Question

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

Answer 1

Let $A=\left[\begin{array}{ccc}1& 1& 2\\ 2& 3& 5\\ 2& 0& 1\end{array}\right]$

We know that $adjA=C^T$

So, we will find out co-factors of each element of $A$.

$C_{11}=(-1{)}^{1+1}\left|\begin{array}{cc}3& 5\\ 0& 1\end{array}\right|$

$\Rightarrow C_{11}=3-0=3$

$C_{12}=(-1{)}^{1+2}\left|\begin{array}{cc}2& 5\\ 2& 1\end{array}\right|$

$\Rightarrow C_{12}=-(2-10)=8$

$C_{13}=(-1{)}^{1+3}\left|\begin{array}{cc}2& 3\\ 2& 0\end{array}\right|$

$\Rightarrow C_{13}=0-6=-6$

$C_{21}=(-1{)}^{2+1}\left|\begin{array}{cc}1& 2\\ 0& 1\end{array}\right|$

$\Rightarrow C_{21}=-(1-0)=-1$

$C_{22}=(-1{)}^{2+2}\left|\begin{array}{cc}1& 2\\ 2& 1\end{array}\right|$

$\Rightarrow C_{22}=1-4=-3$

$C_{23}=(-1{)}^{2+3}\left|\begin{array}{cc}1& 1\\ 2& 0\end{array}\right|$

$\Rightarrow C_{23}=-(0-2)=2$

$C_{31}=(-1{)}^{3+1}\left|\begin{array}{cc}1& 2\\ 3& 5\end{array}\right|$

$\Rightarrow C_{31}=5-6=-1$

$C_{32}=(-1{)}^{3+2}\left|\begin{array}{cc}1& 2\\ 2& 5\end{array}\right|$

$\Rightarrow C_{32}=-(5-4)=-1$

$C_{33}=(-1{)}^{3+3}\left|\begin{array}{cc}1& 1\\ 2& 3\end{array}\right|$

$\Rightarrow C_{33}=3-2=1$

So, the cofactor matrix $C=\left[\begin{array}{ccc}3& 8& -6\\ -1& -3& 2\\ -1& -1& 1\end{array}\right]$

$\Rightarrow C^T=\left[\begin{array}{ccc}3& -1& -1\\ 8& -3& -1\\ -6& 2& 1\end{array}\right]$

Hence, $adjA=\left[\begin{array}{ccc}3& -1& -1\\ 8& -3& -1\\ -6& 2& 1\end{array}\right]$