Question

Find adjoint of the matrix $\left[\begin{array}{ccc}1& -1& 2\\ 2& 3& 5\\ -2& 0& 1\end{array}\right]$

Answer 1

Given Matrix: $\left[\begin{array}{ccc}1& -1& 2\\ 2& 3& 5\\ -2& 0& 1\end{array}\right]$
Minors:
$M_{11}=\left|\begin{array}{cc}3& 5\\ 0& 1\end{array}\right|=3\left(1\right)-0\left(5\right)=3-0=3$
$M_{12}=\left|\begin{array}{cc}2& 5\\ -2& 1\end{array}\right|=2-(-10)=2+10=12$
$M_{13}=\left|\begin{array}{cc}2& 3\\ -2& 0\end{array}\right|=0-(-6)=0+6=6$
$M_{21}=\left|\begin{array}{cc}-1& 2\\ 0& 1\end{array}\right|=-1-\left(0\right)=-1-0=-1$
$M_{22}=\left|\begin{array}{cc}1& 2\\ -2& 1\end{array}\right|=1-(-2)2=1+4=5$
$M_{23}=\left|\begin{array}{cc}1& -1\\ -2& 0\end{array}\right|=0-(-2)(-1)=-2$
$M_{31}=\left|\begin{array}{cc}-1& 2\\ 3& 5\end{array}\right|=5(-1)-6=-11$
$M_{32}=\left|\begin{array}{cc}1& 2\\ 2& 5\end{array}\right|=5-4=1$
$M_{33}=\left|\begin{array}{cc}1& -1\\ 2& 3\end{array}\right|=3-(-2)=5$
Cofactors:
$A_{11}=(-1{)}^{1+1}{M}_{11}=(-1{)}^2\times 3=3$
$A_{12}=(-1{)}^{1+2}{M}_{12}=(-1{)}^3\times \left(12\right)=-12$
$A_{13}=(-1{)}^{1+3}{M}_{13}=(-1{)}^4\times \left(6\right)=6$
$A_{21}=(-1{)}^{2+1}{M}_{21}=(-1{)}^3\times (-1)=1$
$A_{22}=(-1{)}^{2+2}{M}_{22}=(-1{)}^4\times \left(5\right)=5$
$A_{23}=(-1{)}^{2+3}{M}_{23}=(-1{)}^5\times (-2)=2$
$A_{31}=(-1{)}^{3+1}{M}_{31}=(-1{)}^4\times (-11)=-11$
$A_{32}=(-1{)}^{3+2}{M}_{32}=(-1{)}^5\times \left(1\right)=-1$
$A_{33}=(-1{)}^{3+3}{M}_{33}=(-1{)}^6\times \left(5\right)=5$
Now,
$adjA={\left[\begin{array}{ccc}{A}_{11}& A_{12}& A_{13}\\ A_{21}& A_{22}& A_{23}\\ A_{31}& A_{32}& A_{33}\end{array}\right]}^T=\left[\begin{array}{ccc}{A}_{11}& A_{21}& A_{31}\\ A_{12}& A_{22}& A_{32}\\ A_{13}& A_{23}& A_{33}\end{array}\right]$
$\Rightarrow adjA=\left[\begin{array}{ccc}3& 1& -11\\ -12& 5& -1\\ 6& 2& 5\end{array}\right]$