<s>[INST] Finding the determinant of a $2\times 2$ matrix Let $R$ be a commutative ring with 1. Show with the Leibniz signature formula that $\det\biggl(\begin{bmatrix}
\lambda & A_{12}\\
0 & A_{22}\\
\end{bmatrix}\biggl) = \lambda \cdot \det(A_{22})$ applies whereby $\lambda \in R \land A_{22} \in R^{n,n}$.
My idea:
$\sigma_{1} =  \operatorname{id} = 
\begin{pmatrix}
\lambda & A_{12}\\
0 & A_{22}\\
\end{pmatrix}
$ and $\sigma_{2} = \begin{pmatrix}
0 & A_{22}\\
\lambda & A_{12}\\
\end{pmatrix}
$ 
I know that if the number of transpositions are equal then I will get a positive mathematical operator otherwise I will get a negative one. But I think this approach is only useful if I have actual numbers in the matrix. I got the following out of this:
$\det A = \lambda \cdot A_{22} - 0 \cdot A_{12}$
Just to make clear what I mean here is an example where it works:
$A = \begin{pmatrix}
1 & 3&4\\
2&1&7\\
6&7&8
\end{pmatrix}
$ $\sigma_{1} =  \operatorname{id} = 
\begin{pmatrix}
1 & 3&4\\
2&1&7\\
6&7&8
\end{pmatrix}
$ $\Rightarrow$ $\sigma_{2} =  
\begin{pmatrix}
2 & 1&7\\
1&3&4\\
6&7&8
\end{pmatrix}
$ $\Rightarrow$ $\sigma_{3} = 
\begin{pmatrix}
1 & 3&4\\
6&7&8\\
2&1&7
\end{pmatrix}
$
$\Rightarrow$ $\sigma_{4} =  
\begin{pmatrix}
6 & 7&8\\
2&1&7\\
1&3&4
\end{pmatrix}
$ $\Rightarrow$ $\sigma_{5} = 
\begin{pmatrix}
6 & 7&8\\
1&3&4\\
2&1&7
\end{pmatrix}
$ $\Rightarrow$ $\sigma_{6} = 
\begin{pmatrix}
2 & 1&7\\
6&7&8\\
1&3&4
\end{pmatrix}
$ 
so that $\det A = 8 -48 -49 -24 +126 +56 = 69$.
Additional information:
Leibniz signature formula: 
$$\det: R^{n,n} \to R, A = [a_{ij}] \mapsto \det(A):= \sum_{\sigma \in S_{n}}sgn(\sigma) \prod_{i=1}^n a_{i},_{\sigma(i)}$$
How do I show now that $\det\biggl(\begin{bmatrix}
\lambda & A_{12}\\
0 & A_{22}\\
\end{bmatrix}\biggl) = \lambda \cdot \det(A_{22})$?
Could someone show me how to solve this proof in a proper way?
 [/INST] While you are right to be concerned that it's a little more tricky if $A_{22}$ is a larger matrix, since $A$ is only a $2\times 2$ matrix, $A_{22}$ is just a number and your work is correct!
</s>