Home
Random
Log in
Settings
Donate
About Wikiversity
Disclaimers
Search
Finite field/Invertible matrix/Z mod 3/0120/Order/Exercise/Solution
Language
Watch
Edit
<
Finite field/Invertible matrix/Z mod 3/0120/Order/Exercise
We have
(
0
1
2
0
)
2
=
(
0
1
2
0
)
(
0
1
2
0
)
=
(
2
0
0
2
)
,
{\displaystyle {}{\begin{pmatrix}0&1\\2&0\end{pmatrix}}^{2}={\begin{pmatrix}0&1\\2&0\end{pmatrix}}{\begin{pmatrix}0&1\\2&0\end{pmatrix}}={\begin{pmatrix}2&0\\0&2\end{pmatrix}}\,,}
(
0
1
2
0
)
3
=
(
2
0
0
2
)
(
0
1
2
0
)
=
(
0
2
1
0
)
,
{\displaystyle {}{\begin{pmatrix}0&1\\2&0\end{pmatrix}}^{3}={\begin{pmatrix}2&0\\0&2\end{pmatrix}}{\begin{pmatrix}0&1\\2&0\end{pmatrix}}={\begin{pmatrix}0&2\\1&0\end{pmatrix}}\,,}
and
(
0
1
2
0
)
4
=
(
2
0
0
2
)
(
2
0
0
2
)
=
(
1
0
0
1
)
,
{\displaystyle {}{\begin{pmatrix}0&1\\2&0\end{pmatrix}}^{4}={\begin{pmatrix}2&0\\0&2\end{pmatrix}}{\begin{pmatrix}2&0\\0&2\end{pmatrix}}={\begin{pmatrix}1&0\\0&1\end{pmatrix}}\,,}
therefore, the order is
4
{\displaystyle {}4}
.
To the exercise