Wronskian and Homogenous Equation

Question

I am given three functions and an interval which I have to prove linear independence.

$$\begin{align}f_1(x)&=x \\ f_2(x)&=x^2 \\ f_3(x)&=4x-3x^2\end{align}$$

This is when I ask you to bear with me, because this is a new type of math problem I am solving. So starting off I recognize that I have to get a $3\times 3$ determinant, so I will be going up to the second derivative. Another thing I realize is that I have to set $\det(3 \times 3 \ \text{matrix})=0$ If I know its $\ne 0 \therefore$ I can say its linearly independent

My Work

I began my work as follows: $W=\begin{bmatrix}x & x^2 & 4x-3x^2 \\ 1 & 2x & 4-6x \\ \bbox[red]{0} & 2 & -6\end{bmatrix}$

From there I proceeded to get the $\det(W)$, here is where the problem arose I wanted to make use of that zero in red. So my setup went something like this: $\begin{align}\begin{bmatrix}2x & x^2 \\ 4-6x & 4x-3x^2\end{bmatrix}0-\begin{bmatrix}2 &x^2 \\-6 & 4x-3x^2\end{bmatrix}1\end{align}+\begin{bmatrix}2&2x\\-6&4-6x\end{bmatrix}x$

I know this is probably wrong but any help would be appreciated because I would like to make use of that 0, to zero out a $2\times 2\ \text{matrix}.$Is my determinant right?

Answer 1

The Wronskian is zero: $$W=\begin{vmatrix}x & x^2 & 4x-3x^2 \\ 1 & 2x & 4-6x \\ {0} & 2 & -6\end{vmatrix}=x\begin{vmatrix}1 & x & 4-3x\\ 1 & 2x & 4-6x \\ {0} & 2 & -6\end{vmatrix}$$

$$W=x\begin{vmatrix}0 & -x & 3x\\ 1 & 2x & 4-6x \\ {0} & 2 & -6\end{vmatrix}=-x\begin{vmatrix} -x & 3x\\ 2 & -6\end{vmatrix}$$ $$W=0$$ it's linearly dependant. Which is a bit obvious from the beginning...Since $4x-3x^2$ is a linear combination of $(x,x^2)$

Answer 2

The Wronskian is equal to 0:

Using the setup I was trying in the question and LostinSpace answer I was able to complete my attempt and get the same answer.

$$\begin{align} W&=\begin{bmatrix}x&x^2&4x-3x^2\\ 1&2x&4-6x\\ 0&2&-6\end{bmatrix} \\\det(W)&=\begin{bmatrix}x^2&4x-3x^2\\ 2x&4-6x\end{bmatrix}\cdot0-\begin{bmatrix}x^2&4x-3x^2\\ 2&-6\end{bmatrix}\cdot 1+\begin{bmatrix}2x&4-6x\\ 2&-6\end{bmatrix}\cdot x\\ \det(W)&=0 +8x-8x=0\end{align}$$

Therefore both LostinSpace and my answer coincide, and it shows that they are not linearly interdependent.