Matrix - Gaussian elimination

**Paymemoney** · October 13th 2010, 02:44 AM

Hi

Having trouble with solving the equations using Gaussian elimination

1) Solve the following system of equations using Gaussian elimination

$5x-2y+z=-6$
$-2x+z=-4$
$-x-2y+2z=3$

Show the final tableau (augmented matrix) of the elimination and the values of x, y and z.

This is what i have done but it is incorrect according the book's answer.

$\begin{bmatrix} 5 & -2 & 1 & | & -6 \\ -2 & 0 & 1 & | & -4 \\ -1 & -2 & 2 & | & 3 \\ \end{bmatrix} $

$\begin{bmatrix} 5 & -2 & 1 & | & -6 \\ 0 & -4 & -3 & | & -10 \\ -1 & -2 & 2 & | & 3 \\ \end{bmatrix} $
R2-2R3

$\begin{bmatrix} 1 & \frac{-2}{5} & \frac{1}{5} & | & \frac{-6}{5} \\ 0 & -4 & 3 & | & -10 \\ -1 & -2 & 2 & | & 3 \\ \end{bmatrix} $
$\frac{1}{5}R1$

$\begin{bmatrix} 1 & \frac{-2}{5} & \frac{1}{5} & | & \frac{-6}{5} \\ 0 & -4 & 3 & | & -10 \\ 0 & \frac{-12}{5} & \frac{11}{5} & | & \frac{9}{5} \\ \end{bmatrix} $
$R1+R3$

$\begin{bmatrix} 1 & \frac{-2}{5} & \frac{1}{5} & | & \frac{-6}{5} \\ 0 & -4 & 3 & | & -10 \\ 0 & \frac{-10}{5} & \frac{11}{5} & | & \frac{9}{5} \\ \end{bmatrix} $
$\frac{1}{4}R2$

$\begin{bmatrix} 1 & \frac{-2}{5} & \frac{1}{5} & | & \frac{-6}{5} \\ 0 & -4 & 3 & | & -10 \\ 0 & 0 & \frac{7}{10} & | & \frac{-16}{5} \\ \end{bmatrix} $
$\frac{10}{5}R2+R3$

P.S

**Goldberg Variation** · October 13th 2010, 03:07 AM

Would the answer perhaps be:

x = 2-t
y = 1/2t - 8
z = 2t

If not, then I should take another look at the question

If it is, or close, I can post my solution.

**HallsofIvy** · October 13th 2010, 04:42 AM

Originally Posted by Paymemoney

Hi

Having trouble with solving the equations using Gaussian elimination

1) Solve the following system of equations using Gaussian elimination

$5x-2y+z=-6$
$-2x+z=-4$
$-x-2y+2z=3$

Show the final tableau (augmented matrix) of the elimination and the values of x, y and z.

This is what i have done but it is incorrect according the book's answer.

$\begin{bmatrix} 5 & -2 & 1 & | & -6 \\ -2 & 0 & 1 & | & -4 \\ -1 & -2 & 2 & | & 3 \\ \end{bmatrix} $

$\begin{bmatrix} 5 & -2 & 1 & | & -6 \\ 0 & -4 & -3 & | & -10 \\ -1 & -2 & 2 & | & 3 \\ \end{bmatrix} $
R2-2R3

$\begin{bmatrix} 1 & \frac{-2}{5} & \frac{1}{5} & | & \frac{-6}{5} \\ 0 & -4 & 3 & | & -10 \\ -1 & -2 & 2 & | & 3 \\ \end{bmatrix} $
$\frac{1}{5}R1$

$\begin{bmatrix} 1 & \frac{-2}{5} & \frac{1}{5} & | & \frac{-6}{5} \\ 0 & -4 & 3 & | & -10 \\ 0 & \frac{-12}{5} & \frac{11}{5} & | & \frac{9}{5} \\ \end{bmatrix} $
$R1+R3$

$\begin{bmatrix} 1 & \frac{-2}{5} & \frac{1}{5} & | & \frac{-6}{5} \\ 0 & -4 & 3 & | & -10 \\ 0 & \frac{-10}{5} & \frac{11}{5} & | & \frac{9}{5} \\ \end{bmatrix} $
$\frac{1}{4}R2$

$\begin{bmatrix} 1 & \frac{-2}{5} & \frac{1}{5} & | & \frac{-6}{5} \\ 0 & -4 & 3 & | & -10 \\ 0 & 0 & \frac{7}{10} & | & \frac{-16}{5} \\ \end{bmatrix} $
$\frac{10}{5}R2+R3$

P.S

That will give you the correct values for x, y, and z (assuming your arithmetic is correct, I did not check it) but what, exactly, does your text consider the "final tableau"? You have got to "triangular form" and now you can solve for x, y, and z by back substitution. But I notice you got a "1" in the first column but did not get "1"s on the diagonal in the other columns. Some texts, at least, would consider having "1" s all along the diagonal as necessary. Others might expect to see a complete reduction to the identity matrix for the first 3 columns.

**Paymemoney** · October 13th 2010, 01:45 PM

Originally Posted by HallsofIvy

That will give you the correct values for x, y, and z (assuming your arithmetic is correct, I did not check it) but what, exactly, does your text consider the "final tableau"? You have got to "triangular form" and now you can solve for x, y, and z by back substitution. But I notice you got a "1" in the first column but did not get "1"s on the diagonal in the other columns. Some texts, at least, would consider having "1" s all along the diagonal as necessary. Others might expect to see a complete reduction to the identity matrix for the first 3 columns.

actually the answer is not supplied for this question, but when i used the calculator it gave me what you said '1' on the diagonal

$\begin{bmatrix} 1 & \frac{-2}{5} & \frac{1}{5} & | & \frac{-6}{5} \\ 0 & 1 & \frac{-1}{12} & | & \frac{-3}{4} \\ 0 & 0 & 1 & | & \frac{-21}{4} \\ \end{bmatrix}$

**HallsofIvy** · October 14th 2010, 05:08 AM

Originally Posted by Paymemoney

Hi

Having trouble with solving the equations using Gaussian elimination

1) Solve the following system of equations using Gaussian elimination

$5x-2y+z=-6$
$-2x+z=-4$
$-x-2y+2z=3$

Show the final tableau (augmented matrix) of the elimination and the values of x, y and z.

This is what i have done but it is incorrect according the book's answer.

$\begin{bmatrix} 5 & -2 & 1 & | & -6 \\ -2 & 0 & 1 & | & -4 \\ -1 & -2 & 2 & | & 3 \\ \end{bmatrix} $

$\begin{bmatrix} 5 & -2 & 1 & | & -6 \\ 0 & -4 & -3 & | & -10 \\ -1 & -2 & 2 & | & 3 \\ \end{bmatrix} $
R2-2R3

There is an error in this first step. -2 times the second number in R3 is + 4. The second number in R2 now should be 4, not -4.

$\begin{bmatrix} 1 & \frac{-2}{5} & \frac{1}{5} & | & \frac{-6}{5} \\ 0 & -4 & 3 & | & -10 \\ -1 & -2 & 2 & | & 3 \\ \end{bmatrix} $
$\frac{1}{5}R1$

$\begin{bmatrix} 1 & \frac{-2}{5} & \frac{1}{5} & | & \frac{-6}{5} \\ 0 & -4 & 3 & | & -10 \\ 0 & \frac{-12}{5} & \frac{11}{5} & | & \frac{9}{5} \\ \end{bmatrix} $
$R1+R3$

$\begin{bmatrix} 1 & \frac{-2}{5} & \frac{1}{5} & | & \frac{-6}{5} \\ 0 & -4 & 3 & | & -10 \\ 0 & \frac{-10}{5} & \frac{11}{5} & | & \frac{9}{5} \\ \end{bmatrix} $
$\frac{1}{4}R2$

$\begin{bmatrix} 1 & \frac{-2}{5} & \frac{1}{5} & | & \frac{-6}{5} \\ 0 & -4 & 3 & | & -10 \\ 0 & 0 & \frac{7}{10} & | & \frac{-16}{5} \\ \end{bmatrix} $
$\frac{10}{5}R2+R3$

P.S

**Paymemoney** · October 15th 2010, 03:57 AM

ok, i have tried again, this is what i got:

$\begin{bmatrix} 5 & -2 & 1 & | & -6 \\ -2 & 0 & 1 & | & -4 \\ -1 & -2 & 2 & | & 3 \\ \end{bmatrix} $

$\begin{bmatrix} 5 & -2 & 1 & | & -6 \\ 0 & 4 & -3 & | & -10 \\ -1 & -2 & 2 & | & 3 \\ \end{bmatrix} $
R2-2R3

$\begin{bmatrix} 1 & \frac{-2}{5} & \frac{1}{5} & | & \frac{-6}{5} \\ 0 & 4 & -3 & | & -10 \\ -1 & -2 & 2 & | & 3 \\ \end{bmatrix} $
$\frac{1}{5}R1$

$\begin{bmatrix} 1 & \frac{-2}{5} & \frac{1}{5} & | & \frac{-6}{5} \\ 0 & 4 & -3 & | & -10 \\ 0 & \frac{-12}{5} & \frac{11}{5} & | & \frac{9}{5} \\ \end{bmatrix} $
R3+R1

$\begin{bmatrix} 1 & \frac{-2}{5} & \frac{1}{5} & | & \frac{-6}{5} \\ 0 & 4 & -3 & | & -10 \\ 0 & 0 & \frac{2}{10} & | & \frac{-21}{5} \\ \end{bmatrix} $
$R3+\frac{3}{5}R2$

$z=\frac{-105}{10}$
$y=\frac{-10-\frac{315}{10}}{10}$ $= \frac{-415}{40}$
$x=\frac{-53}{10}$

**Goldberg Variation** · October 15th 2010, 04:05 AM

I'm not sure what you are doing, but as I understood it you just have to set Z to t, in my case I set it to 2t to make it more simple counting.

Then it should all fall out, if z = 2t you can find out what x is and then ultimately what y is.

**davemk** · October 16th 2010, 10:48 AM

Originally Posted by Paymemoney

o
$\begin{bmatrix} 1 & \frac{-2}{5} & \frac{1}{5} & | & \frac{-6}{5} \\ 0 & 4 & -3 & | & -10 \\ 0 & 0 & \frac{2}{10} & | & \frac{-21}{5} \\ \end{bmatrix} $
$R3+\frac{3}{5}R2$

$z=\frac{-105}{10}$
$y=\frac{-10-\frac{315}{10}}{10}$ $= \frac{-415}{40}$
$x=\frac{-53}{10}$

Briefly looking at that, you should have a figure of z = -210/10 and not -105/10. Then substitute that value into your other equations to work out the value of y and x.

**Aurum** · October 16th 2010, 04:48 PM

Originally Posted by Paymemoney

$\begin{bmatrix} 1 & \frac{-2}{5} & \frac{1}{5} & | & \frac{-6}{5} \\ 0 & 4 & -3 & | & -10 \\ 0 & \frac{-12}{5} & \frac{11}{5} & | & \frac{9}{5} \\ \end{bmatrix} $
R3+R1

$\begin{bmatrix} 1 & \frac{-2}{5} & \frac{1}{5} & | & \frac{-6}{5} \\ 0 & 4 & -3 & | & -10 \\ 0 & 0 & \frac{2}{10} & | & \frac{-21}{5} \\ \end{bmatrix} $
$R3+\frac{3}{5}R2$

$z=\frac{-105}{10}$
$y=\frac{-10-\frac{315}{10}}{10}$ $= \frac{-415}{40}$
$x=\frac{-53}{10}$

Everything looks fine until the 3rd column in your final matrix $\frac{3(-3)}{5} = \frac{-9}{5}$

So $\frac{11}{5} + \frac{-9}{5} = \frac{2}{5}$

Take $\frac{2}{5}R3$ and the bottom row becomes 0 0 1 | $\frac{-21}{2}$

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