For which $t \in \mathbb{R}$ we have a basis in $\mathbb{R}^{3}$ if the vectors are

$$\begin{pmatrix} 1\\ 2\\ 2t \end{pmatrix}, \begin{pmatrix} 1\\ 2t\\ 0 \end{pmatrix}, \begin{pmatrix} 1\\ t\\ 1+t \end{pmatrix}$$

This is no homework, just another task that could be asked in my linear algebra exam.

I have no idea how to solve such a task. I have tried Gauss but it didn't work (I could get $3$ zeroes in bottom left corner of the matrix but getting $3$ more zeroes in top right seemed impossible).

Then I thought try to insert some value for $t$ such that the determinant $\neq 0$.

So I just took $t = 0$ because the determinant would be $-2$ and thus the vectors would be linearly independent which indicates they are a basis.

But then there would be many solutions... Did I do it correctly anyway? Is the notation correct too ?


I think the easiest way is to use the determinant. Then You can put the polynomial equal to zero.

You should put your vectors in columns of a $3 \times 3$ matrix. If the determinant of a matrix is $0$, then it means that the columns are linearly dependent and they cannot make a basis for $\mathbb{R}^3$. Therefore, you calculate the determinant of the matrix, which is a polynomial with respect to $t$, and put it equal to zero.

What you achieve at the end is


It means that for no value of $t$ the determinant can be $0$ and the vectors are always linearly independent.

  • $\begingroup$ There are things I don't understand here. "Then You can put the polynomial equal to zero" I put $t=0$ ? So then the determinant would equal $-2$. "it would not make a basis for $t=2$". Why not? For $t=2$ the determinant would be $-2$. $\endgroup$ – cnmesr Feb 28 '17 at 14:14
  • $\begingroup$ @cnmesr, has there been changes in the vectors? If it is not the case, I have made a mistake. Still the idea would be the same. I edited the answer. $\endgroup$ – Med Feb 28 '17 at 15:00

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