# Thread: Gaussian Elimination. Question is attached.

1. ## Gaussian Elimination. Question is attached.

When I work out from the piovt rows 2, 3 & 4 work out the same and I not sure about x,y,z &w being non negative, some of my values are negative.

The routes are to be one way which means x,y,z &w are to be non negative.

Attempted Solution =

+1 +2 -1 +1 x 30
+2 -1 +1 +1 y 60 (R2 - R4)
+3 -3 +2 +2 z 70 (R3 - 3R4)
+2 +8 -4 +2 w 80 (R4 - R4)

+1 +2 -1 +1 x 30
+0 -9 -5 -1 y -20
+0 -9 +5 -1 z -20 (R3 - R2)
+0 -9 -5 +1 w 20 (R4 + R3)

+1 +2 -1 +1 x 30
+0 -9 -5 -1 y -20
+0 0 +10 0 z 0 (R3 - R2)
+0 0 0 +2 w 0 (R4 + R3)

2. Hello, den!

. . And you didn't finish the Elimination.

When I work out from the pivot rows 2, 3 & 4 work out the same,
and I not sure about x,y,z,w being nonnegative, some of my values are negative.

We have: .$\displaystyle \left|\begin{array}{cccc|c} 1 & 2 &\text{-}1&1 & 30 \\ 2 &\text{-}1 & 1 & 1 & 60 \\ 3 &\text{-}3&2&2 & 70 \\ 2&8 &\text{-}4 &2 & 80 \end{array}\right|$

. $\displaystyle \begin{array}{cccc} \\ R_2-R_4 \\ R_3-3R_1 \\ R_4-R_2 \end{array} \left|\begin{array}{cccc|c} 1 & 2 & \text{-}1 & 1 & 30 \\ 0 & \text{-}9 & 5 & \text{-}1 & \text{-}20 \\ 0 & \text{-}9 & 5 & \text{-}1 & \text{-}20 \\ 0 & 9 & \text{-}5 & 1 & 20 \end{array}\right|$

. .$\displaystyle \begin{array}{c}\\ \\ R_3-R_2 \\ R_4+R_3 \end{array} \left|\begin{array}{cccc|c}1 & 2 & \text{-}1 & 1 & 30 \\ 0 & \text{-}9 & 5 & \text{-}1 & \text{-}20 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right|$

. . . .$\displaystyle \begin{array}{c} \\ \text{-}\frac{1}{9}R_2 \\ \\ \\ \end{array} \left|\begin{array}{cccc|c} 1 & 2 & \text{-}1 & 1 & 30 \\ 0 & 1 & \text{-}\frac{5}{9} & \frac{1}{9} & \frac{20}{9} \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right|$
. . . . . . . . . . . . . . . . .$\displaystyle \uparrow$
. . . . .
Besides, these are not the x, y, z, w values.

3. ## Gaussian elimination

is it possible to solve vehicle flows because row 3 & 4 have all zero value.