Question

Does a solution x exist for this equation? Is it unique?

Answer 1

\[\left[\begin{matrix}2 & -1 \\ -3 & 3 \\ -1 & 2 \end{matrix}\right] x= \left[\begin{matrix}-1 \\ 0\ \\ -1 \end{matrix}\right] = y\]

Answer 2

Equation 1: 2x+(-1)x=-y
Equation 2: -3x+3x=0
Equation 3: -x+2x=-y

Answer 3

Oh.

Answer 4

well it'd be x1 and x2 right

Answer 5

Then, ignore my last reply.

Answer 6

Yes so I solved so then what would I do next?

Answer 7

The first equation is 2x-x=-1=y
You would get x=-1=y.

Answer 8

The second equation is -3x+3x=0=y
You would get 0=0=y

Answer 9

Wait, is x and y matrix?

Answer 10

Yes they are vectors

Answer 11

Sorry.
I was being stupid...

Answer 12

Let's start over.

Answer 13

\[\left[\begin{matrix}2 & -1 \\ -3 & 3 \\ -1 & 2 \end{matrix}\right] \left[\begin{matrix}x1 \\ x2\ \end{matrix}\right]= \left[\begin{matrix}-1 \\ 0\ \\ -1 \end{matrix}\right]\]

Answer 14

Right.

Answer 15

so...
2(x1)-(x2)=-1
-3(x1)+3(x2)=0
-1(x1)+2(x2)=-1

Answer 16

The second equation is saying that 0=0, so we don't need it.

Answer 17

We now have to solve the system of equation;

2(x1)-(x2)=-1
-(x1)+2(X2)=-1

Answer 18

Do you know how to?

Answer 19

Yeah I got x2 = -2/3 and -1/6?

Answer 20

Wait. You're only supposed to get one answer for each variable.

Answer 21

Oh that's wrong I got -5/6 for x1

Answer 22

not -1/6

Answer 23

Add those 2 equations to get
(x1)+(x2)=-2

Answer 24

Add that equation to the first equation to get

3(x1)=-3
(x1)=-1

Answer 25

So, (x2)=-1

Answer 26

Oh yes my algebra was wrong

Answer 27

for the first part so it carried onto the second my bad

Answer 28

which is expected from x1=x2 in the second equation

Answer 29

Yep.

Answer 30

Okay so there is a unique solution? [1 1]'

Answer 31

I meant [-1 -1]'

Answer 32

Correct!

Answer 33

How would you know if there's not a unique solution?

Answer 34

I'm assuming there wouldn't be a 0 in the y vector?

Answer 35

not necessarily

Answer 36

The existence of 0 in the y vector doesn't mean anything.

Answer 37

Unless there is a rule that I don't know, I believe that you have to solve each one of them to see if there's any solution.

Answer 38

Okay thank you!

Answer 39

yep