Question

Determine whether the given functions are linearly dependent or linearly independent. f1(t) = 4t − 5, f2(t) = 4t2 + 1, f3(t) = 5t2 + t linearly dependent linearly independent Correct: Your answer is correct. If they are linearly dependent, find a linear relation among them. (Use f1 for f1(t), f2 for f2(t), and f3 for f3(t). Enter your answer in terms of f1, f2, and f3. If the system is independent, enter INDEPENDENT.)

Answer 1

Answer:

$f_1(t),f_2(t)$ and $f_3(t)$ linearly dependent.

The required relation is

$\therefore a[f_1(t)-4f_2(t)+5f_3(t)]=0$

where a is a nonzero number.

Step-by-step explanation:

Given that,

$f_1(t)=4t-5$ , $f_2(t)=4t^2+1$ and $f_3(t)=5t^2+t$

We consider a linear combination

$k_1f_1(t)+k_2f_2(t)+k_3f_3(t)=0$

Putting the value of $f_1(t),f_2(t)$ and $f_3(t)$

$\therefore k_1(4t-5)+k_2(4t^2+1)+k_3(5t^2+t)=0$

$\Rightarrow 4k_1t-5k_1+4k_2t^2+k_2+5k_3t^2+k_3t=0$

$\Rightarrow 4k_2t^2+5k_3t^2+k_3t+4k_1t-5k_1+k_2=0$

$\Rightarrow (4k_2+5k_3)t^2+(k_3+4k_1)t-5k_1+k_2=0$

Equating the co-efficient of $t^2$, t and constant terms

$\therefore (4k_2+5k_3)=0$ .......(1)

$\therefore (k_3+4k_1)=0$ ..........(2)

$\therefore -5k_1+k_2=0$ ..........(3)

From (2) we get

$\therefore (k_3+4k_1)=0$

$\therefore k_3=-4k_1$

From (3) we get

$\therefore -5k_1+k_2=0$

$\Rightarrow k_2=5k_1$

Putting the value of $k_2 \ and \ k_3$ in equation (3)

$\therefore (4k_2+5k_3)=0$

$\Rightarrow 4.5k_1+5(-4k_1)=0$

$\Rightarrow 0=0$

Let $k_1=a$ [ a is a non zero number]

Then $k_2= 5a$ and $k_3=-4a$

We get a nonzero value of $k_1$,$k_2 \ and \ k_3$ .

Then $f_1(t),f_2(t)$ and $f_3(t)$ linearly dependent.

The required relation is

$af_1(t)+(-4a)f_2(t)+5af_3(t)=0$

$\Rightarrow a[f_1(t)-4f_2(t)+5f_3(t)]=0$ [a is a nonzero number]