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
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!
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.
$\displaystyle \begin{array}{ccc} x_1 + x_2 + 2x_3 &=& 4 \\
2x_1 + 3x_2 + x_3 &=& 1 \\ -x_1 + 3x_2 + 3x_3 &=& 2 \end{array}$
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: .$\displaystyle \begin{Bmatrix}x_1 &=& 1 \\ x_2 &=& \text{-}1 \\ x_3 &=& 2 \end{Bmatrix}$