Question

Given that det(AB)=det(A)det(B) and det(I)=1, prove that the determinant behaves as a linear function of a given row if the other rows stay the same. In other words, prove that \[\begin{vmatrix}ta & tb \\ c & d\end{vmatrix}=t\begin{vmatrix}a & b \\ c & d\end{vmatrix} \text{ where } a,b,c,d,t \in \mathbb R\] and \[\begin{vmatrix}a+a' & b+b' \\ c & d\end{vmatrix}=\begin{vmatrix}a & b \\ c & d\end{vmatrix}+\begin{vmatrix}a' & b' \\ c & d\end{vmatrix} \text{ where } a,a',b,b',c,d \in \mathbb R\]

Answer 1

I is the identity matrix and A and B are assumed to be compatible.

Answer 2

(d(ta)-c(tb)) = t[(ad-bc)]
t(ad-bc)
therefore they are the same


[(a+a')d-(b+b')c] = (ad-bc)+(a'd-b'c)
d(a+a')-c(b+b')
therefore they are the same

Answer 3

all you have to know how to is find a determinant of a two by two
which is
\[\begin{vmatrix}A & B \\ C & D\end{vmatrix}=AD-BC\]

Answer 4

I know that, let's just assume you only know the properties I gave... what I put were two generic two-by-two determinants... I know this property holds for all square matrices, so I'm looking for a connection between the first two statements and these last two. I've seen the determinant defined by just those two properties that I outlined.

Answer 5

Look how the determinant is defined here: http://mathprelims.wordpress.com/2009/06/13/determinants/