# Thread: A 3 x 3 determinant proof

1. ## A 3 x 3 determinant proof

Can someone please prove the determinant proof shown in the image with steps? I can't figure it out for the life of me.

BTW I'm not allowed to directly do it by the "a1(b2c3-c2b3)..." rule. I have to use some primitive(not to mention crazy) rows/columns arithmetic(addition/subtraction) method where you have to write r1'=r1-r2, c2'=c2+c3...stuff like that after every relevant step.

2. $\displaystyle \begin{vmatrix}x^2 & yz & zx+z^2\\x^2+xy & y^2 & zx\\xy & y^2+yz & z^2\end{vmatrix}=$ (factor x from the first column, y from the second, z from the third)

$\displaystyle =xyz\begin{vmatrix}x & z & x+z\\x+y & y & x\\y & y+z & z\end{vmatrix}=$ (add all rows to the first row)

$\displaystyle =xyz\begin{vmatrix}2(x+y) & 2(y+z) & 2(x+z)\\x+y & y & x\\y & y+z & z\end{vmatrix}=$ (factor 2 from the first row)

$\displaystyle =2xyz\begin{vmatrix}x+y & y+z & x+z\\x+y & y & x\\y & y+z & z\end{vmatrix}=$ (substract the first row from the others two)

$\displaystyle =2xyz\begin{vmatrix}x+y & y+z & x+z\\0 & -z & -z\\-x & 0 & -x\end{vmatrix}=$ (factor -z from the second row and -x from the third)

$\displaystyle =2x^2yz^2\begin{vmatrix}x+y & y+z & x+z\\0 & 1 & 1\\1 & 0 & 1\end{vmatrix}=$ (substract the second column from the third)

$\displaystyle =2x^2yz^2\begin{vmatrix}x+y & y+z & x-y\\0 & 1 & 0\\1 & 0 & 1\end{vmatrix}=2x^2yz^2\begin{vmatrix}x+y & x-y\\1 & 1\end{vmatrix}=4x^2y^2z^2$