Linear Algebra: Solving a system with free variables

[Ryry013]

Here is a picture of the problem.

Can anyone give me some hints on the problem? I've looked in my textbook, but I don't know what "s" means. I found stuff on the parametric vector form, and it gives me the equation x = su + tv, but I don't see any "t"'s in this problem.

I first tried setting up an augmented matrix with

[ 1 1 0 0 | -3 ]
[ 0 1 1 0 | -3 ]
[ 0 0 1 1 | -4 ]
[ 1 0 0 1 | -4 ]

And I solved for the four variables, like:
x₁ = -3 - x₂
x₂ = -3 - x₃
x₃ = -4 - x₄
x₄ = -4 - x₁

And now I'm stumped. I didn't think these would do anything, but setting the first and second rows equal to each other, and the third and fourth rows equal to each other simply gets x₁ - x₃ = 0 for both of them.

I didn't think this would do anything either, but I tried also back substituting the x₁ equation into the x₄ equation, and then the x₂ equation into x₄, and so on, and that ended up with x₄ = x₄.

 

[Chas3down]

edit: never my idea didn't work

 

[Ryry013]

edit: never my idea didn't work

Thanks for the try though.

 

[Mark44]

Here is a picture of the problem.

Can anyone give me some hints on the problem? I've looked in my textbook, but I don't know what "s" means. I found stuff on the parametric vector form, and it gives me the equation x = su + tv, but I don't see any "t"'s in this problem.

I first tried setting up an augmented matrix with

[ 1 1 0 0 | -3 ]
[ 0 1 1 0 | -3 ]
[ 0 0 1 1 | -4 ]
[ 1 0 0 1 | -4 ]

And I solved for the four variables, like:
x₁ = -3 - x₂
x₂ = -3 - x₃
x₃ = -4 - x₄
x₄ = -4 - x₁

It looks like you didn't actually do anything with your augmented matrix. Use row reduction to get the matrix in reduced row echelon form. When it's completely reduced you can solve for x₁, x₂, and x₃ in terms of x₄. Your solution will look like a vector + a parameter (that's s) times another vector.

And now I'm stumped. I didn't think these would do anything, but setting the first and second rows equal to each other, and the third and fourth rows equal to each other simply gets x₁ - x₃ = 0 for both of them.

I didn't think this would do anything either, but I tried also back substituting the x₁ equation into the x₄ equation, and then the x₂ equation into x₄, and so on, and that ended up with x₄ = x₄.

 

[HallsofIvy]

I agree with Mark44. When I row reduce the augmented matrix, I get all "0"s in the last row, non-zero values in the third column and fourth columns.
Something like this:
\begin{bmatrix} 1 & 0 & 0 & a_1 & b_1 \\ 0 & 1 & 0 & a_2 & b_2 \\ 0 & 0 & 1 & a_3 & b_3 \\ 0 & 0 & 0 & 0 & 0\end{bmatrix}
(with, of course, specific numbers for the "a"s and "b"s but I wanted to leave that fun to you!)

That is equivalent to the equation x_1+ a_1x_4= b_1, x_2+ a_2x_4= b_4, and x_3+ a_3x_4= b_4 which can be solved as "a_i equals a number plus a multiple of x_4".