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 & 1\\
-1 & -1
\end{pmatrix} RREF
\begin{pmatrix}
1 & 1\\
0 & 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?
Best Answer
The first equation and the differentiated equation give a nontrivial element $(\sin(x),\cos(x))$ of the kernel of the matrix
$$\begin{bmatrix} r & s \\ -s & r \end{bmatrix}.$$ 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.