3
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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 $C=1$.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.