Explanation Let me give several different explanations, all of which are equally valid.

Well, a result in the text (which I also hope to mention in class) declares that any collection of more than n vectors in Rⁿ is linearly dependent. Here we have 5 vectors in R² and surely 5>2. So the set {u₁, u₂, u₃, u₄, u₅} is linearly dependent.
The vector equation ∑_j=1⁵c_ju_j=0 represents a homogeneous system of 2 linear equations in 5 unknowns. Again, the textbook declares that such a homogeneous system (more unknowns than equations) must have non-trivial solutions. So the set {u₁, u₂, u₃, u₄, u₅} is linearly dependent. (This explanation and the preceding one are really the same, just phrased in different "languages".)
The coefficient matrix of the linear system ∑_j=1⁵c_ju_j=0 is:
```
[ 1  4  5  3  2 ]
[ 1  5  8  6  3 ]
```
and subtracting row 1 from row 2 and replacing row 2 by this gives:
```
[ 1  4  5  3  2 ]
[ 0  1  3  3  1 ]
```
and now subtracting 4 times row 2 from row 1 and replacing row 1 by this gives:
```
[ 1  0 -7 -9 -2 ]
[ 0  1  3  3  1 ]
```
so this equivalent system has 2 basic variables and 3 free variables. There are many non-trivial solutions. Further discussion is not necessary, but if you wish, we can even "exhibit" such a solution: c=7, c=–3, c=1, c₄=0, and c₅=0.
There are probably other ways of doing this problem! (Hey, I just noticed that 2u₁+u₅ is the same as u₂, for example, so that 2u₁+u₅–u₂=0!)

So we certainly have verified that the set {u₁, u₂, u₃, u₄, u₅} is linearly dependent.