I need to find two more $b$'s other than $b=(2,5,7)$, such that the equation can be solved and two more such that the equation can't be solved. $$u\begin{bmatrix}1\\2\\3\end{bmatrix}+v\begin{bmatrix}1\\0\\1\end{bmatrix}+w\begin{bmatrix}1\\3\\4\end{bmatrix}=b$$
How do I find those values? What should I do to find both, values of b that make the system solvable and values that make the system unsolvable?
I tried to set $b$ as $(b_1,b_2,b_3)$, but I ended up with this matrix:
$$\left[\begin{array}{l}1&1&1&b_1\\0&2&-1&2b_1-b_2\\0&0&0&b_1-b_3+b_2\end{array}\right]$$
Sorry if it is something obvious, I'm knew with linear algebra.
Best Answer
After you apply your row reduction, look at the last row of your matrix.
It shows you that
$0u + 0v + 0 w = b_1 - b_3 + b_2$
This is only possible if $b_1 - b_3 + b_2 = 0$
So if you choose $b_1, b_2, b_3$ that satisfy this condition, your matrix has a solution (in this case infinite), otherwise it has no solutions.
for example if $b = (2, 5, 7)$ as you suggested, then it is solvable because $2 - 7 + 5 = 0$
Since you are new to linear algebra I did not mention determinants, dimensions, ranks etc. You don't need them in this case, just recall how a matrix represents a system of equations.