If $S=\begin{bmatrix}\frac{\sqrt{3}-1}{2\sqrt{2}}&\frac{\sqrt{3}+1}{2\sqrt{2}}\\-(\frac{\sqrt{3}+1}{2\sqrt{2}})&\frac{\sqrt{3}-1}{2\sqrt{2}}\end{bmatrix},A=\begin{bmatrix}1&0\\-1&1\end{bmatrix}$ and $P=S (adj.A) S^T$, then find matrix $S^T P^{10} S$.
Answer & explanation
Correct answer: option 2
$S=\begin{bmatrix}\frac{\sqrt{3}-1}{2\sqrt{2}}&\frac{\sqrt{3}+1}{2\sqrt{2}}\\-(\frac{\sqrt{3}+1}{2\sqrt{2}})&\frac{\sqrt{3}-1}{2\sqrt{2}}\end{bmatrix}$
$=\begin{bmatrix}\sin 15°&\cos 15°\\-\cos 15°&\sin 15°\end{bmatrix}$
$∴SS^T = S^T S=I$
Now,
$S^T P^{10} S = S^T (S (adj. A) S^T)^{10}S$
$=S^TS (adj. A) S^T(S (adj. A) S^T)^9S$
$=I (adj. A)S^T(S (adj. A) S^T)^9S$
$= (adj. A)S^TS (adj. A) S^T(S (adj. A) S^T)^8S$
$= (adj. A)^2 S^T(S (adj. A) S^T)^8S$
....
....
$= (adj. A)^{10}$
$A=\begin{bmatrix}1&0\\-1&1\end{bmatrix}$
$∴adj. A=\begin{bmatrix}1&0\\1&1\end{bmatrix}$
$∴(adj. A)^2=\begin{bmatrix}1&0\\1&1\end{bmatrix}\begin{bmatrix}1&0\\1&1\end{bmatrix}=\begin{bmatrix}1&0\\2&1\end{bmatrix}$
$(adj. A)^2=\begin{bmatrix}1&0\\3&1\end{bmatrix}$
and so on.
$∴(adj. A)^{10}=\begin{bmatrix}1&0\\10&1\end{bmatrix}=S^T P^{10} S$