Prove that the set $\{\sin(x),\cos(x)\}$ is linearly independent using derivative. Prove that the set $\{\sin(x),\cos(x)\}$ is linearly independent using matricies.

Method 1

Let $r\sin(x)+s\cos(x)=0$

If $x=0$, this means that $s=0$

If $x=\pi/2$, this means that $r=0$.

Therefore the set is linearly independent.

Method 2 - using derivatives

$$r\sin(x)+s\cos(x)=0 \tag 0$$ 
Differentiating once,

$$r\cos(x)-s\sin(x)=0 \tag 1$$
Differentiating again,

$$-r\sin(x)-s\cos(x)=0 \tag 2$$ 

Take equation $0$ and $2$ in a matrix.

\begin{pmatrix}
1 &amp; 1\\
-1 &amp; -1
\end{pmatrix} RREF 
\begin{pmatrix}
1 &amp; 1\\
0 &amp; 0
\end{pmatrix}

So this still brings us back to our original $r\sin(x)+s\cos(x)=0$. Does this mean that it cannot be solved with a matrix, and we must solve it manually?
The first equation and the differentiated equation give a nontrivial element $(\sin(x),\cos(x))$ of the kernel of the matrix

$$\begin{bmatrix} r &amp; s \\ -s &amp; r \end{bmatrix}.$$
Note that the determinant of this matrix is $r^2+s^2=0$, so $r=s=0$. Note that this method proves more: that there is no interval on which the sine and cosine functions are constant multiples of one another.
Following your argument, you have two equations:
$$r\sin(x)+s\cos(x)=0 \tag 0$$
$$r\cos(x)-s\sin(x)=0 \tag 1$$
In matrix notation:
$$\begin{pmatrix}r &amp; s\\ -s &amp; r\end{pmatrix}
\begin{pmatrix}
\sin x \\
\cos x 
\end{pmatrix}=\begin{pmatrix}
0 \\
0
\end{pmatrix}$$
for all $x$. This implies that
$$\begin{pmatrix}r &amp; s\\ -s &amp; r\end{pmatrix}
\begin{pmatrix}
x \\
y 
\end{pmatrix}=\begin{pmatrix}
0 \\
0
\end{pmatrix}$$
for all x and y. It follows that
$$\begin{pmatrix}r &amp; s\\ -s &amp; r\end{pmatrix}=0$$