Systems of Linear Equations

• Oct 10th 2008, 11:16 AM
Yan
Systems of Linear Equations
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
• Oct 10th 2008, 06:11 PM
watchmath
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. (Rofl)
• Oct 10th 2008, 07:52 PM
Chris L T521
Quote:

Originally Posted by Yan
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... (Nod)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 $\left(w,x,y,z\right)=\color{red}\boxed{\left(\tfra c{1}{2}-t,~\tfrac{1}{2}-2t,~\tfrac{5}{2}-4t,~t\right)}$

(Whew)

Does this make sense?

--Chris
• Oct 10th 2008, 08:02 PM
Chris L T521
Quote:

Originally Posted by watchmath
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.

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)] (Worried)

--Chris