How can the vector functions \(X_1(t)=\begin{pmatrix} 1\\0\\1\end{pmatrix}\) and \(X_2(t)=\begin{pmatrix} t\\0\\t\end{pmatrix}\) be linearly independant when I find a relation between them $$X_2=tX_1?$$

You have a relation between \(X_1\) and \(X_2\) but the relation is NOT a linear combination.

A linear combination of \(X_1\) and \(X_2\) is in the form \(c_1X_1+c_2X_2\) where \(c_1\) and \(c_2\) are **scalars**. \(c_1\) and \(c_2\) are constant numbers and don't depend on \(t\).

Solve \(c_1X_1+c_2X_2=0\): It implies that \(c_1+c_2t=0\) or \(c_1=-c_2t\).

The left hand side of the equality, \(c_1\) is a contant. The right handside of the equality \(c_2t\) will be a constant only if \(c_2=0\), then \(c_1=0\) too.

There is exactly 1 linear combination between the vector functions \(X_1\) and \(X_2\) which is \(0X_1+0X_2=0\), the vectors are linearly independent.

When can I use the determinant to determine whether vectors are linearly independent?

You can use the determinant to check if vectors are linearly independant in the following cases:

- n constant vectors of dimension n
- n vector functions of dimension n if they are solution to the same homogeneous system X'=AX. In this case the determinant is called the Wronskian.

If you are not in the cases mentioned above, I would recommend you go back to the definition. Does the equation

$$\lambda_1X_1+\lambda_2X_2+\cdots\lambda_nX_n=0$$ have a unique solution?

if yes, they are linearly independent.

If no, they are linearly dependent.

Remember that the coefficients \(\lambda_i\) are constant, They are not variable functions.