Can you row reduce the Wronskian with functions? UPDATED BELOW
Recently, I came across a problem asking me to exploit the Wronskian to determine the independence of a few functions. The functions were $\sin(x), \cos(x), x\sin(x)$, and $x\cos(x)$. I immediately noticed it would be a $4\times 4$ matrix with (probably?) some cyclical functions.
I absolutely refused to "brute force" the matrix -- but I still wanted to solve it by hand. I noticed in this case if I could use row reduction I could get some zero entries, expand the determinate across that row, or down that column and lessen the work. I don't immediately see a problem with this, if you do row reduction aren't you stating that the two systems are equivalent?
I asked my professor if this is a valid approach and he stated he has never thought about it. He then further challenged me to use his solutions which were done "long and painfully" and compare what I would do if row reduction is valid.
It seems we got the same answer -- but I'm not sure if that will always happen.
TL;DR: Can you do row reduction after setting up the Wronskian, but before you take the determinate? Is this always valid? How can one prove either way? Furthermore, can you do column reduction?
Problem Statement

My work:

Professor's Solution: $W=4$
Update: Guidance from a previous professor. Credit Glenn Lahodony, UTSA mathematics.
"This is a very interesting question!
The short answer to your question is yes, you can perform row reduction on a matrix of functions.  However, we may have to be careful when performing row reduction on such a matrix depending on the problem of interest.  The matrix you are considering in your problem has entries involving sin(x), cos(x), xsin(x), and xcos(x).  The domain of these functions (and their derivatives) is all real numbers.  However, if we perform certain row reduction operations such as multiplying by 1/sin(x), we have restrictions on the domain (i.e. x is not an integer multiple of pi).  If we then find the Wronskian is nonzero, the set of functions is still linearly independent on any interval not containing integer multiples of pi.  However, if we need to prove linear independence on every interval, we would have to consider a separate case where x is an integer multiple of pi.  Depending on the row operations performed, we may also end up multiplying a row by zero (i.e. multiplying by sin(x) amounts to multiplying a row by zero if x is an integer multiple of pi) which could result in a zero Wronskian for certain values.
We can perform row reduction operations to columns of a matrix as well, but the applications are very limited as the system of algebraic equations is not equivalent if column operations are performed.  The only case I can think that column operations would be needed is in studying properties of the transpose of a matrix."
 A: Yes, you can definitely use row-reduction to evaluate a Wronskian. Just remember that

*

*Adding a multiple of one row to another row does not change the value of the determinant.


*Swapping two rows reverses the sign of the determinant.


*Removing a common factor from a row divides the determinant by that factor.
Consider the following example.
Evaluate the Wronskian to determine the independence of $\{e^x,e^{-x},\cos x\}$
$W=\begin{vmatrix}
 e^x & e^{-x} & \cos x \\ 
 e^x & -e^{-x} & -\sin x \\ 
 e^x & e^{-x} & -\cos x
\end{vmatrix}$
Perform the following row operations on rows 2 and 3.
$-R_1+R_2\to R_2$,$\quad-R_1+R_3\to R_3$
Adding a multiple of one row to another row does not change the value of $W$, so
$W=\begin{vmatrix}
 e^x & e^{-x} & \cos x \\ 
 0 & -2e^{-x} & -\cos x -\sin x \\ 
 0 & 0 & -2\cos x
\end{vmatrix}$
This is an upper triangular determinant, so the value equals the product of the main diagonal.
$W=4\cos x$
Thus the functions are linearly independent.
A: Another thing you can do.  Once you get the determinant, substitute a value of $x$.  (Presumably one that makes the determinant simple.)  If that determinant is non-zero, your functions are linearly independent.
Here, take $x=0$ to get
$$
W(0) = \left|
\begin{array} 1
0 & 1 & 0 & 0
\\
1 & 0 & 0 & 1
\\
0 & -1 & 2 & 0
\\
1 & 0 & 0 & -3
\end{array}
\right| = 4 .
$$
Easier to evaluate than the original functional Wronskian.  This proves linear independence.  (Of course, getting $0$ for one $x$ would not suffice to prove dependence.)
