Determinant of a 3x3 matrix via row reduction

[sooyong94]

Homework Statement

Show that the determinant of

is (a-b)(b-c)(c-a)

Homework Equations

Row reduction, determinants

The Attempt at a Solution

Apparently I got a (a-b)^2 instead of (a-b) when I multiplied them up. It would be grateful if someone can point me out where the mistakes are.

 

[member 587159]

The problem is this. You seem to believe that when you perform a row operation: ##(a-c)R_2 - (a-b)R_3 \rightarrow R3##, the determinant remains unchanged.

To show you an easy example that this is not true:

## A = \begin{pmatrix}
1 & 2 \\
2 & 3
\end{pmatrix}##

It's obvious that this matrix has a determinant equal to ##-1##

Now perform: ##2R_1 + 3R_2 \rightarrow R_2##

Then, we obtain a new matrix A':

## A' = \begin{pmatrix}
1 & 2 \\
8 & 13
\end{pmatrix}##

And this matrix has a determinant equal to ##-3##

 

[sooyong94]

So any ideas to work it out then?

 

[member 587159]

So any ideas to work it out then?

You were on the right track. I will give you this hint:

When you perform a row (kolom) operation:

##R_a + k*R_b \rightarrow R_a##, the determinant remains unchanged.
##l*R_a + R_b \rightarrow R _a##, the determinant is multiplied by ##l##.
##l*R_a + k*R_b \rightarrow R_a##, the determinant is multiplied by ##l##.

Where ##R_a## and ##R_b## are the a'th and the b'th row and ##k,l \in \mathbb{R}##

Now, keep in mind that you had something of the form ##l*R_a + k*R_b \rightarrow R_a##, so your determinant is multiplied with ##-(a-b)##. To make sure that the equality will still hold, multiply the determinant with the factor ##\frac{-1}{a-b}##.

 

[PeroK]

So any ideas to work it out then?

Why bother with row operations? Why not evaluate the determinant as it is and simplify?

 

[member 587159]

Why bother with row operations? Why not evaluate the determinant as it is and simplify?

I supppose that's the exercise since it's in the title...

 

[PeroK]

mistakes are.

If you got this far, why not just take out the factors of ##(a-b)## and ##(c-a)## and you're nearly finished.