# Factorise a matrix using the factor theorem

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. ## 2 Answers 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$C=1$. • 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. – 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.

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