Can someone check this please? $$ \begin{vmatrix} x&y&z\\ x^2&y^2&z^2\\ x^3&y^3&z^3\\ \end{vmatrix}$$ $$C_2=C_2-C_1\implies\quad \begin{vmatrix} x&y-x&z\\ x^2&y^2-x^2&z^2\\ x^3&y^3-x^3&z^3\\ \end{vmatrix}$$ $$(y-x) \begin{vmatrix} x&1&z\\ x^2&y+x&z^2\\ x^3&y^2+xy+x^2&z^3\\ \end{vmatrix}$$ $$(y-x)(z-x) \begin{vmatrix} x&1&1\\ x^2&y+x&z+x\\ x^3&y^2+xy+x^2&z^2+xz+x^2\\ \end{vmatrix}$$ $$R_2=R_2-xR_1\implies\quad (y-x)(z-x) \begin{vmatrix} x&1&1\\ 0&y&z\\ x^3&y^2+xy+x^2&z^2+xz+x^2\\ \end{vmatrix}$$ $$R_3=R_3-x^2R_1\implies\quad (y-x)(z-x) \begin{vmatrix} x&1&1\\ 0&y&z\\ 0&y^2+xy&z^2+xz\\ \end{vmatrix}$$ factor $x$$$\implies\quad x(y-x)(z-x) \begin{vmatrix} 1&1&1\\ 0&y&z\\ 0&y^2+xy&z^2+xz\\ \end{vmatrix}$$ $$\implies\quad x(y-x)(z-x)(yz^2-zy^2)$$ $$\implies\quad xyz(y-x)(z-x)(z-y)$$ Also I'd like practical tips on using the factor theorem for these types of questions. My understanding is that the determinant is $f(x,y,z)$ so if we hold $y$ and $z$ constant we could apply it somehow to $f(x)$ alone. I'm not that great spotting difference of squares etc and want a more fail safe alternative. Thanks in advance.


What you did is correct. But there is an easier way. Remember that for polynomial $p(x)$, if $p(a)=0$ then $(x-a)$ is a factor of $p(x)$.

Denote the determinant by $\Delta$. It is obviously a polynomial in $x,\ y$ and $z$. Now, note that:

  • $x=0\implies \Delta = 0$, so $x$ is a factor of $\Delta$. Same for $y = 0$ and $z=0$.
  • $x=y\implies \Delta = 0$, so $(x-y)$ is a factor of $\Delta$. Similarly for $y=z$ and $z=x$

Finally note that $\Delta$ is degree $6$ polynomial. So it cannot have more than $6$ linear factors, and we have listed all of them above. Clearly $$\Delta=Cxyz(x-y)(y-z)(z-x)$$ where $C$ is some constant. Taking some values (eg. $x=1,\ y=2,\ z=3$), we get $C=1$.

  • $\begingroup$ Thank you for your answer. Really liked the elegance of this method. I have a couple of follow up questions but I believe the correct thing to do is to post a new question as per meta. $\endgroup$ – Karl Jan 5 '15 at 19:26

you are not factoring the matrix. you are using the properties of the determinants to simplify. for example, you could write your first step as $$AE = \pmatrix{x & y & z \\ x^2 & y^2 & z^2 \\ x^3 & y^3 & z^3} \pmatrix{1 & -1 & 0\\0&1&0\\0&0&1} = \pmatrix{x & y-x & z \\ x^2 & y^2-x^2 & z^2 \\ x^3 & y^3 -x^3& z^3} = B $$

the $E$ matrices are called elementary column matrices and their determinants usually is the product of the diagonals. now use the product rule of the determinants $det(AE) = det(A) det(E)$ to conclude that $det(A) = det(B)$ now you start with $B$ and do further reductions.

  • $\begingroup$ Thanks.I never thought of using a matrix and I can see it tidied workings nicely. $\endgroup$ – Karl Jan 5 '15 at 19:27

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