Development:
We have to find the complementary minors of the elements of the main diagonal, that is, of the elements $$a_{11}, a_{22}, a_{33}, a_{44}$$.
$$$a_{11}=\left(\begin{matrix} \xcancel{1} & \cancel{0} & \cancel{1} & \cancel{0} \\ \bcancel{1} & 1 & 0 & 2\\ \bcancel{1} & 1 & 1 & 0 \\ \bcancel{1} & 2 & 3 & 1 \end{matrix}\right)=\begin{matrix} \left|\begin{matrix} 1 & 0 & 2 \\ 1 & 1 & 0\\ 2 & 3 & 1 \end{matrix}\right| \\ \begin{matrix} 1 & 0 & 2 \\ 1 & 1 & 0 \end{matrix}\end{matrix}=$$$
$$$=1\cdot1\cdot1+1\cdot3\cdot2+\cancel{2\cdot0\cdot0}-2\cdot1\cdot2-\cancel{0\cdot3\cdot1}-\cancel{1\cdot0\cdot1}=1+6-4=3$$$
Using the formula of Sarrus to calculate the determinant $$3\times3$$.
The rest of minors must be calculated in the same way
$$$a_{22}=\left(\begin{matrix} 1 & \cancel{0} & 1 & 0 \\ \bcancel{1} & \xcancel{1} & \bcancel{0} & \bcancel{2}\\ 1 & \cancel{1} & 1 & 0 \\ 1 & \cancel{2} & 3 & 1 \end{matrix}\right)=\left|\begin{matrix} 1 & 1 & 0 \\ 1 & 1 & 0\\ 1 & 3 & 1 \end{matrix}\right| = 0 $$$
(since there are repeated rows)
$$$a_{33}=\left(\begin{matrix} 1 & 0 & \cancel{1} & 0 \\ 1 & 1 & \cancel{0} & 2\\ \bcancel{1} & \bcancel{1} & \xcancel{1} & \bcancel{0} \\ 1 & 2 & \cancel{3} & 1 \end{matrix}\right)=\left|\begin{matrix} 1 & 0 & 0 \\ 1 & 1 & 2\\ 1 & 2 & 1 \end{matrix}\right| =$$$
$$$=1\cdot1\cdot1+1\cdot2\cdot0+1\cdot0\cdot2-0\cdot1\cdot1-2\cdot2\cdot1-1\cdot0\cdot1=-3$$$
$$$a_{44}=\left(\begin{matrix} 1 & 0 & 1 & \cancel{0} \\ 1 & 1 & 0 & \cancel{2}\\ 1 & 1 & 1 & \cancel{0} \\ \bcancel{1} & \bcancel{2} & \bcancel{3} & \xcancel{1} \end{matrix}\right)=\left|\begin{matrix} 1 & 0 & 1 \\ 1 & 1 & 0\\ 1 & 1 & 1 \end{matrix}\right| =$$$
$$$=1\cdot1\cdot1+1\cdot1\cdot1+1\cdot0\cdot0-1\cdot1\cdot1-0\cdot1\cdot1-1\cdot0\cdot1=1$$$
Solution:
$$a_{11}=3, a_{22}=0, a_{33}=-3, a_{44}=1$$
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