Hey,

Im having trouble on the following question but i get stuck after the first step i dont know how to work out the -R2 and so on.

Question:

x1 + x2 + 2x3 = 4

2x1 + 3x2 + x3 = 1

−x1 + 3x2 + 3x3 = 2

Printable View

- May 28th 2009, 01:46 AMmikel03Gaussian elimination, linear equationsHey,

Im having trouble on the following question but i get stuck after the first step i dont know how to work out the -R2 and so on.

Question:

x1 + x2 + 2x3 = 4

2x1 + 3x2 + x3 = 1

−x1 + 3x2 + 3x3 = 2

- May 28th 2009, 02:30 AMOnkelTom
Hey man,

Theres alot of ways to "think" about how to do it but this is mine

You have

"Row 1" x1 + x2 + 2x3 = 4

"Row 2" 2x1 + 3x2 + x3 = 1

"Row 3" −x1 + 3x2 + 3x3 = 2

And you are aming for

x1+x2+2x3=4

Ax2 +Bx3=C

Dx3=E

Where A,B,C,D,E are constants that will come up along the way

So the first step is to eliminate the x1's from the 2nd and 3rd equation. That is to say,

"Row 4=Row1" ie, the top line is unchanged

"Row 5=Row 2 (plus or minus) (some multiple of)Row 1"

"Row 6=Row 3 (plus or minus) (some multiple of)Row 1"

So to help you out, the first 'iteration' would be

"Row 5=Row 2 -2 times Row 1"

"Row 6=Row 3+ 1 times Row 1"

Giving you

"Row4" x1+x2+2x3=4

"Row5" x2 -3x3=-7

"Row6" 4x2+5x3=6

So you need to do one more process like this to get to what you want, which then allows you to use a process of "backwards substitution" to find x3,x2 and x1 in turn.

Hope thats clear and helps

Btw mikel, im sorry if theres any little errors in there, rushing out, hope that helps mate! - May 28th 2009, 08:29 AMSoroban
Hello, mikel03!

If, by "Gaussian Elimination", you mean a matrix method,

. . I don't see where there would be any difficulty.

Every step is clearly indicated.

Quote:

$\displaystyle \begin{array}{ccc} x_1 + x_2 + 2x_3 &=& 4 \\

2x_1 + 3x_2 + x_3 &=& 1 \\ -x_1 + 3x_2 + 3x_3 &=& 2 \end{array}$

We have: .$\displaystyle \begin{array}{|ccc|c|}

1 &1&2&4 \\ 2&3&1&1 \\ \text{-}1&3&3&2 \end{array}$

$\displaystyle \begin{array}{c} \\ R_2-2R_1 \\ R_3 + R_1\end{array} \begin{array}{|ccc|c|} 1&1&2&4 \\ 0&1&\text{-}3&\text{-}7 \\ 0&4&5&6 \end{array}$

$\displaystyle \begin{array}{c}R_1-R_2 \\ \\ R_3-4R_2\end{array} \begin{array}{|ccc|c|} 1&0&0&11 \\ 0&1&\text{-}3 & \text{-}7 \\ 0&0&17&34 \end{array}$

. . . $\displaystyle \begin{array}{c}\\ \\ \frac{1}{17}R_3\end{array} \begin{array}{|ccc|c|} 1&0&5&11 \\ 0&1&\text{-}3&\text{-}7 \\ 0&0&1&2 \end{array}$

$\displaystyle \begin{array}{c}R_1-5R_3 \\ R_2 + 3R_3 \\ \\ \end{array} \begin{array}{|ccc|c|} 1&0&0&1 \\ 0&1&0&\text{-}1 \\ 0&0&1&2 \end{array}$

. . Therefore: .$\displaystyle \begin{Bmatrix}x_1 &=& 1 \\ x_2 &=& \text{-}1 \\ x_3 &=& 2 \end{Bmatrix}$

- May 29th 2009, 02:20 AMOnkelTom
Soroban, i like that very last step to get that identity matrix, ive never seen/thought of doing it like that!

- May 17th 2010, 07:58 PMdwsmith