# Find a 2x2 matrix, not the identity, s.t. A^4 = I

As stated. I'm not sure where to start. I've been looking at matrices if the determinant = 1,-1, but that has been clearly the wrong way to go about this. Any tips?

• Try a rotation matrix – Dark Malthorp Apr 2 at 23:31
• make a matrix that means a rotation of $90^\circ$ in the plane – Will Jagy Apr 2 at 23:31
• Well, you're looking in the right place -- the matrix you seek must have determinant $\pm 1$. But try thinking of matrices as representing transformations --- is there some operation you can do four times and have it be the same as doing nothing? – John Hughes Apr 2 at 23:31
• OH wow. Thanks a ton. Sometimes you get so caught up in the semantics and equations, you forget to just visualize it sometimes. :) – Steve Apr 3 at 0:03

Consider the diagonal matrix $$D= \left[\begin{array}{rr} 1 & 0 \\ 0 & -1 \end{array}\right]$$ Note that $$D^4=I$$.
Now, if $$P$$ is invertible, then $$A=PDP^{-1}$$ satisfies $$A^4=PD^4P^{-1}=PIP^{-1}=PP^{-1}=I$$ For instance, taking $$P=\left[\begin{array}{rr} 1 & 5 \\ 2 & 11 \end{array}\right]$$ gives $$A =\overset{P}{\left[\begin{array}{rr} 1 & 5 \\ 2 & 11 \end{array}\right]}\overset{D}{\left[\begin{array}{rr} 1 & 0 \\ 0 & -1 \end{array}\right]}\overset{P^{-1}}{\left[\begin{array}{rr} 11 & -5 \\ -2 & 1 \end{array}\right]}=\left[\begin{array}{rr} 21 & -10 \\ 44 & -21 \end{array}\right]$$