Using Determinant Properties to Simplify a 3x3 Matrix

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Homework Help Overview

The problem involves using properties of determinants to show that a specific 3x3 matrix determinant equals the product of linear factors involving variables x, y, and z. The subject area is linear algebra, specifically focusing on determinants of matrices.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of determinant properties, with one attempting a transformation of the matrix and expressing confusion about the next steps. Questions arise regarding the expansion of the product (x-y)(x-z)(y-z) and its relation to the determinant.

Discussion Status

The discussion is ongoing, with participants exploring the relationship between the expanded form of the product and the determinant. Some guidance is provided regarding the grouping of terms, but no consensus or resolution has been reached.

Contextual Notes

Participants note the challenge of relating the expanded polynomial to the determinant's cofactor expansion, indicating potential assumptions or missing connections in their reasoning.

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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:
sharks said:
Hi SammyS

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.
Not quite right.

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

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