Using Determinant Properties to Simplify a 3x3 Matrix

[DryRun]

Homework Statement

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)

Homework Equations

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

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.

 

[SammyS]

Homework Statement

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)

Homework Equations

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

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.

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

 

[DryRun]

Hi SammyS

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

The expansion gives: x^2y-xy^2-2xyz+y^2z-x^2z-xz^2-yz^2But 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) -2xyzBut it's different, as the actual cofactor expansion is: x^2(y-z) -y^2(x-z) +z^2 (x-y)

 

[SammyS]

Hi SammyS

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

Not quite right.

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

 

[DryRun]

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.