I have a final next week and am trying to practice linear independence which is the first step I believe for determining the basis but I'm stuck because my answer key says $a, c$ and $d$ are bases for $\mathbb{R}^2$ but I can not get $a_1, a_2 = 0$
a. $\begin{bmatrix}1\\3\end{bmatrix},\begin{bmatrix}2\\1\end{bmatrix}$
b. $\begin{bmatrix}3\\2\end{bmatrix},\begin{bmatrix}-6\\-4\end{bmatrix}$
c. $\begin{bmatrix}-2\\1\end{bmatrix},\begin{bmatrix}3\\1\end{bmatrix}$
d. $\begin{bmatrix}1\\1\end{bmatrix},\begin{bmatrix}-5\\3\end{bmatrix}$
My work:
a. $\begin{bmatrix}1&2&0\\3&1&0\end{bmatrix}\to\begin{bmatrix}1&2&0\\0&-5&0\end{bmatrix}\to\begin{cases}a_1+2a_2=0\\-5a_2=0\end{cases}\to(-2k,-5k),\,k=a_2$
Aren't these non-trivial answers so it is linearly dependent?
b. $\begin{bmatrix}3&-6&0\\2&-4&0\end{bmatrix}\to\begin{bmatrix}1&-2&0\\0&0&0\end{bmatrix}\to\begin{cases}a_1=2a_2\end{cases}\to(2k,k),\,k=a_2$
c. $\begin{bmatrix}-2&3&0\\1&1&0\end{bmatrix}\to\begin{bmatrix}1&1&0\\0&5&0\end{bmatrix}\to\begin{cases}a_1=-a_2\\5a_2=0\end{cases}\to(-k,5k),\,k=a_2$
d. $\begin{bmatrix}1&-5&0\\1&3&0\end{bmatrix}\to\begin{bmatrix}1&-5&0\\0&-8&0\end{bmatrix}\to\begin{cases}a_1-5a_2=0\\-8a_2=0\end{cases}\to(5k,-8k),\,k=a_2$
Cleary I'm doing something wrong. Thanks for the help
In $a)$ you get that $-5a_2 = 0$. So $a_2 = 0$. From there you get that $a_1 = -2a_2 = 0$. There are similar mistakes in the others.
Another thing that might help: Two vectors $u,v$ are linearly dependent if and only if $u = cv$ for some $c \in \mathbb{R}$. That is, two vectors are linearly dependent if and only if they are multiples of each other.