Question 332650
<font face="Garamond" size="+2">


First put your equations into standard form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ Ax\ +\ By\ +\ Cz\ =\ D]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ y\ \ \ \ \ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x\ \ \ \ \ -z\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ \ \ y\ +\ 2z\ =\ 2]


Then create a 3 X 4 matrix of the coefficients.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left|1\ \ 1\ \ \ 0\ \ 1\cr2\ \ 0\ -1\ \ 0\cr0\ \ 1\ \ \ 2\ \ 2\right|]


Manipulate the matrix to obtain:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left|1\ \ 0\ \ \ 0\ \ x\cr0\ \ 1\ \ \ 0\ \ y\cr0\ \ 0\ \ \ 1\ \ z\right|]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -2R1\ +\ R2\ \rightarrow\ R2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left|1\ \ \ 1\ \ \ 0\ \ 1\cr0\ -2\ -1\ -2\cr0\ \ \ 1\ \ \ 2\ \ 2\right|]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{R2}{-2}\ \rightarrow\ R2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left|1\ \ \ 1\ \ \ 0\ \ 1\cr0\ \ \ 1\ \ \  \frac{1}{2}\ \ 1\cr0\ \ \ 1\ \ \ 2\ \ 2\right|]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -R2\ +\ R3\ \rightarrow\ R3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left|1\ \ \ 1\ \ \ 0\ \ 1\cr0\ \ \ 1\ \ \  \frac{1}{2}\ \ 1\cr0\ \ \ 0\ \ \ \frac{3}{2}\ \ 1\right|]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2\cdot R3}{3}\ \rightarrow\ R3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left|1\ \ \ 1\ \ \ 0\ \ 1\cr0\ \ \ 1\ \ \  \frac{1}{2}\ \ 1\cr0\ \ \ 0\ \ \ 1\ \ \frac{2}{3}\right|]


The last two steps are:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{R3}{-2}\ +\ R2\ \rightarrow\ R2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -R2\ +\ R1\ \rightarrow\ R1]


I'll leave the application of the last two steps to the matrix in your now very capable hands.



John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
</font>