Find a number tt such that

(3,1,4),(2,3,5),(5,9,t)(3,1,4), (2,-3,5), (5,9,t)

Is not linearly independent in R3\mathbf R^3


We want (for some ak0a_k \ne 0)

a1(3,1,4)+a2(2,3,5)+a3(5,9,t)=0a_1(3,1,4) + a_2(2,-3,5) + a_3(5,9,t) = 0

If some ak0a_k \ne 0, we must have a30a_3 \ne 0 since (3,1,4)(3,1,4) and (2,3,5)(2,-3,5) are linearly independent. Therefor we can divide by a3a_3, relabel and rearrange to get

(5,9,t)=b1(3,1,4)+b2(2,3,5)(5,9,t) = b_1 (3,1,4) + b_2(2,-3,5)

We'll pick b1,b2b_1, b_2 to satisfy the first two constraints (the first coordinates must be equal) then pick tt to satisfy the last one.

{5=3b1+2b29=1b13b2    b1=3,b2=2\begin{cases} 5 = 3b_1 + 2b_2 \\ 9 = 1b_1 - 3b_2 \end{cases} \implies b_1 = 3, b_2 = -2

Which gives t=3(4)2(5)=2t = 3(4) - 2(5) = 2, which is our solution since

1(5,9,2)+3(3,1,4)2(2,3,5)=0-1(5,9,2) + 3(3,1,4) - 2(2,-3,5) = 0

Implying they are linearly dependent.

(I could have skipped the derivation and jumped to this part, but that makes it harder to understand)