1. ## Solution Space

I am currently learning about linear transforms and came across this problem:

Given a matrix: A =
[ 4 3 -1 7 ]
[ 1 1 0 2 ]
[ 2 5 3 7 ]

We are told that to find all the values t such that b = [ 1 t t^2 ]^T is solvable.

My attempt: I augmented the matrix to,

[ 4 3 -1 7 | 1 ]
[ 1 1 0 2 | t ]
[ 2 5 3 7 | t^2]

and ended up with:

[ 1 0 0 0 | t^2 -6t +1 ]
[ 0 1 2 0 | t^2 + t - 1 ]
[ 0 0 1 -1| t^2 - 3t ]

I'm not sure how to proceed from here. I understand that x_1 = t^2 - 6t + 1, that x_2 + 2 x_3 = t^2 + t - 1, and that x_3 - x_4 = t^2 - 3t. But how do those equations affect the constraints on t? It seems that no matter what I plug in for my free variable x_4, I can plug in whatever I want for t.

2. ## Re: Solution Space

Originally Posted by jsndacruz
I am currently learning about linear transforms and came across this problem:

Given a matrix: A =
[ 4 3 -1 7 ]
[ 1 1 0 2 ]
[ 2 5 3 7 ]

We are told that to find all the values t such that b = [ 1 t t^2 ]^T is solvable.

My attempt: I augmented the matrix to,

[ 4 3 -1 7 | 1 ]
[ 1 1 0 2 | t ]
[ 2 5 3 7 | t^2]

and ended up with:

[ 1 0 0 0 | t^2 -6t +1 ]
[ 0 1 2 0 | t^2 + t - 1 ]
[ 0 0 1 -1| t^2 - 3t ]

I'm not sure how to proceed from here. I understand that x_1 = t^2 - 6t + 1, that x_2 + 2 x_3 = t^2 + t - 1, and that x_3 - x_4 = t^2 - 3t. But how do those equations affect the constraints on t? It seems that no matter what I plug in for my free variable x_4, I can plug in whatever I want for t.
You need to double check your work.

After a few row operations I get the matrix

$\displaystyle \begin{bmatrix}1 & 1 & 0 & 2 & t \\ 0 & -1 & -1 & -1 & 1-4t \\ 0 & 0 & 0 & 0 & t^2-14t+3 \end{bmatrix}$

To have a solution we must have $\displaystyle t^2-14t+3=0$

3. ## Re: Solution Space

Aha - silly mistake. I always seem to make them when doing reef. Thanks so much for your help!