TITLE: Find $2^A$ with $A$ is a matrix
QUESTION [6 upvotes]: Let
$$A=\begin{bmatrix} -1 &-2 &-2 \\1&2&1  \\-1&-1&0  \end{bmatrix}$$
How to find $2^A$ ?
I find out that $A^2=I$ so it would be simple if they ask me how to find a power of $A$, but not. So could you help me?

REPLY [2 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Note that $\ds{\textsf{A}^{1} = \textsf{A}\,,\ \textsf{A}^{2} = \textsf{1}  \implies
\expo{x\textsf{A}} = \mrm{p}\pars{x}\textsf{1} + \mrm{q}\pars{x}\textsf{A}}$
  
  with $\ds{\mrm{p}\pars{0} = 1\,,\mrm{q}\pars{0} = 0}$.

\begin{align}
&\left.\begin{array}{rcl}
\ds{A\expo{x\textsf{A}}} & \ds{=} &
\ds{\mrm{p}\pars{x}'\textsf{1} + \mrm{q}'\pars{x}\textsf{A}}
\\
\ds{A\expo{x\textsf{A}}} & \ds{=} &
\ds{\mrm{p}\pars{x}\textsf{A} + \mrm{q}\pars{x}\textsf{1}}
\end{array}\right\}
\implies
\left\{\begin{array}{rcl}
\ds{\mrm{p}'\pars{x}} & \ds{=} & \ds{\mrm{q}\pars{x}}
\\
\ds{\mrm{q}'\pars{x}} & \ds{=} & \ds{\mrm{p}\pars{x}}
\end{array}\right.
\\[5mm] &\ \implies
\mrm{p}'\pars{x} + \mrm{q}'\pars{x} =
\mrm{p}\pars{x} + \mrm{q}\pars{x}\,,\quad
\mrm{p}'\pars{x} - \mrm{q}'\pars{x} =
-\bracks{\mrm{p}\pars{x} - \mrm{q}\pars{x}}
\\[5mm] &\ \implies
\mrm{p}\pars{x} + \mrm{q}\pars{x} = \expo{x}\,,\qquad\qquad\quad\,\,\,\,\,
\mrm{p}\pars{x} - \mrm{q}\pars{x} = {1 \over \expo{x}}
\\[5mm] &
\implies
\mrm{p}\pars{\ln\pars{2}} + \mrm{q}\pars{\ln\pars{2}}= 2\,,\qquad\quad\,\,
\mrm{p}\pars{\ln\pars{2}} - \mrm{q}\pars{\ln\pars{2}}= {1 \over 2}
\\[5mm] &
\implies
\mrm{p}\pars{\ln\pars{2}} = {5 \over 4}\,,\quad
\mrm{q}\pars{\ln\pars{2}} = {3 \over 4}
\implies
\bbx{2^{\textsf{A}} = {5 \over 4}\,\textsf{1} + {3 \over 4}\textsf{A}}
\end{align}

$$
\bbx{2^{\textsf{A}} =
\pars{\begin{array}{rrr}
\ds{1 \over 2} & \ds{-\,{3 \over 2}} & \ds{-\,{3 \over 2}}
\\[1mm]
\ds{3 \over 4} & \ds{11 \over 4} & \ds{3 \over 4}
\\[1mm]
\ds{-\,{3 \over 4}} & \ds{-\,{3 \over 4}} & \ds{5 \over 4}
\end{array}}}
$$