TITLE: How to prove this property of determinant
QUESTION [1 upvotes]: From the answer to this question, how can one prove that 
$$\begin{vmatrix} a-\lambda & b \\
   c & d-\lambda \end{vmatrix} = 
  \begin{vmatrix} a & b \\
   c & d-\lambda \end{vmatrix} + 
  \begin{vmatrix} -\lambda & b \\
   0 & d-\lambda \end{vmatrix}$$
I mean it is not adding a multiple of a row / column to another row / column, and it is clear that the equality holds, but how to prove this ? Which proper of the determinant is used in here ?
The user who wrote the answer told me to check out this question, but I still don't get it.

REPLY [2 votes]: You can just multiply it out using the definition of the determinant:
The left hand side is
$$(a-\lambda)(d-\lambda)-bc = ad-bc-(a+d)\lambda + \lambda^{2}$$
The right hand side is
$$a(d-\lambda)-bc + (-\lambda)(d-\lambda) = ad-bc -(a+d)\lambda +\lambda^{2}$$

REPLY [2 votes]: Let
$$v_1=(a,c) $$
$$v_2=(-\lambda,0) $$
$$v_3=(b,d-\lambda) $$
we just use the fact that
$$det (v_1+v_2,v_3)=det (v_1,v_3)+det (v_2,v_3) $$
It is a bilinear form.
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