#### Such-that.com # Linear Systems

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.