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?
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!