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}
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 $$ W(z_i) = \begin{vmatrix} z_{1} & \cdots & z_{n} \\ \vdots & & \vdots \\ z_{1}^{(n-1)} & \cdots & z_{n}^{(n-1)} \end{vmatrix} = \det (Z) = \det (Y^T A^T) = \det (A) \det(Y) = \det(A)\ W(y_i) $$