# Thread: Prove a 3x3 determinant

1. ## 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 .

2. 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

\displaystyle \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 $\displaystyle 0 = \begin{vmatrix}x&x^2&1+x^3\\ 1&y+x&y^2+xy+x^2\\ 1&z+x&z^2+xz+x^2\end{vmatrix}.$

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.

3. 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

Thread closed.

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