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Given the vectors $u_1$, $u_2$ $u_3 \in \mathbb{R}^3$ with $u_1=\left(\begin{array}{c} 1 \\ 2 \\ 1 \end{array}\right)$, $u_2=\left(\begin{array}{c} 4 \\ 5 \\ 2 \end{array}\right)$ , $u_3=\left(\begin{array}{c} 2 \\ 6t-1 \\ 7t \end{array}\right)$ determine $t$ so $u_1,u_2,u_3$ are lineary independent

I know that for those three vectors to be independent, $$\lambda u_1 + µu_2+\phi u_3=\mathbf{0}$$ may only be true for $$\lambda,µ,\phi=0$$ So to solve the task I need to determine a value for $t$ so there's inequality in the first equation for all scalars != 0.
How do I find a value for $t$ so all this is satisfied?

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Is $v_1, v_2, v_3$ a typo? If not, you should clarify which vectors are – Jorge Jul 4 '13 at 11:22
typo, sorry for that – Rickyfox Jul 4 '13 at 11:26
What are $u_1$ and $u_2$? – Daniel R Jul 4 '13 at 11:38
added the given vectors – Rickyfox Jul 4 '13 at 11:42
1… – Tomas Jul 4 '13 at 11:45
up vote 1 down vote accepted

These three linear vectors are independent iff the determinant of the matrix built with them is non zero. Hence we study

$$\det \begin{pmatrix}1&4&2\\2&5&6t-1\\1&2&7t\end{pmatrix} = -9t-4.$$

This determinant is non-zero iff $t\ne -\frac 49$.

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Yeah following Tomas's comment I already figured that out, but thanks for putting it in an answer. – Rickyfox Jul 4 '13 at 12:27

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