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Finding the order of $2 \times 2 $ matrix.

$$\begin{pmatrix}1&-1\\1&0\end{pmatrix}$$

Is there an easier way to find the order of this matrix? I have been multiplying the matrix for five times or so, and I still haven't found the order yet.

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    $\begingroup$ You miss one time :): $A^6 = I_2$ $\endgroup$ – Surb Mar 7 '16 at 20:20
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    $\begingroup$ What is the cube of it, exactly? $\endgroup$ – Will Jagy Mar 7 '16 at 20:20
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Not that long:

$$A^2=\begin{pmatrix}1&\!-1\\1&0\end{pmatrix}\begin{pmatrix}1&\!-1\\1&0\end{pmatrix}=\begin{pmatrix}0&\!-1\\1&\!-1\end{pmatrix}$$

$$A^4=\begin{pmatrix}0&\!-1\\1&\!-1\end{pmatrix}\begin{pmatrix}0&\!-1\\1&\!-1\end{pmatrix}=\begin{pmatrix}-1&1\\\!-1&0\end{pmatrix}$$

$$A^6=\begin{pmatrix}0&\!-1\\1&\!-1\end{pmatrix}\begin{pmatrix}-1&1\\\!-1&0\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}$$

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Since $A^3=-I$ for this matrix $A$ we know that $A^6=I$. But then the order of $A$ must divide $6$, so can be $1,2,3,6$. Obviously it is not $1,2,3$, so it must be $6$.

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Once you notice $$A^3=\begin{pmatrix}1&\!-1\\1&0\end{pmatrix}\begin{pmatrix}1&\!-1\\1&0\end{pmatrix}\begin{pmatrix}1&\!-1\\1&0\end{pmatrix}=\begin{pmatrix}-1&\!0\\0&\!-1\end{pmatrix}$$ It is quite obvious that $$A^6=I$$

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You can also do this using the Cayley-Hamilton theorem which states that every matrix must satisfy it's own characteristic polynomial if you do not like matrix multiplication:

The characteristic polynomial of $A=\begin{pmatrix} 1&-1\\1& 0 \end{pmatrix}$ is $x^2-x+1$. So $A^2=A-1$, clearly not the identity matrix. Using this relation,

$$A^3=A(A^2)=A(A-1)=A^2-A =-1.$$

So, $A^6=(A^3)^2=(-1)^2=1$. Thus the order must divide $6$ and we already know that $A^2$ and $A^3$ are not $1$ so the order is 6.

(Unless the characteristic of your field is $2$, in which case $3$ is the order because in this world $-1=1$).

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