# Prove a 3x3 determinant

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• Nov 2nd 2009, 09:37 AM
aaas
Prove a 3x3 determinant
hello,
please help me with this question

If the determinant of this matrix
x x^2 1+x^3
y y^2 1+y^3
z z^2 1+z^3
equals zero , then prove that xyz = -1, where x y and z are unequal.

I tried solving it directly but I couldn't end with xyz so i can prove it.
thanks .
• Nov 2nd 2009, 12:47 PM
Opalg
Quote:

Originally Posted by aaas
please help me with this question

If the determinant of this matrix
x x^2 1+x^3
y y^2 1+y^3
z z^2 1+z^3
equals zero , then prove that xyz = -1, where x y and z are unequal.

Use elementary row operations. If you add a multiple of one row to another row then the determinant is unchanged. So subtract row 1 from row 2 and from row 3, and you get

\begin{aligned}0 = \begin{vmatrix}x&x^2&1+x^3\\ y&y^2&1+y^3\\ z&z^2&1+z&3\end{vmatrix} &= \begin{vmatrix}x&x^2&1+x^3\\ y-x&y^2-x^2&y^3-x^3\\ z-x&z^2-x^2&z^3-x^3\end{vmatrix} \\ &= \begin{vmatrix}x&x^2&1+x^3\\ y-x&(y-x)(y+x)&(y-x)(y^2+xy+x^2)\\ z-x&(z-x)(z+x)&(z-x)(z^2+xz+x^2)\end{vmatrix}.\end{aligned}

Now you know that y–x is not zero, so you can divide through row 2 by y–x; and you can divide through row 3 by z–x. That gives you $0 = \begin{vmatrix}x&x^2&1+x^3\\ 1&y+x&y^2+xy+x^2\\ 1&z+x&z^2+xz+x^2\end{vmatrix}.$

Continue to simplify the determinant by similar manoeuvres. For example, you could now subtract row 2 from row 3, and subtract x times row 2 from row 1. By then, the determinant should look simple enough that you can work it out completely.
• Nov 2nd 2009, 03:49 PM
mr fantastic
Quote:

Originally Posted by aaas
hello,
please help me with this question

If the determinant of this matrix
x x^2 1+x^3
y y^2 1+y^3
z z^2 1+z^3
equals zero , then prove that xyz = -1, where x y and z are unequal.

I tried solving it directly but I couldn't end with xyz so i can prove it.
thanks .

Asked here: http://www.mathhelpforum.com/math-he...lue-1-xyz.html

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