Question

How to solve 2 equations with 4 unknowns?

Answer 1

\[3t - 4u + v = 2\]
\[t + u - 2v + 3w = 1\]

Answer 2

What do you need to do? I don't think solving each variable is possible.

Answer 3

yes I agree

Answer 4

It has an infinite set of solutions. And there's an answer key on the book.

Answer 5

\[( c_1 - (12/7) c_2 + (6/7), c_1 - (9/7)c_2 + (1/7), c_1, c_2) \]

Answer 6

What is full question?

Answer 7

solve each system of equations by the Gaussian elimination method.

Answer 8

It's in the topic of matrices so I tried solving it using an augmented matrix. But it was too hard.

Answer 9

which part is hard ?

Answer 10

The part where I try to make sense of it all

Answer 11

say you're given only 1 equation wih 4 unknowns to solve

would it make sense to you ?

Answer 12

2 equations actually

Answer 13

Ah, I give up. It'll be better to use my time to study other things xD

Answer 14

\[\left[\begin{matrix}
3&-4&1&0&|&2\\
1&1&-2&3&|&1
\end{matrix}\right]
\]

Answer 15

is that your augmented matrix ?

Answer 16

yes

Answer 17

lets row reduce and find `particular` and `null` solutions


heard of them before ?

Answer 18

I've only been given the basics so, nope.

Answer 19

\[\left[\begin{matrix}
3&-4&1&0&|&2\\
1&1&-2&3&|&1
\end{matrix}\right]
\]

R2-(1/3)R1 :
\[\left[\begin{matrix}
3&-4&1&0&|&2\\
0&7/3&-7/3&3&|&1/3
\end{matrix}\right]
\]

Answer 20

its easy, notice that you have two pivots. so there will be two free variables in the general solution

Answer 21

Ahh, if that's what you're talking about then, I know.

Answer 22

yes you just need to know finding null and particular solutions, thats all

Answer 23

lets find the null solution first

Answer 24

you want to solve below for null solution :

\[
\left[\begin{matrix}
3&-4&1&0\\
0&7/3&-7/3&3
\end{matrix}\right]
\begin{pmatrix}
t\\
u\\
v\\
w
\end{pmatrix} =
\left[\begin{matrix}
0\\
0
\end{matrix}\right]
\]

Answer 25

know how to find the null solution ?

Answer 26

I'm searching to see if I do.

Answer 27

let me tell you the strategy quick :
1) set v=0, w=1 and find one solution
2) set v=1, w=0 and find another solution

the combination of above solutions is the complete null solution

Answer 28

v = 0, w = 1 :
\[
\left[\begin{matrix}
3&-4&1&0\\
0&7/3&-7/3&3
\end{matrix}\right]
\begin{pmatrix}
t\\
u\\
0\\
1
\end{pmatrix} =
\left[\begin{matrix}
0\\
0
\end{matrix}\right]
\]

Answer 29

can you solve t and u ?

Answer 30

Ok, gimme some time if it's ok

Answer 31

Okay ill show you how to solve one solution, maybe you can try the other solution

Answer 32

v = 0, w = 1 :
\[
\left[\begin{matrix}
3&-4&1&0\\
0&7/3&-7/3&3
\end{matrix}\right]
\begin{pmatrix}
t\\
u\\
0\\
1
\end{pmatrix} =
\left[\begin{matrix}
0\\
0
\end{matrix}\right]
\]

3t - 4u = 0
7/3u +3 = 0

solving you get :
u = -9/7
t = -12/7

Answer 33

so one null solution vector is :
\[\begin{pmatrix}
-12/7\\
-9/7\\
0\\
1\\
\end{pmatrix}\]

Answer 34

see if you can find the other null solution by letting v=1 and w=0,

you need to solve below equation :

\[\left[\begin{matrix}
3&-4&1&0\\
0&7/3&-7/3&3
\end{matrix}\right]
\begin{pmatrix}
t\\
u\\
1\\
0
\end{pmatrix} =
\left[\begin{matrix}
0\\
0
\end{matrix}\right]\]

Answer 35

3t -4u +1 = 0
7/3 u - 7/3 = 0

u = 1
t = 1

Answer 36

yes so the second null solution vector is

\[\begin{pmatrix}
1\\
1\\
1\\
0\\
\end{pmatrix}\]

Answer 37

Can we say the complete `nullspace` is :

\[

c_1\begin{pmatrix}
1\\
1\\
1\\
0\\
\end{pmatrix}
+ c_2
\begin{pmatrix}
-12/7\\
-9/7\\
0\\
1\\
\end{pmatrix}
\]

?

Answer 38

\[\left(\begin{matrix}-12/7 \\ -9/7\\ 1\\ 1 \end{matrix}\right)\]?

Answer 39

no don't combine, null solution is a linear combination of the two solution vectors we have found earlier

Answer 40

you need to find a particular solution and add it to null solution for the final general solution

Answer 41

lets find the particular solution quick

Answer 42

for particular solution, you swt all the free variables to 0 and solve the original given equation

Answer 43

solve below for particular solution :

\[
\left[\begin{matrix}
3&-4&1&0\\
0&7/3&-7/3&3
\end{matrix}\right]
\begin{pmatrix}
t\\
u\\
0\\
0
\end{pmatrix} =
\left[\begin{matrix}
2\\
1/3
\end{matrix}\right]

\]

Answer 44

3t -4u = 2
7/3 u = 1/3

u - 1/7
t = 6/7

I really have no idea.

Answer 45

thats right!

so a particular solution is
\[
\begin{pmatrix}
6/7\\
1/7\\
0\\
0\\
\end{pmatrix}
\]

Answer 46

we're done! we have `particular` and `null` solutions, so we can write the final `general` solution

Answer 47

`general` = `particular` + `null`

Answer 48

general solution :

\[
\begin{pmatrix}
6/7\\
1/7\\
0\\
0\\
\end{pmatrix}

+

c_1\begin{pmatrix}
1\\
1\\
1\\
0\\
\end{pmatrix}
+ c_2
\begin{pmatrix}
-12/7\\
-9/7\\
0\\
1\\
\end{pmatrix}
\]

Answer 49

see if that makes more or less sense..

Answer 50

It made more sense than what I could have made. at least.

Answer 51

Thanks a lot for the help.

Answer 52

basically it all boils down to just finding `particular` and `null` solutions

Answer 53

you're welcome!
watch this video when u have time
http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/video-lectures/lecture-8-solving-ax-b-row-reduced-form-r/

Answer 54

Thanks, I have a test on it tomorrow anyway.

Answer 55

good luck!