)^T} = $ ${\left( {\begin{array}{c} {{C_{11}}}&{{C_{21}}}&{{C_{31}}} \\ {{C_{12}}}&{{C_{22}}}&{{C_{32}}} \\ {{C_{13}}}&{{C_{23}}}&{{C_{33}}} \end{array}} \right)}$
Where ${C_{ij}}$ denotes the cofactor of ${a_{ij}}$ in $A$
Question 11.
Find the adjoint of the matrix $A = \left[ {\begin{array}{c} 1&2&3 \\ 0&5&0 \\ 2&4&3 \end{array}} \right]$.
Solution:
If $C$ be the matrix of cofactors of the element in $\left| A \right|$
then$C = \left( {\begin{array}{c} {{C_{11}}}&{{C_{12}}}&{{C_{13}}} \\ {{C_{21}}}&{{C_{22}}}&{{C_{23}}} \\ {{C_{31}}}&{{C_{32}}}&{{C_{33}}} \end{array}} \right)$
Here ,
\[{C_{11}} = \left| {\begin{array}{c} 5&0 \\ 4&3 \end{array}} \right| = 15 \]
\[{C_{12}} = - \left| {\begin{array}{c} 0&0 \\ 2&3 \end{array}} \right| = 0 \]
\[{C_{13}} = \left| {\begin{array}{c} 0&5 \\ 2&4 \end{array}} \right| = - 10 \]
\[{C_{21}} = - \left| {\begin{array}{c} 2&3 \\ 4&3 \end{array}} \right| = 6 \]
\[ {C_{22}} = \left| {\begin{array}{c} 1&3 \\ 2&3 \end{array}} \right| = - 3 \]
\[{C_{23}} = \left| {\begin{array}{c} 1&2 \\ 2&4 \end{array}} \right| = 0 \]
\[{C_{31}} = \left| {\begin{array}{c} 2&3 \\ 5&0 \end{array}} \right| = - 15 \]
\[{C_{32}} = - \left| {\begin{array}{c} 1&3 \\ 0&0 \end{array}} \right| = 0 \]
\[ {C_{33}} = \left| {\begin{array}{c} 1&2 \\ 0&5 \end{array}} \right| = 5 \]
\[ \therefore C = \left( {\begin{array}{c} {15}&0&{ - 10} \\ 6&{ - 3}&0 \\ { - 15}&0&5 \end{array}} \right) \\ \\ \therefore AdjA = {C^T} = \left( {\begin{array}{c} {15}&6&{ - 15} \\ 0&{ - 3}&0 \\ { - 10}&0&5 \end{array}} \right)\]
Question 12.
If $A = \left[ {\begin{array}{c} 3&{ - 3}&4 \\ 2&{ - 3}&4 \\ 0&{ - 1}&1 \end{array}} \right]$
then find $Adj\left( {AdjA} \right)$
Solution:
We know that $Adj\left( {AdjA} \right) = {\left| A \right|^{n - 2}}A$
Since $A = \left[ {\begin{array}{c} 3&{ - 3}&4 \\ 2&{ - 3}&4 \\ 0&{ - 1}&1 \end{array}} \right]$
Here $n$=3
$\therefore \left| A \right| = \left| {\begin{array}{c} 3&{ - 3}&4 \\ 2&{ - 3}&4 \\ 0&{ - 1}&1 \end{array}} \right|$
$ = 3\left( 1 \right) + 3\left( {2 - 0} \right) + 4\left( { - 2 - 0} \right)$
$ = 1 \ne 0$
$\therefore A$ is non singular
$\therefore Adj\left( {AdjA} \right) = {\left| A \right|^{3 - 2}}.A$
$ = {\left| A \right|^{}}.A$
$ = I.A$
$ = A$
$\therefore A = \left[ {\begin{array}{c} 3&{ - 3}&4 \\ 2&{ - 3}&4 \\ 0&{ - 1}&1 \end{array}} \right]$
