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Show that if $p(x)=a+bx+cx^2$ is a 2rd degree polynomial such that $p(1)=p(2)=p(3)=0$ then $p(x)=0$ (i.e. $a=b=c=0$), using the determinant of the matrix: $\left(\begin{array}{l}1&x&x^2\\1&y&y^2\\1&z&z^2\end{array}\right)$

Hey everyone. So I managed to prove that the determinant of the given matrix is $(y-x)(z-x)(z-y)$ and therefore the matrix is invertible whenever $x\neq z \lor y\neq x \lor z\neq y$

I've proved this claim using basic algebra but am confused on how to show it using the determinant. I tried showing it by plugging in the values of the polynomial by order:

$\left|\begin{array}{l}a&b&c\\a&2b&4c\\a&3b&9c\end{array}\right|$ (Column operations) $ =\left|\begin{array}{l}a&b&c+b+a\\a&2b&4c+2b+a\\a&3b&9c+3b+a\end{array}\right|$ $(p(1)=p(2)=p(3)=0)$$ = \left|\begin{array}{l}a&b&0\\a&2b&0\\a&3b&0\end{array}\right|=0$ but this does not help me. I would love to get some help on this question. Thanks in advance.

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    $\begingroup$ Yes: $(1-2)(2-3)(3-1)\neq0,$ which gives $a=b=c=0.$ $\endgroup$ Feb 24 '18 at 9:56
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    $\begingroup$ It's really a pity to use determinants on this one. I would simply say that $(x-1)(x-2)(x-3)$ is a factor of $p(x)$, end of story. $\endgroup$ Feb 24 '18 at 9:56
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    $\begingroup$ It is of course a 2nd degree polynomial to begin with ... $\endgroup$ Feb 24 '18 at 9:58
  • $\begingroup$ @MichaelRozenberg Thank you! I guess I was too confused to see that :) $\endgroup$
    – Noy
    Feb 24 '18 at 9:58
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Note that\begin{align}p(1)=p(2)=p(3)=0&\iff\left\{\begin{array}{l}a+b\times1+c\times1^2=0\\a+b\times2+c\times2^2=0\\a+b\times3+c\times3^2=0\end{array}\right.\\&\iff\begin{pmatrix}1&1&1\\1&2&2^2\\1&3&3^2\end{pmatrix}.\begin{pmatrix}a\\b\\c\end{pmatrix}=0.\end{align}So, if $(a,b,c)\neq(0,0,0)$, the determinant of that matrix will be $0$. But it isn't.

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