Q.Answer both

(a) verified using $\cos\alpha=\tfrac{1-t^2}{1+t^2},\sin\alpha=\tfrac{2t}{1+t^2}$ with $t=\tan\tfrac\alpha2$; (b) the determinant is $0$, so $x=y=0$. (OR: $x=2,y=1,z=3$.)

Main (a). Let $t=\tan\tfrac{\alpha}{2}$, so $A=\begin{bmatrix}0&-t\\t&0\end{bmatrix}$, $I-A=\begin{bmatrix}1&t\\-t&1\end{bmatrix}$, and $I+A=\begin{bmatrix}1&-t\\t&1\end{bmatrix}$. Use the half-angle forms $\cos\alpha=\dfrac{1-t^{2}}{1+t^{2}}$, $\sin\alpha=\dfrac{2t}{1+t^{2}}$. Compute

$(I-A)\begin{bmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{bmatrix}=\begin{bmatrix}\cos\alpha+t\sin\alpha&-\sin\alpha+t\cos\alpha\\-t\cos\alpha+\sin\alpha&t\sin\alpha+\cos\alpha\end{bmatrix}.$

Entry $(1,1)$: $\dfrac{1-t^{2}}{1+t^{2}}+t\cdot\dfrac{2t}{1+t^{2}}=\dfrac{1+t^{2}}{1+t^{2}}=1$. Entry $(1,2)$: $-\dfrac{2t}{1+t^{2}}+t\cdot\dfrac{1-t^{2}}{1+t^{2}}=\dfrac{-t-t^{3}}{1+t^{2}}=-t$. Entry $(2,1)$: $-t\cdot\dfrac{1-t^{2}}{1+t^{2}}+\dfrac{2t}{1+t^{2}}=\dfrac{t+t^{3}}{1+t^{2}}=t$. Entry $(2,2)$: $t\cdot\dfrac{2t}{1+t^{2}}+\dfrac{1-t^{2}}{1+t^{2}}=1$. This is exactly $\begin{bmatrix}1&-t\\t&1\end{bmatrix}=I+A$. Proved.

Main (b). Expand along the first row:

$\begin{vmatrix}6i&-3i&1\\4&3i&-1\\20&3&i\end{vmatrix}=6i\begin{vmatrix}3i&-1\\3&i\end{vmatrix}+3i\begin{vmatrix}4&-1\\20&i\end{vmatrix}+1\begin{vmatrix}4&3i\\20&3\end{vmatrix}.$ …