Homework Help: Matrix question

niteshadw · Sep 20, 2005

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 · Sep 20, 2005

In matrix-form, it looks like this:

[tex]\left( {\begin{array}{*{20}c}
1 & 1 & 1 & 0 & 2 \\
2 & 0 & 2 & 2 & 2 \\
1 & 2 & 2 & 0 & 1 \\
2 & 2 & 0 & 1 & 2 \\
\end{array}} \right)[/tex]

Do you know how Gaussian elimination works?

saltydog · Sep 20, 2005

How about this:

[tex]\left(
\begin{array}{cccc|c}1 & 1 & 1 & 0 & 2\\
2 & 0 & 2 & 2 & 2 \\
1 & 2 & 2 & 0 & 2 \\
2 & 2 & 0 & 1 & 1
\end{array}\right)
[/tex]

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:

[tex]
\left(
\begin{array}{cccc|c}1 & 1 & 1 & 0 & 2 \\
0 & -2 & 0 & 2 & -2 \\
1 & 2 & 2 & 0 & 2 \\
2 & 2 & 0 & 1 & 1
\end{array}\right)
[/tex]

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

HallsofIvy · Sep 20, 2005

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

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

niteshadw · Sep 20, 2005

saltydog said:

How about this:

[tex]\left(
\begin{array}{cccc|c}1 & 1 & 1 & 0 & 2\\
2 & 0 & 2 & 2 & 2 \\
1 & 2 & 2 & 0 & 2 \\
2 & 2 & 0 & 1 & 1
\end{array}\right)
[/tex]

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:

[tex]
\left(
\begin{array}{cccc|c}1 & 1 & 1 & 0 & 2 \\
0 & -2 & 0 & 2 & -2 \\
1 & 2 & 2 & 0 & 2 \\
2 & 2 & 0 & 1 & 1
\end{array}\right)
[/tex]

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,

[tex]\left(
\begin{array}{cccc|c}1 & 0 & 0 & 0 & 3\\
0 & 1 & 0 & 0 & -1 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & -2
\end{array}\right)
[/tex]

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 · Sep 21, 2005

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

[tex]\left\{ \begin{array}{l}
x_1 = 3 \\
x_2 = - 1 \\
x_3 = 0 \\
x_4 = - 2 \\
\end{array} \right[/tex]