Answer
$A_{21}=-5$; $A_{22}=1$ and $A_{23}=3$
Work Step by Step
Use the formula $A_{ij}=(-1)^{i+j}M_{ij}$ for $i=2$ and $j=1$ to obtain $A_{21}$.
First, we will find the minors for element $a_{21},a_{22}$ and $a_{23}$.
$M_{21}=\left|\begin{array}{ll}
-1 & 2\\ -3 & 1 \end{array}\right|=(-1)(1)-(2)(-3)=5$ and $M_{21}=\left|\begin{array}{ll}
1 & 2\\ 0 & 1 \end{array}\right|=(1)(1)-(2)(0)=1$ and $M_{23}=\left|\begin{array}{ll}
1 & -1\\ 0 & -3 \end{array}\right|=(1)(-3)-(-1)(0)=-3$
Now, $A_{21}=(-1)^{2+1}M_{21}=(-1)(5)=-5$; $A_{22}=(-1)^{2+2}M_{22}=(1)(1)=1$ and $A_{23}=(-1)^{2+3}M_{23}=(-1)(-3)=3$
Therefore, $A_{21}=-5$; $A_{22}=1$ and $A_{23}=3$
