# Properties of determinants

1. May 20, 2012

### sharks

1. The problem statement, all variables and given/known data
Use the properties of the determinant of a matrix to show that$$\begin{vmatrix}1+x^2 & x & 1 \\ 1+y^2 & y & 1 \\ 1+z^2 & z & 1\end{vmatrix}=(x-y)(x-z)(y-z)$$

2. Relevant equations
Properties of determinants. There's 10 of them, according to my notes.

3. The attempt at a solution
I used the property where the scalar multiple of -1 of the third column added to the first column gives:$$\begin{vmatrix}x^2 & x & 1 \\ y^2 & y & 1 \\ z^2 & z & 1\end{vmatrix}$$
And then i'm stuck.

2. May 20, 2012

### SammyS

Staff Emeritus
What do you get when you expand $(x-y)(x-z)(y-z)\ ?$

3. May 20, 2012

### sharks

Hi SammyS
The expansion gives: $$x^2y-xy^2-2xyz+y^2z-x^2z-xz^2-yz^2$$But i'm not sure how to relate this to the determinant.

I grouped the squared terms, as it seemed to me that they formed the cofactor expansion by the first column of the determinant:
$$x^2(y-z) -y^2(x-z) -z^2 (x+y) -2xyz$$But it's different, as the actual cofactor expansion is: $$x^2(y-z) -y^2(x-z) +z^2 (x-y)$$

Last edited: May 20, 2012
4. May 20, 2012

### SammyS

Staff Emeritus
Not quite right.

There's a xyz and a -xyz which cancel .

5. May 20, 2012

### sharks

You are correct. So, to write the solution, meaning how i got to the product of the 3 factors, i just trace back my steps from the factors' expansion.

Thanks, SammyS.