Proving Det(B) = k*Det(A): A Fundamental Property of Determinants?

[asap9993]

Homework Statement

I need help in proving that if B is the matrix that results when a single row or column of A is multiplied by a scalar k then det(B) = k*det(A).

Homework Equations

The Attempt at a Solution

The only way I could think of is setting up a general n x n matrix where all the entries on one of the rows or columns is multiplied by k. Then I could say that if one uses cofactor expansion along that particular row or column then the k factors out from all the cofactors.
However, the weakness to this method is that I would have to show that it also holds if one decides to expand along a row or column besides the one multiplied by k. I'm not sure if I could do that for every nth row or column.

 

[vela]

Don't you already know that it doesn't matter which row or column you expand on, or has that not been proved in your course yet?

 

[asap9993]

Of course everyone knows that. But how do you prove it for EVERY SINGLE POSSIBLE nth row or column for EVERY SINGLE POSSIBLE SQUARE MATRIX? Neither my teacher or my textbook proves it.
I can't just write "Take my word that it doesn't matter which row or column you expand upon" or "the textbook says so" on my proof.
If you could give some me help or tell me where I could find it, I would greatly appreciate it.

 

[Ryker]

What definition of the determinant are you going off? Have you defined what happens to the determinant when you perform an elementary row operation on it?

 

[vela]

I'll disagree a bit. If it's in your textbook and you've already covered it, it's fair game to use in a subsequent proof.

When you take math courses, there are all sorts of facts you're using without proof, like the properties of the real numbers. You're not expected to prove real numbers have all these properties before you use them in a linear algebra proof. So part of succeeding in the course is to figure out what you know and what you don't know, in the sense you can use this knowledge without necessarily proving it first.

When I took linear algebra, it was established pretty early that it didn't matter what row or column you expanded on to calculate the determinant, precisely to avoid the issue you're worried about now. That's what I meant by my question, whether it's already considered known to you.

 

[Stardust*]

For what I know, this fact for a row directly derives from the properties of the determinant. It is defined:
a) to be a linear application in the rows,
b) to give 0 if two rows are equal,
c) to give 1 if you calculate det(I_n), with I_n the identity matrix.
In particular the answer comes from a).

Instead for a column, you can think that the determinant of A calculated along the i-th column is the same as the determinant calculated along the i-th row of the transpose of the matrix, ^tA. It is well known (and not so long to show) that det(A)=det(^tA). So the situation is reduced to the one explained before.