determinant of matrix in polynomial Suppose $X=\sqrt{k^2 - k_1^2}, Y= \sqrt{k^2 - k_2^2},Z=\sqrt{k^2 - k_3^2}, S=\sqrt{k^2 - k_4^2}, T= \sqrt{k^2 - k_5^2}$. \
And there is $5 \times 5 $ matrix as given below.
\begin{equation}
\begin{vmatrix}
a_{11} & a_{12}  & a_{13} & a_{14}*S & a_{15}*T\\
a_{21}*X & a_{22}*Y  & a_{23}*Z & a_{24} & a_{25}\\
a_{31}*X & a_{32}*Y  & a_{33}*Z & 0 & 0\\
a_{41} & a_{42}  & a_{43} & a_{44}*S & a_{45}*T\\
a_{51}*X & a_{52}*Y  & a_{53}*Z & a_{54}& a_{55}\\
\end{vmatrix} = 0,
\end{equation}
I have to find the determinant in terms of k with inter powers. By usual exapnsion method (squaring the terms), finding the expression for k is getting complex. Is there any other approach to find the expression for above determinant. And that expression should not contain fraction powers of k, means expression should have $X^2, Y^2,Z^2,S^2,T^2$ instead of $X, Y,Z,S,T$.

Here is a special case of your matrix
$$
\begin{vmatrix}
1 & 0  & 0 & 0 & 0\\
0& 0  & 0 & 1 & 0\\
0 & 0  &  Z & 0 & 0\\
0 & 1  & 0 & 0 & 0\\
0 & 0  & 0 & 0& 1\\
\end{vmatrix}.
$$
It evaluates to $Z=\sqrt{k^2 - k_3^2}$. (Or is it $-Z$? No matter.) You can't achieve what you are asking for.
