Solving Gaussian Elimination Algorithm Problem (Matrices)

[adc85]

I tried to solve this Gaussian elimination algorithm problem (matrices) but for some reason when I plug in the x variables it doesn't work. The problem is:

[ 3 10 4 : 7 ]
[ 2 7 3 : 5 ]
[ 1 3 2 : 2 ]

Alright so the first thing I did was divide the 1st row by 1/3 (scaling). Then I made the entries below the first pivot equal to 0 using:

Row2 = Row2 - Second Row, First Column * Row 1
Row3 = Row3 = Third Row, First Column * Row 1

Then I repeat this algorithm for the submatrix created afterwards (ignoring the first row and first column). Afterwards, I used backwards substitution (even tried using reduced echelon form). But I am not getting quite the right answers (very close for row 3 and the other two rows are fine though). Any input appreciated.

Also, I am having trouble understanding what my professor is saying when he says to solve a certain problem like this using 10^-3 precision for example. Do you just use the same method except placing decimal places at the end of each number or whatever?

Then he has this other weird problem that goes like:

[1 1 1 : 0]
[3 4 8 : 1]
[4 5 c^2 : c - 2]

What value of c would make this inconsistent (in other words, no solution)? I'm thinking that x3 could equal anything by making the last row full of zeros. Not sure though.

Thanks for any help.

 

[cronxeh]

\left(\begin{array}{x1x2x3k}<br /> 3 &amp; 10 &amp; 4 &amp; 7\\ <br /> 2 &amp; 7 &amp; 3 &amp; 5\\ <br /> 1 &amp; 3 &amp; 2 &amp; 2<br /> \end{array}\right)

Gaussian Elimination or any other fancy name is simply an attempt to reduce this matrix into a \left(\begin{array}{x1x2x3k}<br /> 1 &amp; 0 &amp; 0 &amp; k1\\ <br /> 0 &amp; 1 &amp; 0 &amp; k2\\ <br /> 0 &amp; 0 &amp; 1 &amp; k3<br /> \end{array}\right) form.

I will use notation R1, R2, and R3 to indicate Rows 1, 2, and 3. For example (R2 = R2 + R1 means you add row 1 into row 2, and after that you should have Row 1 as it was, and Row 2 will be sum of Row 1 and Row 2 from before.

Now in your case:

\left(\begin{array}{x1x2x3k}<br /> 3 &amp; 10 &amp; 4 &amp; 7\\ <br /> 2 &amp; 7 &amp; 3 &amp; 5\\ <br /> 1 &amp; 3 &amp; 2 &amp; 2<br /> \end{array}\right) R1 = R1 - R2 -> \left(\begin{array}{x1x2x3k}<br /> 1 &amp; 3 &amp; 1 &amp; 2\\ <br /> 2 &amp; 7 &amp; 3 &amp; 5\\ <br /> 1 &amp; 3 &amp; 2 &amp; 2<br /> \end{array}\right) R2 = R2 - 2*R3 -> \left(\begin{array}{x1x2x3k}<br /> 1 &amp; 3 &amp; 1 &amp; 2\\ <br /> 0 &amp; 1 &amp; -1 &amp; 1\\ <br /> 1 &amp; 3 &amp; 2 &amp; 2<br /> \end{array}\right). R3 = R3 - R1 -> \left(\begin{array}{x1x2x3k}<br /> 1 &amp; 3 &amp; 1 &amp; 2\\ <br /> 0 &amp; 1 &amp; -1 &amp; 1\\ <br /> 0 &amp; 0 &amp; 1 &amp; 0<br /> \end{array}\right).


\left(\begin{array}{x1x2x3k}<br /> 1 &amp; 3 &amp; 1 &amp; 2\\ <br /> 0 &amp; 1 &amp; -1 &amp; 1\\ <br /> 0 &amp; 0 &amp; 1 &amp; 0<br /> \end{array}\right) R1 = R1 - 3R2 -> \left(\begin{array}{x1x2x3k}<br /> 1 &amp; 0 &amp; 4 &amp; -1\\ <br /> 0 &amp; 1 &amp; -1 &amp; 1\\ <br /> 0 &amp; 0 &amp; 1 &amp; 0<br /> \end{array}\right). R1 = R1 - 4R3 -> \left(\begin{array}{x1x2x3k}<br /> 1 &amp; 0 &amp; 0 &amp; -1\\ <br /> 0 &amp; 1 &amp; -1 &amp; 1\\ <br /> 0 &amp; 0 &amp; 1 &amp; 0<br /> \end{array}\right). R2 = R2 + R3 -> \left(\begin{array}{x1x2x3k}<br /> 1 &amp; 0 &amp; 0 &amp; -1\\ <br /> 0 &amp; 1 &amp; 0 &amp; 1\\ <br /> 0 &amp; 0 &amp; 1 &amp; 0<br /> \end{array}\right)

So you end up with this:
x1 = -1, x2 = 1, x3 = 0

Your answer is: (-1, 1, 0) \epsilon \mathbb{R}^3

 

[M98Ranger]

Hi there,

Firstly, it's great that you are actively trying to solve this Gaussian elimination algorithm problem and seeking help when you encounter difficulties. It shows determination and a willingness to learn.

Regarding the first problem, it seems like you have a good understanding of the steps involved in Gaussian elimination. However, it's important to double-check your calculations, as even small errors can lead to incorrect solutions. It's also possible that you may have missed a step or made a mistake in the backwards substitution process. I would recommend checking your work and perhaps seeking assistance from your professor or classmates if you are still having trouble.

As for the precision question, using 10^-3 precision means that your final solutions should be rounded to three decimal places. This is to ensure that your answers are as accurate as possible without being overly complicated.

For the second problem, you are correct in thinking that there is no value of c that will make this system consistent. This is because the third row has an equation that is dependent on c, meaning that it can take on any value and still satisfy the system. This is known as a free variable and results in an infinite number of solutions.

I hope this helps and good luck with your future problem-solving endeavors!