Systems of Linear Equations

Yan · #1 $Old$ October 10th, 2008, 12:16 PM

Solve the given system of equations using either Gaussian or Gauss-Jordan elimination.

a)2w+3x-y+4z=0
3w-x+z=1
3w-4x+y-z=2

b) -w+3x-2y+4z=0
2w-6x+y-2z=-3
w-3x+4y-8z=2

watchmath · #2 $Old$ October 10th, 2008, 07:11 PM

In my opinion people won't response your question. The main reason is that it is messy to right matrix in latex. The second one elimination problem is kind of routine so no body has no interest to answer.

Show us how much you have done with this problem and we (should I say I) will be happy to assist you.

Chris L T521 · #3 $Old$ October 10th, 2008, 08:52 PM

Quote:

Originally Posted by Yan $View Post$

Solve the given system of equations using either Gaussian or Gauss-Jordan elimination.

a)2w+3x-y+4z=0
3w-x+z=1
3w-4x+y-z=2

I will only do one, since these will take a while...

$\begin{bmatrix}2&3&-1&4&0\\3&-1&0&1&1\\3&-4&1&-1&2\end{bmatrix}$

--------------------------------

$\tfrac{1}{2}R_1\rightarrow R_1$

$\begin{bmatrix}1&\frac{3}{2}&-\frac{1}{2}&2&0\\3&-1&0&1&1\\3&-4&1&-1&2\end{bmatrix}$

--------------------------------

$-3R_1+R_2\rightarrow R_2$
$-3R_1+R_3\rightarrow R_3$

$\begin{bmatrix}1&\frac{3}{2}&-\frac{1}{2}&2&0\\0&-\frac{11}{2}&\frac{3}{2}&-5&1\\0&-\frac{17}{2}&\frac{5}{2}&-7&2\end{bmatrix}$

--------------------------------

$-\tfrac{2}{11}R_2\rightarrow R_2$

$\begin{bmatrix}1&\frac{3}{2}&-\frac{1}{2}&2&0\\0&1&-\frac{3}{11}&\frac{10}{11}&-\frac{2}{11}\\0&-\frac{17}{2}&\frac{5}{2}&-7&2\end{bmatrix}$

--------------------------------

$\tfrac{17}{2}R_2+R_3\rightarrow R_3$

$\begin{bmatrix}1&\frac{3}{2}&-\frac{1}{2}&2&0\\0&1&-\frac{3}{11}&\frac{10}{11}&-\frac{2}{11}\\0&0&\frac{2}{11}&\frac{8}{11}&\frac{5}{11}\end{bmatrix}$

--------------------------------

$\tfrac{11}{2}R_3\rightarrow R_3$

$\begin{bmatrix}1&\frac{3}{2}&-\frac{1}{2}&2&0\\0&1&-\frac{3}{11}&\frac{10}{11}&-\frac{2}{11}\\0&0&1&4&\frac{5}{2}\end{bmatrix}$

--------------------------------

We have 3 equations with 4 unknowns. Let us introduce a parameter, say $z=t$

Using Gaussian Elimination, we can now back substitute:

$\color{red}\boxed{z=t}$

$y+4z=\tfrac{5}{2}\implies \color{red}\boxed{y=\tfrac{5}{2}-4t}$

$x-\tfrac{3}{11}y+\tfrac{10}{11}z=-\tfrac{2}{11}\implies x=\tfrac{3}{11}\left[\tfrac{5}{2}-4t\right]-\tfrac{10}{11}t-\tfrac{2}{11}\implies \color{red}\boxed{x=\tfrac{1}{2}-2t}$

$w+\tfrac{3}{2}x-\tfrac{1}{2}y+2z=0\implies w=-\tfrac{3}{2}\left[\tfrac{1}{2}-2t\right]+\tfrac{1}{2}\left[\tfrac{5}{2}-4t\right]-2t\implies \color{red}\boxed{w=\tfrac{1}{2}-t}$

Therefore, the solution set is $\left(w,x,y,z\right)=\color{red}\boxed{\left(\tfrac{1}{2}-t,~\tfrac{1}{2}-2t,~\tfrac{5}{2}-4t,~t\right)}$

Does this make sense?

--Chris

Chris L T521 · #4 $Old$ October 10th, 2008, 09:02 PM

Quote:

Originally Posted by watchmath $View Post$

In my opinion people won't response your question. The main reason is that it is messy to right matrix in latex.

That is a possibility...but there is a neat way of creating matrices without using

Code:

\left[\begin{array}{cccc} [insert matrix here] \end{array}\right]

You can use

Code:

\begin{bmatrix} [insert matrix entries here] \end{bmatrix}

This saves me a lot of time.

For example, instead of using

Code:

\left[\begin{array}{cccc} 1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1 \end{array}\right]

to generate $\left[\begin{array}{cccc} 1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1 \end{array}\right]$ , you can use

Code:

\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1 \end{bmatrix

to generate $\begin{bmatrix} 1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1 \end{bmatrix}$

Quote:

The second one elimination problem is kind of routine so no body has no interest to answer.

It may be this, or just that using eliminations on this will lead to ugly looking fraction elements!!! [as seen in my response to part a)]

--Chris