$\{\vec{a}, \vec{b}\}$ is linearly independent then $\{4\vec{a}-\vec{b}, 2\vec{a}+3\vec{b}\}$ is linearly independent

Question

if $\{\vec{a}, \vec{b}\}$ is linearly independent how would you determine if $\{4\vec{a}-\vec{b}, 2\vec{a}+3\vec{b}\}$ is linearly independent

I have have managed to show through manipulation that for $C_1, C_2 \in \Bbb{R}$ $(4C_1+C_2)\vec{a}+(3C_2-C_1)\vec{b}=\vec{0}$ Is it enough to say that since $\{\vec{a}, \vec{b}\}$ is linearly independent then as a result since $(4C_1+C_2),(3C_2-C_1)\in \Bbb{R}$ then $\{4\vec{a}-\vec{b}, 2\vec{a}+3\vec{b}\}$ is linearly independent?

Answer 1

Because $$4\cdot3-2\cdot(-1)\neq0.$$

Answer 2

$\begin{pmatrix} a \\ b \end{pmatrix}=\begin{pmatrix} 4 & -1 \\ 2 & 3 \end{pmatrix}\begin{pmatrix} a \\ b \end{pmatrix}$ and \begin{pmatrix} 4 & -1 \\ 2 & 3 \end{pmatrix} is invertible

Answer 3

You are on the right track, but you need to say more and be clear about why you are concluding linear independence. (I think) what you did was begin by assuming

$$ C_1(4a - b) + C_2(2a + 3b) = 0 $$

for some $C_1,C_2 \in \mathbb{R}$. Now, your goal is to show that $C_1 = C_2 = 0$ (Why?). You can rearrange the above equation to get

$$ (4C_1 + 2C_2)a + (3C_2 - C_1)b = 0 $$

Because $a$ and $b$ are linearly independent we conclude $4C_1 + 2C_2 = 0$ and $3C_2 - C_1 = 0$. From this, what do you conclude about $C_1$ and $C_2$. And what does that mean in the context of your problem?

Answer 4

Suppose $(4c_1 + c_2)a + (-c_1 + 3c_2)b = 0$

then:

$4c_1 + c_2 = 0\\ -c_1 + 3c_2 = 0$

This system has a unique solution, $c_1= c_2 = 0$

If the only solution is the trivial solution, your vectors are independent.