<s>[INST] Calculate the determinant of this $5 \times 5$ matrix 
Calculate the determinant of the matrix $$A=\begin{pmatrix} \sin\alpha
  &  \cos\alpha   & a\sin\alpha     & b\cos\alpha    &  ab     \\ 
  -\cos\alpha   &  \sin\alpha   & -a^2\sin\alpha  & b^2\cos\alpha  &  a^2b^2 \\  0             
  &  0            & 1               & a^2      
  &  b^2    \\  0             &  0            & 0               & a     
  &  b      \\  0             &  0            & 0               & -b    
  &  a \end{pmatrix} \text{ with  } (\alpha,a,b \in \mathbb{R})$$

I have trouble solving the determinant.. But what is immediately visible are those zeroes in the matrix, just one more zero is needed such that this matrix is a triangular matrix (the element $a_{54}$ must be zero for this but it is $-b$ instead). If it was zero we could just multiply the diagonal and the product would be our determinant.
I have tried various ways to form this matrix such that $a_{54}$ is zero but the way I formed harmed the matrix and I got a wrong determinant as solution : /
As example, I have multiplied row $4$ with $b$, multiply row $5$ with $a$ and then add row $4$ to row $5$. Because I multiplied row $5$ with $a$, I need to divide the determinant by $a$ at the end.
So then I have the matrix
$$\begin{pmatrix}
\sin\alpha & \cos\alpha & a\sin\alpha & b\cos\alpha & ab\\ 
-\cos\alpha & \sin\alpha & -a^2\sin\alpha & b^2\cos\alpha & a^2b^2\\ 
0 & 0 & 1 & a^2 & b^2\\ 
0 & 0 & 0 & a & b\\ 
0 & 0 & 0 & 0 & a^2+b^2
\end{pmatrix}$$
$$\text{Thus }\det = \frac{\sin\alpha \cdot \sin\alpha \cdot 1 \cdot a \cdot (a^2+b^2)}{a}=\sin^2\alpha \cdot (a^2+b^2)$$
But this is wrong and I don't see how to get the correct determinant...?
 [/INST] Using determinant of block matrix
$$\det\begin{pmatrix}B&C\\ 0& D\end{pmatrix}=\det(B)\det(D)$$
we get that the desired determinant is
$$\det\begin{pmatrix}\sin\alpha&\cos\alpha\\ -\cos\alpha& \sin\alpha\end{pmatrix}\det(1)\det\begin{pmatrix}a&b\\ -b& a\end{pmatrix}=a^2+b^2$$
</s>