Answer:The statement is true
Step-by-step explanation:
This is an implication, then if we by starting from the premise we can get to the conclusion, then the implication is true. First let us remember that a given set of n vectors, say [tex]\{u_1,\ldots, u_n\}[/tex] is l.i (linearly independent), by definition if the only solution to [tex]\alpha_1\cdot u_1 + \cdots + \alpha_n\cdot u_n = \textbf{0}, is (\alpha_1, \ldots, \alpha_n)= \textbf{0}\in \mathcal {R}^n, \mathcal {R}[/tex] the set of real numbers, that is the only linear combination equal to the null vector is the null combination, all the scalars must be zero. And since it is a definition it is  if and only if.
Then if we say that the premise is true, then u4 is not a linear combination of {u1,u2,u3}, this means there do not exist [tex]\alpha_1, \alpha_2, \alpha_3 \in \mathcal{R}[/tex] not all zero, such that, [tex]\alpha_1\cdot u_1+ \alpha_2\cdot u_2+ \alpha_3\cdot u_3=u_4[/tex] which is equivalent to say that for every [tex]\alpha_1, \alpha_2, \alpha_3,\alpha_4 \in \mathcal{R}[/tex] with
[tex]\alpha_1\cdot u_1 + \alpha_2\cdot u_2+ \alpha_3\cdot u_3+ \alpha_4\cdot u_4 = \textbf{0}[/tex] implies [tex]\alpha_1= \alpha_2= \alpha_3= \alpha_4=0[/tex], then the given set is linearly independent as by definition and just as it is stated in the conclusion, therefore the affirmation is true.
