MHB 12.6 linearly dependent or linearly independent?

[karush]

Are the vectors
$$v_1=x^2+1
,\quad v_2=x+2
,\quad v_3=x^2+2x$$
linearly dependent or linearly independent?
if
$$c_1(x^2+1)+c_2(x+2)+c_3(x^2+2x)=0$$
is the system
$$\begin{array}{rrrrr}
&c_1 & &c_3 = &0\\
& &c_2 &2c_3= &0\\
&c_1 &2c_2& = &0
\end{array}$$
I presume at this point observation can be made that this linear dependent
but also...
$$\left[ \begin{array}{ccc|c} 1 & 0 & 1 & 0 \\ 0 & 1 & 2 & 0 \\ 1 & 2 & 0 & 0 \end{array} \right]
\sim
\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array} \right]$$

 

[Ackbach]

With a set of functions, you normally use the Wronskian to determine linear independence.

 

[HallsofIvy]

"

I presume at this point observation can be made that this linear dependent"
Why would you presume that?
​

Perhaps it is just the way I learned systems of equations but I never want to change to matrices to solve systems of equations!

To determine whether or not v_1= x^2+ 1, v_2= x+ 2, and v_3= x^2+ 2x are independent or dependent we need to decide if there exist numbers, a, b, and c, no all 0, such that av_1+ bv_2+ cv_3= a(x^2+ 1)+ b(x+ 2)+ c(x^2+ 2x)= (a+ c)x^2+ (b+ 2c)x+ (a+ 2b)= 0.

In order that a polynomial be 0 for all x, all coefficients must be 0 so we must have
a+ c= 0
b+ 2c= 0
a+ 2b= 0

From the first equation c= -a so the second equation can be written as b+ 2(-a)= b- 2a= 0. Then b= 2a so the third equation is a+ 2(2a)= 5a= 0. a= 0 so b= 2(0)= 0 and c= -0= 0. The only solution is a= b= c= 0 so the vectors are independent.

 

[karush]

With a set of functions, you normally use the Wronskian to determine linear independence.

ok we haven't done that yet