Show that these matrices are anticommuative. I am told to show that these matrices are anticommutative (AB = -BA)$$
\beta=\begin{bmatrix}I_2&0\\0&-I_2\end{bmatrix}, \alpha_x=\begin{bmatrix}0&\sigma_1\\\sigma_1&0\end{bmatrix},\alpha_y=\begin{bmatrix}0&\sigma_2\\\sigma_2&0\end{bmatrix}, \alpha_z=\begin{bmatrix}0&\sigma_3\\\sigma_3&0\end{bmatrix}$$ where $\sigma_1=\begin{bmatrix}0&1\\1&0\end{bmatrix}, \sigma_2=\begin{bmatrix}0&-i\\i&0\end{bmatrix}, \sigma_3=\begin{bmatrix}1&0\\0&-1\end{bmatrix}$ (the pauli spin matrices). I could just use brute force and plug every matrix in for B and A, but my professor is very hard on us when it comes to proofs and only gives full credit for "elegant" proofs. The only problem is, I have no idea how I would even start to prove this; any help or hints would be greatly appreciated!!!
 A: Just observe that for any (2D)matrices $\rm A,B$ (and zero matrix $\rm O$):
$$\begin{align}\begin{pmatrix}\rm A & \rm O \\ \rm O & -\rm A\end{pmatrix}\begin{pmatrix}\rm O & \rm B \\ \rm B & \rm O\end{pmatrix}~=~&\begin{pmatrix}\rm O & \rm AB \\ \rm -AB & \rm O\end{pmatrix}\\[1ex] =~&-\begin{pmatrix}\rm O & \rm -AB \\ \rm AB & \rm O\end{pmatrix}\\[1ex] =~& -\begin{pmatrix}\rm O & \rm B \\ \rm B & \rm O\end{pmatrix}\begin{pmatrix}\rm A & \rm O \\ \rm O & \rm -A\end{pmatrix}\end{align}$$
So if $\rm A=I_2$ and $\rm B\in\{\sigma_1, \sigma_2, \sigma_3\}$ you've proven three pairs are anti-commutative.
The rest follows from showing $\sigma_1, \sigma_2, \sigma_3$ are themselves anticommuative and showing if $\rm B,C$ are then $(\begin{smallmatrix}\rm O & \rm B \\ \rm B & \rm O\end{smallmatrix})$ and $(\begin{smallmatrix}\rm O & \rm C \\ \rm C & \rm O\end{smallmatrix})$ must be too.

PS: $\sigma_2, \sigma_3$ are not.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\left.\alpha_{i}\alpha_{j}\right\vert_{\ i\ \not=\ j} & =
\pars{\begin{array}{cc}
\ds{0} & \ds{\sigma_{i}}
\\
\ds{\sigma_{i}} & \ds{0}
\end{array}}
\pars{\begin{array}{cc}
\ds{0} & \ds{\sigma_{j}}
\\
\ds{\sigma_{j}} & \ds{0}
\end{array}}
=
\pars{\begin{array}{cc}
\ds{\sigma_{i}\sigma_{j}} & \ds{0}
\\
\ds{0} & \ds{\sigma_{i}\sigma_{j}}
\end{array}}
=
\pars{\begin{array}{cc}
\ds{-\sigma_{j}\sigma_{i}} & \ds{0}
\\
\ds{0} & \ds{-\sigma_{j}\sigma_{i}}
\end{array}}
=
\bbx{\ds{-\alpha_{j}\alpha_{i}}}
\end{align}
