Solve Linear Equations Using Gauss-Jordan in C

[niteshadw]

Use the Gauss-Jordan algorithm to find all solutions of the following system of linear equations in C:

x1 + x2 + x3 [] = 2
2x1 [] + 2x3 + 2x4 = 2
x1 + 2x2 + 2x3 [] = 1
2x1 + 2x2 [] + x4 = 2

[] signify a blank space in the equation. How do you even proceed to do this, I have never seen it. Any help would be appreciated. Thank you.

 

[TD]

In matrix-form, it looks like this:

\left( {\begin{array}{*{20}c}<br /> 1 &amp; 1 &amp; 1 &amp; 0 &amp; 2 \\<br /> 2 &amp; 0 &amp; 2 &amp; 2 &amp; 2 \\<br /> 1 &amp; 2 &amp; 2 &amp; 0 &amp; 1 \\<br /> 2 &amp; 2 &amp; 0 &amp; 1 &amp; 2 \\<br /> \end{array}} \right)

Do you know how Gaussian elimination works?

 

[saltydog]

How about this:

\left(<br /> \begin{array}{cccc|c}1 &amp; 1 &amp; 1 &amp; 0 &amp; 2\\<br /> 2 &amp; 0 &amp; 2 &amp; 2 &amp; 2 \\<br /> 1 &amp; 2 &amp; 2 &amp; 0 &amp; 2 \\<br /> 2 &amp; 2 &amp; 0 &amp; 1 &amp; 1 <br /> \end{array}\right)<br />

That's the augumented matrix Niteshadw.

So, reduce it. Here, I'll do the first part. I'll multiply the top row by -2 and then add it to the second row yielding:

<br /> \left(<br /> \begin{array}{cccc|c}1 &amp; 1 &amp; 1 &amp; 0 &amp; 2 \\<br /> 0 &amp; -2 &amp; 0 &amp; 2 &amp; -2 \\<br /> 1 &amp; 2 &amp; 2 &amp; 0 &amp; 2 \\<br /> 2 &amp; 2 &amp; 0 &amp; 1 &amp; 1 <br /> \end{array}\right)<br />

Can you continue doing this until you get it to row-reduced form?

 

[HallsofIvy]

Typo! The first matrix in Saltydog's response should be
\left(\begin{array}{cccc|c}1 &amp; 1 &amp; 1 &amp; 0 &amp; 2\\ 2 &amp; 0 &amp; 2 &amp; 2 &amp; 2 \\ 1 &amp; 2 &amp; 2 &amp; 0 &amp; 1 \\ 2 &amp; 2 &amp; 0 &amp; 1 &amp; 2 \end{array}\right)

Except for separating out the "augmenting" part, that's just what TD said.

 

[niteshadw]

How about this:

\left(<br /> \begin{array}{cccc|c}1 &amp; 1 &amp; 1 &amp; 0 &amp; 2\\<br /> 2 &amp; 0 &amp; 2 &amp; 2 &amp; 2 \\<br /> 1 &amp; 2 &amp; 2 &amp; 0 &amp; 2 \\<br /> 2 &amp; 2 &amp; 0 &amp; 1 &amp; 1 <br /> \end{array}\right)<br />

That's the augumented matrix Stunner.

So, reduce it. Here, I'll do the first part. I'll multiply the top row by -2 and then add it to the second row yielding:

<br /> \left(<br /> \begin{array}{cccc|c}1 &amp; 1 &amp; 1 &amp; 0 &amp; 2 \\<br /> 0 &amp; -2 &amp; 0 &amp; 2 &amp; -2 \\<br /> 1 &amp; 2 &amp; 2 &amp; 0 &amp; 2 \\<br /> 2 &amp; 2 &amp; 0 &amp; 1 &amp; 1 <br /> \end{array}\right)<br />

Can you continue doing this until you get it to row-reduced form?

Hmm..looks simple enough, just did not know what to do with those blank spots...Ok, I got this,

\left(<br /> \begin{array}{cccc|c}1 &amp; 0 &amp; 0 &amp; 0 &amp; 3\\<br /> 0 &amp; 1 &amp; 0 &amp; 0 &amp; -1 \\<br /> 0 &amp; 0 &amp; 1 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 1 &amp; -2 <br /> \end{array}\right)<br />

so um,
x_4 = -2x_4
x_3 = 0
x_2 = -x_2
x_1 = 3x_1

are those the solutions?
Thank you for the help thus far...

 

[TD]

I think you row-reduced fine but the last column refers to the constants, not to an unknown. So the solution should be:

\left\{ \begin{array}{l}<br /> x_1 = 3 \\ <br /> x_2 = - 1 \\ <br /> x_3 = 0 \\ <br /> x_4 = - 2 \\ <br /> \end{array} \right