- #1

- 1,395

- 0

(1,0,1)

(2,-1,0)

(0,-1,-2)

(1,1,3)

i tried:

i can guess (x+2y+0*z+t,0*x -y-z+t,x+0-2*z +3*t)=x(1,0,1) +y(2,-1,0) +z(0,-1,-2)+t(1,1,3)

but what i got is a big vector

how i get the equations?

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- Thread starter transgalactic
- Start date

- #1

- 1,395

- 0

(1,0,1)

(2,-1,0)

(0,-1,-2)

(1,1,3)

i tried:

i can guess (x+2y+0*z+t,0*x -y-z+t,x+0-2*z +3*t)=x(1,0,1) +y(2,-1,0) +z(0,-1,-2)+t(1,1,3)

but what i got is a big vector

how i get the equations?

- #2

- 1,395

- 0

x+2y+0*z+t=0

etc..

is it correct??

i cant understand why its correct?

- #3

- 96

- 0

How does one usually go about solving a system of Linear equations to find the solution vectors?

It represents (In some equivalent row or column form) a matrix equation Ax=y, with x and y vectors.

Are you sure this is a system of linear equations? Because if it is, then you aren't going to have four different solution vectors, ever, for one system of linear equations. So all you should have to do is write a set of dependent equations.

Unless the question is as I suspect, find a set of linear equations such that each of these vectors independently is a solution.

It represents (In some equivalent row or column form) a matrix equation Ax=y, with x and y vectors.

Are you sure this is a system of linear equations? Because if it is, then you aren't going to have four different solution vectors, ever, for one system of linear equations. So all you should have to do is write a set of dependent equations.

Unless the question is as I suspect, find a set of linear equations such that each of these vectors independently is a solution.

Last edited:

- #4

- 1,395

- 0

i was told to equalize each part to zero

x+2y+0*z+t=0

etc..

is it correct??

x+2y+0*z+t=0

etc..

is it correct??

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