1) v1 and v2 are linear dependent if exist two non zero numbers a1 and a2 such that

`a_1*v_1+a_2*v_2=0 =>`

`a_1(2,1)+a_2(1,-1)=(0,0)=>`

`2a_1+a_2=0 and a_1-a_2=0=>` the second equation yield

`a_1=a_2 ` , substitue in the first yields that a1=a2=0.

**Hence the vectors are linear independent.**

2) Fof this example I will use a similar approach, but instead of adding them to zero, I will see if I can rewirte u3 in term of u1 and u2.

`k(1,2,-1)+j(5,1,4)=(6,1,5)`

`k+5j=6`

`2k+j=1`

I will use the third one to check.

multiply the first equation by 2, then subtract you obtain

`(2k+10j=12)-(2k+j=1)=>`

`9j=11=>j=11/9`

substitue in the first original you obtain

`k+5(11/9)=6=>k=6-55/9=(54-55)/9=-1/9`

let's check our answers to see if we are going to obtain the 3rd coordinate of the vector.

`-1/9*(-1)+11/9*4=1/9+44/9=45/9=5`

Thus we can rewrite `u_3=-1/9*u_1+11/9*u_2`

**Thus the vectors are linear dependent.**

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