Answer
$A_{21}=2$; $A_{22}=-6$ and $A_{23}=4$
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 $a_{23}$.
$M_{21}=\left|\begin{array}{ll}
0 &1\\ 2 & 1 \end{array}\right|=(0)(1)-(1)(2)=-2$ and $M_{21}=\left|\begin{array}{ll}
-2 &1\\ 4 & 1 \end{array}\right|=(-2)(1)-(1)(4)=-6$ and $M_{23}=\left|\begin{array}{ll}
0 &1\\ 2 & 1 \end{array}\right|=(0)(1)-(1)(2)=-2$
Now, $A_{21}=(-1)^{2+1}M_{21}=(-1)(-2)=2$; $A_{22}=(-1)^{2+2}M_{22}=(1)(-6)=-6$ and $A_{23}=(-1)^{2+3}M_{23}=(-1)(-4)=4$
Therefore, $A_{21}=2$; $A_{22}=-6$ and $A_{23}=4$
