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 $S^T P^{10} S$.

$\begin{bmatrix}1&0\\10&1\end{bmatrix}$

$S=\begin{bmatrix}\frac{\sqrt{3}-1}{2\sqrt{2}}&\frac{\sqrt{3}+1}{2\sqrt{2}}\\-(\frac{\sqrt{3}+1}{2\sqrt{2}})&\frac{\sqrt{3}-1}{2\sqrt{2}}\end{bmatrix}$

$=\begin{bmatrix}\sin 15°&\cos 15°\\-\cos 15°&\sin 15°\end{bmatrix}$

$∴SS^T = S^T S=I$

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.

$∴(adj. A)^{10}=\begin{bmatrix}1&0\\10&1\end{bmatrix}=S^T P^{10} S$