I have a question that goes as follows:
Let $\alpha = \frac{\pi}{5}$ and $A= \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \\ \end{bmatrix} $ and let $B=A^4-A^3+A^2-A$
Then which of the following is true? (i)B is singular (ii) B is non singular (iii) B is symmetric (iv) The determinant $|B| =1$
There may be multiple correct options.
Now here's what I did, I evaluated $A^2$ and $A^3$ and then it is pretty easy to conclude by induction that $$A^n = \begin{bmatrix} \cos n\alpha & \sin n\alpha \\ -\sin n\alpha & \cos n\alpha \\ \end{bmatrix}$$
From this, I tried to find out $B$ by putting the values of $A^4, A^3, A^2, A$ and then applied the transformation formulae $\bigg($like $\sin C + \sin D=$ $2\sin\bigg( \frac{C+D}{2} \bigg)\cos\bigg( \frac{C-D}{2}$ $\bigg)$, $...$ etc$\bigg)$
By simplifying, and using the fact that $\alpha = \frac{\pi}{5}$ I got the matrix B as follows:
$$B = \begin{bmatrix} -4\cos \alpha \sin \frac{\alpha}{2} & 0 \\ 0 & -4\cos \alpha \sin \frac{\alpha}{2} \\ \end{bmatrix}$$
So, clearly, the matrix is symmetric, which makes (iii) true. But, how do I check whether the matrix is non-singular or not, and how do I check whether $|B|=1$? Using the manipulations that I performed, I seem to not be able to check those options, any help would be nice.
The answer is given as (ii), (iii), (iv) are true.
One way to think of this is that you are essentially computing a geometric sum associated to $-A$. As usual, we have that $(-A - 1)B = -A^5 + A = 1 + A$. Since $-A - 1$ is non-singular (as you can compute by finding the determinant, by computing its eigenvalues, or just by explicitly writing down the inverse of a $2\times2$ matrix), we see that $B = -1$. Sure enough, this is symmetric and has determinant $1$.
It may help in this case to think that we are essentially computing with complex numbers; the ring of complex numbers $\mathbb C$ embeds in the ring $\operatorname M_{2\times2}(\mathbb R)$ of $2\times2$ real matrices by $a + b i \mapsto \begin{pmatrix} a & b \\ -b & a \end{pmatrix}$ for all $a, b \in \mathbb R$. In particular, you are really computing a sum inside $\mathbb C$ involving $z = \cos(\alpha) + i\sin(\alpha) = e^{i\alpha}$. In this context, the determinant on $\operatorname M_{2\times2}(\mathbb R)$ pulls back to the squared-norm on $\mathbb C$, in the sense that $$ \det \begin{pmatrix} a & b \\ -b & a \end{pmatrix} = a^2 + b^2 = \lvert a + b i\rvert^2 $$ for all $a, b \in \mathbb R$; and the transpose on $\operatorname M_{2\times 2}(\mathbb R)$ pulls back to complex conjugation on $\mathbb C$, in the sense that $$ a + b i \mapsto \begin{pmatrix} a & b \\ -b & a \end{pmatrix} \quad\text{and}\quad \overline{a + b i} = a - b i \mapsto \begin{pmatrix} a & -b \\ b & a \end{pmatrix} = \begin{pmatrix} a & b \\ -b & a \end{pmatrix}^{\mathsf T} $$ for all $a, b \in \mathbb R$ (and remember that $\overline{e^{i\theta}}$ equals $e^{-i\theta}$ for all $\theta \in \mathbb R$); so asking whether the alternating sum of matrices is symmetric, respectively has determinant $1$, is the same as asking whether the corresponding alternating sum of complex numbers is real, respectively lies on the unit circle centred at the origin. Finally, since $\mathbb C$ is a field, every non-$0$ matrix in the image of $\mathbb C$ is invertible (with the inverse of the image of $z = a + b i$ being the image of $z^{-1} = \frac a{a^2 + b^2} - \frac b{a^2 + b^2}i$ for all $(a, b) \in \mathbb R^2 \setminus \{(0, 0)\}$); this is yet another way to see that $-A - I_2$ is non-singular.