Find $2^A$ with $A$ is a matrix
\begin{align}2^A&=e^{\log(2)A}\\&=\operatorname{Id}+\log(2)A+\frac{\log(2)^2A^2}{2!}+\frac{\log(2)^3A^3}{3!}+\cdots\\&=\operatorname{Id}+\log(2)A+\frac{\log(2)^2\operatorname{Id}}{2!}+\frac{\log(2)^3A}{3!}+\cdots\\&=\left(1+\frac{\log^2(2)}{2!}+\frac{\log^4(2)}{4!}+\cdots\right)\operatorname{Id}+\left(\log(2)+\frac{\log^3(2)}{3!}+\cdots\right)A\end{align}
Hint. One has $$ e^A=I+A+\frac{A^2}{2!}+\frac{A^3}{3!}+\cdots+\frac{A^n}{n!}+\cdots $$ then use that $$ A^2=I,\,A^3=A,\,\cdots. $$
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
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}
$$ \bbx{2^{\textsf{A}} = \pars{\begin{array}{rrr} \ds{1 \over 2} & \ds{-\,{3 \over 2}} & \ds{-\,{3 \over 2}} \\[1mm] \ds{3 \over 4} & \ds{11 \over 4} & \ds{3 \over 4} \\[1mm] \ds{-\,{3 \over 4}} & \ds{-\,{3 \over 4}} & \ds{5 \over 4} \end{array}}} $$