Question : If x, y, z are all different and given that

$\displaystyle

\begin{vmatrix}

x & x^2 & 1 + x^3 \\

y & y^2 & 1 + y^3 \\

z & z^2 & 1 + z^3

\end{vmatrix} = 0,

$

Determine the value of (1 + xyz)

Printable View

- Nov 1st 2009, 10:52 PMzorroDetermine the value of (1 + xyz)
Question : If x, y, z are all different and given that

$\displaystyle

\begin{vmatrix}

x & x^2 & 1 + x^3 \\

y & y^2 & 1 + y^3 \\

z & z^2 & 1 + z^3

\end{vmatrix} = 0,

$

Determine the value of (1 + xyz) - Nov 1st 2009, 11:09 PMJodles
What have you done, and where are you stuck?

- Nov 1st 2009, 11:16 PMmath2009
$\displaystyle \det\begin{bmatrix} x&x^2&1+x^3\\ y&y^2&1+y^3\\ z&z^2&1+z^3 \end{bmatrix}=\cdots=(z-x)(z-y)(y-x)(1+xyz)=0$, if you have enough patience.

$\displaystyle \because x\neq y\neq z\neq x \ \therefore 1+xyz=0$ - Nov 2nd 2009, 02:50 PMmr fantastic
Also asked and replied to here: http://www.mathhelpforum.com/math-he...terminant.html

- Dec 1st 2009, 12:29 PMzorroCould u please provide the end result

I have seen the reply post in the other thread but cold u please provide me with the end answer, just to be sure that i have done it right or no.At present i am trying to do as provided by the prev thread but it would be good if u could provide with the end result

ie, $\displaystyle (1 + xyz)$ - Dec 2nd 2009, 12:38 AMmr fantastic
- Dec 22nd 2009, 04:45 PMzorroIs this correct?
- Dec 23rd 2009, 01:15 AMmr fantastic
- Dec 23rd 2009, 01:17 AMzorro
- Dec 23rd 2009, 01:27 AMmr fantastic
- Dec 25th 2009, 06:23 PMsimplependulum

I think some marks will be deducted in your solution .

You should mention that iff two of the elements are the same , the factor

$\displaystyle xy^2 - xz^2 - yx^2 + zx^2 + yz^2 - zy^2 = 0 $

because you have only showed that

$\displaystyle xy^2 - xz^2 - yx^2 + zx^2 + yz^2 - zy^2 = 0$

OR

$\displaystyle 1 + xyz = 0 $

so $\displaystyle 1 + xyz $ is not 100% zero

Factorising it you will get full mark ! (Happy)