Question

Gaussian Elimination Method, help!

Answer 1

What's your specific problem?

Answer 2

here L

Answer 3

I tried to add to second row -2R1
that gives me 0 0 0 1 1 0
and then for third row, I added 3R1
which gives me 0 0 0 -4 -122 0 if I did the math right

Answer 4

How do we solve that system? should we set parameters and is it okay to have 3 zeros beginning in the last two rows?

Answer 5

Please correct me if I incorrectly transcribed your matrix, I'm just not sure about the 3rd row, third column entry.
\[\left[\begin{array}{rrrrr|r}
3&-1&1&-5&-1&0\\
6&-2&2&-9&1&0\\
-9&3&-3&11&-1&0
\end{array}\right]\]You have some mistakes in your arithmetic. Adding \(-2R_1+R_2\to R_2\) actually gives
\[\left[\begin{array}{rrrrr|r}
3&-1&1&-5&-1&0\\
0&0&0&1&\color{red}3&0\\
-9&3&-3&11&-1&0
\end{array}\right]\]Then adding \(3R_1+R_3\to R_3\) gives
\[\left[\begin{array}{rrrrr|r}
3&-1&1&-5&-1&0\\
0&0&0&1&3&0\\
0&0&0&-4&\color{red}{-4}&0
\end{array}\right]\]The fact that rows 2 and 3 start with three zeros is not a problem, and is something you should expect from this matrix. (Why this is so might be beyond your current curriculum if you haven't taken a course in linear algebra yet.) All it means is that there isn't a unique solution to three of these variables.

Answer 6

Okiedokiee. Thanks! @HolsterEmission so how do we solve for the system? My professor taught us about using free variables and setting them equal to specific variables (s and t) but how do we deal with that given that we for row 2 and 3 we have 1 and -4 on x4 and x5 ?

Answer 7

you use the 1 in row 2 as the next "pivot"
multiply the 2nd row by 4 and add to the 3rd row
you should get
000 0 8

Answer 8

Continuing from the last matrix, you can use the operations \(-\dfrac{1}{4}R_3\to R_3\) followed by \(R_3-R_2\to R_3\) and \(-\dfrac{1}{2}R_3\to R_3\) to get
\[\left[\begin{array}{rrrrr|r}
3&-1&1&-5&-1&0\\
0&0&0&1&3&0\\
0&0&0&0&1&0
\end{array}\right]\]which is the row reduced form of the initial matrix.

At this point you can now solve for two variables directly, \(x_4\) and \(x_5\). Clearly, \(x_5=0\), which means \(x_4+3x_5=x_4=0\) as well.

From the first row, you can set \(x_2=s\) and \(x_3=t\), which gives \(3x_1-x_2+x_3-5x_4-x_5=3x_1-s+t=0\), or \(x_1=\dfrac{s-t}{3}\).