Using Determinant Properties to Simplify a 3x3 Matrix

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DryRun
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Homework Statement


Use the properties of the determinant of a matrix to show that[tex]\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)[/tex]

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:[tex]\begin{vmatrix}x^2 & x & 1 \\ y^2 & y & 1 \\ z^2 & z & 1\end{vmatrix}[/tex]
And then I'm stuck.
 
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sharks said:

Homework Statement


Use the properties of the determinant of a matrix to show that[tex]\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)[/tex]

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:[tex]\begin{vmatrix}x^2 & x & 1 \\ y^2 & y & 1 \\ z^2 & z & 1\end{vmatrix}[/tex]
And then I'm stuck.
What do you get when you expand [itex](x-y)(x-z)(y-z)\ ?[/itex]
 
Hi SammyS
SammyS said:
What do you get when you expand [itex](x-y)(x-z)(y-z)\ ?[/itex]

The expansion gives: [tex]x^2y-xy^2-2xyz+y^2z-x^2z-xz^2-yz^2[/tex]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:
[tex]x^2(y-z) -y^2(x-z) -z^2 (x+y) -2xyz[/tex]But it's different, as the actual cofactor expansion is: [tex]x^2(y-z) -y^2(x-z) +z^2 (x-y)[/tex]
 
Last edited:
You are correct. :redface: 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.