Considering the following vectors $u,v,w\in R^5$
$u = (0,1,1,-1,1/2)$ $v = (t,1,1,0,1)$ $w = (1,3,3,-2,-2)$
Then, they are linearly independent:
a) for each value of $t$ b) if and only if $t=0$ c) if $t=1$ d) if and only if $t\neq 1$
I think the best way to solve the exercise, is to find a combination of elements that is equal to the zero vector. So I go on in this way:
$\alpha(0,1,1,-1,1/2)+\beta(t,1,1,0,1)+\gamma(1,3,3,-2,-2) = (0,0,0,0,0)$
finding ways for which $\alpha,\beta,\gamma$ are equal to $0$, but I cannot get the result through the calculations. Any advices?
The vectors are linearly independent if and only if the matrix $$ \begin{bmatrix} 0 & t & 1 \\ 1 & 1 & 3 \\ 1 & 1 & 3 \\ -1 & 0 & -2 \\ 1/2 & 1 & -2 \end{bmatrix} $$ has rank $3$.
We can do Gaussian elimination: swap the first and second row and eliminate in the first column, getting $$ \begin{bmatrix} 1 & 1 & 3 \\ 0 & t & 1 \\ 0 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1/2 & -7/2 \end{bmatrix} $$ Now swap the second and fourth column; then eliminate in the second column: $$ \begin{bmatrix} 1 & 1 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 1-t \\ 0 & 0 & 4 \end{bmatrix} $$ Then you have the answer.