# Gauss-Jordan Elimination

Trying to solve this using Gauss-Jordan Elimination.

$x_1 + x_2 - x_3 = -4$

$-x_1 - 2x_2 + x_3 = 3+t$

$2x_1 + x_2 + (s-3)x_3 = st-9$

$2x_1 + (s-3)x_3 = st+2t-10$

I came across: $t=0$ And $s\neq 1$

Need to find the values of s and t that the equations will have: no solution, 1 solution, unlimited solutions.

$\left[\begin{array}{ccc|c}1&1&-1&-4\\-1&-2&1&3+t\\2&1&s-3&st-9\\2&0&s-3&st+2t-10\end{array}\right]$

$\left[\begin{array}{ccc|c}1&1&-1&-4\\0&1&0&1-t\\0&0&s-1&t(s-1)\\0&1&0&1-2t\end{array}\right]$

$\left[\begin{array}{ccc|c}1&1&-1&-4\\0&1&0&1-t\\0&0&1&t\\0&0&0&-t\end{array}\right]$

$\left[\begin{array}{ccc|c}1&0&0&2t-5\\0&1&0&1-t\\0&0&1&t\\0&0&0&-t\end{array}\right]$

• It is not clear what the initial system is and where your work begins. I have begun the work of improving your formatting, but you should improve it further to make it absolutely clear what is being asked. Are all four equations given as hypothesis and we are to draw conclusions based on that? – JMoravitz Nov 20 '16 at 20:34
• You shoul write down your calculations. – MrYouMath Nov 20 '16 at 20:35
• Visit this page to learn how to write with MathJax and $\LaTeX$ so you can write matrices and such correctly. – JMoravitz Nov 20 '16 at 20:48
• Ok. Thanks for the edit. I explanied the question better – Asking ps Nov 20 '16 at 21:35

## 1 Answer

From your final system, $0=-t$, hence if $t \neq 0$, there is no solution.

If $t=0$, we can drop the very last equation as it is just $0=0$.

The $3 \times 3$ matrix on the LHS is non-singular, hence the system has a unique solution.

• That I understood. But does the Elimination is correct? – Asking ps Nov 21 '16 at 5:35
• seems fine. Remark: our final forms matches. Is there any particular step of your working that you are doubtful of? It would be great if you can type out what operations did you do from step to step. – Siong Thye Goh Nov 21 '16 at 6:12