Suppose that
\begin{equation}
A
=
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
\end{equation}
for some complex numbers $a, b, c, d \in \mathbb{C}$. Result
R5518: Complex arithmetic expression for the determinant of a 2-by-2 complex square matrix shows that
\begin{equation}
\text{Det} A
= a d - b c
\end{equation}
Our goal is to show that the cofactor partitions equal this value. To this end, we have the cofactors
\begin{equation}
\begin{split}
C_{1, 1} & = (-1)^{1 + 1} \begin{bmatrix} d \end{bmatrix} = d \\
C_{1, 2} & = (-1)^{1 + 2} \begin{bmatrix} c \end{bmatrix} = -c \\
C_{2, 1} & = (-1)^{2 + 1} \begin{bmatrix} b \end{bmatrix} = -b \\
C_{2, 2} & = (-1)^{2 + 2} \begin{bmatrix} a \end{bmatrix} = a \\
\end{split}
\end{equation}
and the values
\begin{equation}
\begin{split}
A_{1, 1} & = a \\
A_{1, 2} & = b \\
A_{2, 1} & = c \\
A_{2, 2} & = d \\
\end{split}
\end{equation}
Let's go through each case for all possible values of $i, j \in \{ 1, 2 \}$. Assume first that $i = j = 1$. Then
\begin{equation}
A_{i, 1} C_{j, 1} + A_{i, 2} C_{j, 2}
= A_{1, 1} C_{1, 1} + A_{1, 2} C_{1, 2}
= a d + b (-c)
= ad - b c
= \text{Det} A
\end{equation}
and
\begin{equation}
A_{1, i} C_{1, j} + A_{2, i} C_{2, j}
= A_{1, 1} C_{1, 1} + A_{2, 1} C_{2, 1}
= a d + c (-b)
= ad - b c
= \text{Det} A
\end{equation}
Assume next that $i = j = 2$. Then
\begin{equation}
A_{i, 1} C_{j, 1} + A_{i, 2} C_{j, 2}
= A_{2, 1} C_{2, 1} + A_{2, 2} C_{2, 2}
= c (-b) + d a
= a d - b c
= \text{Det} A
\end{equation}
and
\begin{equation}
A_{1, i} C_{1, j} + A_{2, i} C_{2, j}
= A_{1, 2} C_{1, 2} + A_{2, 2} C_{2, 2}
= b (-c) + d a
= a d - b c
= \text{Det} A
\end{equation}
Suppose then that $i = 1$ and $j = 2$. Then
\begin{equation}
A_{i, 1} C_{j, 1} + A_{i, 2} C_{j, 2}
= A_{1, 1} C_{2, 1} + A_{1, 2} C_{2, 2}
= a (-b) + b a
= a b - a b
= 0
\end{equation}
and
\begin{equation}
A_{1, i} C_{1, j} + A_{2, i} C_{2, j}
= A_{1, 1} C_{1, 2} + A_{2, 1} C_{2, 2}
= a (-c) + c a
= a c - ac
= 0
\end{equation}
Finally, suppose that $i = 2$ and $j = 1$. Then
\begin{equation}
A_{i, 1} C_{j, 1} + A_{i, 2} C_{j, 2}
= A_{2, 1} C_{1, 1} + A_{2, 2} C_{1, 2}
= c d + d (-c)
= c d - c d
= 0
\end{equation}
and
\begin{equation}
A_{1, 2} C_{1, 1} + A_{2, 2} C_{2, 1}
= b d + d (-b)
= b d - bd
= 0
\end{equation}
Since all of the possible cases work out, the proof is complete. $\square$
