Find the matrix $A^{15}$. Let $I=
\begin{pmatrix}
1 & 0 \\
0 & 1 \\
\end{pmatrix}$ and  $O=\begin{pmatrix}
0 & 0 \\
0 & 0 \\
\end{pmatrix}$.

1.Let $A=\begin{pmatrix}
1 & 3 \\
3 & 5 \\
\end{pmatrix}$ and  $ B=\begin{pmatrix}
x & 3 \\
3 & 6 \\ 
\end{pmatrix}$. Find the value of $x$ which satisfies $AB=BA$.
$AB=\begin{pmatrix}
1 & 3 \\
3 & 5 \\
\end{pmatrix}\cdot\begin{pmatrix}
x & 3 \\
3 & 6 \\
\end{pmatrix}=\begin{pmatrix}
x+9 & 21 \\
3x+15 & 39 \\
\end{pmatrix}$
$BA=\begin{pmatrix}
x & 3 \\
3 & 6 \\
\end{pmatrix}\cdot\begin{pmatrix}
1 & 3 \\
3 & 5 \\
\end{pmatrix}=\begin{pmatrix}
x+9 & 3x+15 \\
21 & 39 \\
\end{pmatrix}$
.So that we get $3x+15=21 \Rightarrow x=2$

2.Let $A=\begin{pmatrix}
1 & 2 \\
2 & 4 \\
\end{pmatrix}$ and  $ B=\begin{pmatrix}
-2 & x \\
4 & y \\
\end{pmatrix}$.Find the values of x and y which satify $BA=O$.
$BA=\begin{pmatrix}
-2 & x \\
4 & y \\
\end{pmatrix}\cdot\begin{pmatrix}
1 & 2 \\
2 & 4 \\
\end{pmatrix}=\begin{pmatrix}
-2+2x & 0 \\
4+2y & 8+4y \\
\end{pmatrix}=\begin{pmatrix}
0 & 0 \\
0 & 0 \\
\end{pmatrix}$
S0 that we can get $x=1,y=-2$.

3.Let $A$ satisfying $A^2=A-I$. Find $A^{15}$.
Please help to show me about this. Thank you in advance!

You may proceed as follows:
From the given condition we have
$$A-A^2 = A(I-A) =I$$
It follows
\begin{eqnarray} A^{15}
& = & A(A^2)^7 \\
& = & A(A-I)^7 \\
& = & -(A-I)^6 \\
& = & -(A^2-2A + I)^3 \\
& = & -(-A)^3 \\
& = & A(A-I) \\
& = & -I
\end{eqnarray}
Another way would be seeing
$$A^3+I = (A+I)(A^2-A+I)=O \Rightarrow A^3 = -I$$
