How is Wronskian used for linear system of equations? When came across the  "Advanced Mathematical Methods in Science and Engineering[1]", I was stuck here $W(z_i)=\det{[a_{ij}]}W(y_i)$ for $W(z_i)$. Since the determinant is just a scalar, it is obviously true $z_i=\det{[a_{ij}]}y_i$...
It seems not consistently following the Wronskian's definition
$W(f_{1} \cdots  f_{n}):=$ \begin{vmatrix}
f_{1} & \cdots & f_{n} \\ 
\vdots  &  & \vdots \\ 
f_{1}^{(n-1)} & \cdots & f_{n}^{(n-1)} 
\end{vmatrix}

 A: take it as linear transformation
$$
z = A y
$$
where $z= (z_1,...,z_n)^T$, $y= (y_1,...,y_n)^T$ and $T$ is the materix with elements $a_{ij}$
then likewise for the derivatives
$$
z^{(k)} = A y^{(k)};\qquad k\in \{0,...,n-1\} 
$$
we can combinte all of these equation in one matrix equatiuon by denoting the following square matrices
$$
Z=\begin{pmatrix}z&z^{(1)}&...&z^{(n-1)}\end{pmatrix} \qquad Y=\begin{pmatrix}y&y^{(1)}&...&y^{(n-1)}\end{pmatrix}
$$
in the equation
$$
Z = A Y
$$
taking the transpose
$$
Z^T = \begin{pmatrix}
z_{1} & \cdots & z_{n} \\ 
\vdots  &  & \vdots \\ 
z_{1}^{(n-1)} & \cdots & z_{n}^{(n-1)} 
\end{pmatrix} = \left[\begin{pmatrix}
a_{11} & \cdots & a_{1n} \\ 
\vdots  &  & \vdots \\ 
a_{n1} & \cdots & a_{nn} 
\end{pmatrix}\begin{pmatrix}
f_{1} & \cdots & f_{n} \\ 
\vdots  &  & \vdots \\ 
f_{1}^{(n-1)} & \cdots & f_{n}^{(n-1)} 
\end{pmatrix}\right]^T = Y^TA^T
$$
I think you know that $\det(M) = \det(M^T)$, then
$$
W(z_i) = \begin{vmatrix}
z_{1} & \cdots & z_{n} \\ 
\vdots  &  & \vdots \\ 
z_{1}^{(n-1)} & \cdots & z_{n}^{(n-1)} 
\end{vmatrix} = \det (Z) = \det (Y^T A^T) = \det (A) \det(Y) = \det(A)\ W(y_i)
$$
