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

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...

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

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

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

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

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

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

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

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

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

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

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

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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