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…?
Best Answer
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
$$\det\begin{pmatrix}\sin\alpha&\cos\alpha\\ -\cos\alpha& \sin\alpha\end{pmatrix}\det(1)\det\begin{pmatrix}a&b\\ -b& a\end{pmatrix}=a^2+b^2$$