Question

The functions `f_1(x)=3x-1`, `f_2(x)=4x` and `f_3(x)=2x` are linearly dependent. Show this by finding values of `c_1`, `c_2` and `c_3`, not all zeros, such that `c_1f_1(x)+c_2f_2(x)+c_3f_3(x)=0`.

Answer

`f_1(x)=3x-1`, `f_2(x)=4x`, `f_3(x)=2x`.

The task is to find such constants `c_1`, `c_2`, `c_3` not equal zero simultaneously such that

`c_1f_1(x)+c_2f_2(x)+c_3f_3(x)=0`.

`c_1(3x-1)+c_2*4x+c_3*2x=0`

`x(3c_1+4c_2+2c_3)-c_1=0`.

From this following system is obtained:

`{(3c_1+4c_2+2c_3=0),(c_1=0):}`.

From second equation `c_1=0` and first equation gives `4c_2+2c_3=0` or `c_3=-2c_2`.

So, if `c_1=0`, `c_2=t`, then `c_3=-2t` `(t in R)` and `c_1f_1(x)+c_2f_2(x)+c_3f_3(x)=0`.

In particular, we can take `c_1=0`, `c_2=1`, `c_3=-2`.