Method of the Gauss-Jordan

[Apprentice123]

Solving the systems:

1)
x-y+2z-w=-1
2x+y-2z-w=-2
-x+2y-4z+w=1
3x-3w=-3


2)
v+3w-2x=0
2u+v-4w+3x=0
2u+3v+2w-x=0
-4u-3v+5w-4x=0


They have a solution?

 

[soroban]

Hello, Apprentice123!

These do not have a unique solution . . .

\(\displaystyle 1)\;\;\begin{array}{ccc}x-y+2z-w &=& \text{-}1 \\ 2x+y-2z-w &=& \text{-}2 \\ -x+2y-4z+w &=& 1 \\ 3x\qquad\qquad-3w& =& \text{-}3 \end{array}\)

\(\displaystyle \text{We have: }\;\begin{array}{|cccc|c|} 1& \text{-}1 & 2 & \text{-}1 & \text{-}1 \\ 2 & 1 & \text{-}2 & \text{-}1 & \text{-}2 \\ \text{-}1 & 2 & \text{-}4 & 1 & 1 \\ 3 & 0 & 0 & \text{-}3 & \text{-}3 \end{array}\)


\(\displaystyle \begin{array}{c}\\ R_2-2R_1 \\ R_3 + R_1 \\ R_4 - 3R_1 \end{array} \begin{array}{|cccc|c|}1 & \text{-}1 & 2 & \text{-}1 & \text{-}1 \\ 0 & 3 & \text{-}6 & 1 & 0 \\ 0 & 1 & \text{-}2 & 0 & 0 \\ 0 & 3 & \text{-}6 & 0 & 0 \end{array}\)


\(\displaystyle \begin{array}{c}R_1+R_3 \\ R_2-3R_3 \\ \\ R_4 - R_2 \end{array} \begin{array}{|cccc|c|} 1 & 0 & 0 & \text{-}1 & \text{-}1 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & \text{-}2 & 0 & 0 \\ 0 & 0 & 0 & \text{-}1 & 0 \end{array}\)


\(\displaystyle \begin{array}{c}R_1+R_2 \\ \\ \\ R_4+R_2\end{array} \begin{array}{|cccc|c|} 1 & 0 & 0 & 0 & \text{-}1 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & \text{-}2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\)


\(\displaystyle \text{This translates to: }\;\begin{array}{c} x \:=\:\text{-}1 \\ w \:=\:0 \\ y-2z \:=\:0 \end{array}\)


\(\displaystyle \text{So we have: }\:\begin{array}{c}x \:=\:\text{-}1 \\ y \:=\:2z \\ z \:=\:z \\ w \:=\:0 \end{array}\)

\(\displaystyle \text{On the right side, replace }z \text{ with a parameter }t\!:\;\;\begin{Bmatrix}x &=& -1 \\ y &=& 2t \\ z &=& t \\ w &=&0 \end{Bmatrix}\)


We have an infinite number of solutions, one for every value of \(\displaystyle t.\)

 

[Apprentice123]

thank you