To compute the minor \(A_{1,3}\text{,}\) we remove the first row and third column of \(A\) then take the determinant.
\begin{align*}
A \amp = \bbm 1\amp 2\amp 3\\4\amp 5\amp 6\\7\amp 8\amp 9\ebm\Rightarrow\bbm\mathbf{\cancel{1}}\amp \mathbf{\cancel{2}}\amp \mathbf{\cancel{3}}\\4\amp 5\amp \mathbf{\cancel{6}}\\7\amp 8\amp \mathbf{\cancel{9}}\ebm\Rightarrow\bbm 4\amp 5\\7\amp 8\ebm\\
A_{1,3} \amp = \bvm 4\amp 5\\7\amp 8\evm = 32-35 = -3\text{.}
\end{align*}
The corresponding cofactor, \(C_{1,3}\text{,}\) is
\begin{equation*}
C_{1,3} = (-1)^{1+3}A_{1,3} = (-1)^4(-3) = -3\text{.}
\end{equation*}
The minor \(A_{3,2}\) is found by removing the third row and second column of \(A\) then taking the determinant.
\begin{align*}
A \amp = \bbm 1\amp 2\amp 3\\4\amp 5\amp 6\\7\amp 8\amp 9\ebm\Rightarrow\bbm 1\amp \mathbf{\cancel{2}}\amp 3\\4\amp \mathbf{\cancel{5}}\amp 6\\ \mathbf{\cancel{7}}\amp \mathbf{\cancel{8}}\amp \mathbf{\cancel{9}}\ebm\Rightarrow \bbm 1\amp 3\\4\amp 6\ebm\\
A_{3,2} \amp = \bvm 1\amp 3\\4\amp 6\evm = 6-12 = -6\text{.}
\end{align*}
The corresponding cofactor, \(C_{3,2}\text{,}\) is
\begin{equation*}
C_{3,2} = (-1)^{3+2}A_{3,2} = (-1)^5(-6) = 6\text{.}
\end{equation*}
The minor \(A_{3,2}\) is found by removing the third row and second column of \(A\) then taking the determinant.
\begin{align*}
A \amp = \bbm 1\amp 2\amp 3\\4\amp 5\amp 6\\7\amp 8\amp 9\ebm\Rightarrow\bbm 1\amp \mathbf{\cancel{2}}\amp 3\\4\amp \mathbf{\cancel{5}}\amp 6\\ \mathbf{\cancel{7}}\amp \mathbf{\cancel{8}}\amp \mathbf{\cancel{9}}\ebm\Rightarrow \bbm 1\amp 3\\4\amp 6\ebm\\
A_{3,2} \amp = \bvm 1\amp 3\\4\amp 6\evm = 6-12 = -6\text{.}
\end{align*}
The corresponding cofactor, \(C_{3,2}\text{,}\) is
\begin{equation*}
C_{3,2} = (-1)^{3+2}A_{3,2} = (-1)^5(-6) = 6\text{.}
\end{equation*}
The minor \(B_{2,1}\) is found by removing the second row and first column of \(B\) then taking the determinant.
\begin{align*}
B \amp = \bbm 1\amp 2\amp 0\amp 8\\-3\amp 5\amp 7\amp 2\\-1\amp 9\amp -4\amp 6\\1\amp 1\amp 1\amp 1\ebm \Rightarrow \bbm\mathbf{\cancel{1}}\amp 2\amp 0\amp 8\\ \mathbf{\cancel{-3}}\amp \mathbf{\cancel{5}}\amp \mathbf{\cancel{7}}\amp \mathbf{\cancel{2}}\\ \mathbf{\cancel{-1}}\amp 9\amp -4\amp 6\\ \mathbf{\cancel{1}}\amp 1\amp 1\amp 1\ebm \Rightarrow \bbm 2\amp 0\amp 8\\9\amp -4\amp 6\\1\amp 1\amp 1\ebm\\
B_{2,1} \amp = \bvm 2\amp 0\amp 8\\9\amp -4\amp 6\\1\amp 1\amp 1\evm \stackrel{!}{=}\text{?}
\end{align*}
We’re a bit stuck. We don’t know how to find the determinant of this \(3\times 3\) matrix. We’ll come back to this later. The corresponding cofactor is
\begin{equation*}
C_{2,1} = (-1)^{2+1}B_{2,1} = -B_{2,1}\text{,}
\end{equation*}
whatever this number happens to be. The minor \(B_{4,3}\) is found by removing the fourth row and third column of \(B\) then taking the determinant.
\begin{align*}
B \amp = \bbm 1\amp 2\amp 0\amp 8\\-3\amp 5\amp 7\amp 2\\-1\amp 9\amp -4\amp 6\\1\amp 1\amp 1\amp 1\ebm \Rightarrow \bbm 1\amp 2\amp \mathbf{\cancel{0}}\amp 8\\-3\amp 5\amp \mathbf{\cancel{7}}\amp 2\\-1\amp 9\amp \mathbf{\cancel{-4}}\amp 6\\ \mathbf{\cancel{1}}\amp \mathbf{\cancel{1}}\amp \mathbf{\cancel{1}}\amp \mathbf{\cancel{1}}\ebm \Rightarrow \bbm 1\amp 2\amp 8\\-3\amp 5\amp 2\\-1\amp 9\amp 6\ebm\\
B_{4,3} \amp = \bvm 1\amp 2\amp 8\\-3\amp 5\amp 2\\-1\amp 9\amp 6\evm \stackrel{!}{=}\text{?}
\end{align*}
Again, we’re stuck. We won’t be able to fully compute \(C_{4,3}\text{;}\) all we know so far is that
\begin{equation*}
C_{4,3} = (-1)^{4+3}B_{4,3} = (-1)B_{4,3}\text{.}
\end{equation*}
Once we learn how to compute determinants for matrices larger than \(2\times 2\) we can come back and finish this exercise.