= Solution
The <Pauli matrix multiplication law> is $\sigma_i\sigma_j=\delta_{ij}I_2+i\epsilon_{ijk}\sigma_k$, so $\{\sigma_i,\sigma_j\}=2\delta_{ij}I_2$. Block multiplication of the <chiral gamma-matrix representation> gives
$$
(\gamma^0)^2=I_4,\qquad \gamma^0\gamma^i=\begin{pmatrix}-\sigma_i&0\\0&\sigma_i\end{pmatrix},\qquad
\gamma^i\gamma^0=\begin{pmatrix}\sigma_i&0\\0&-\sigma_i\end{pmatrix},
$$
and
$$
\gamma^i\gamma^j=\begin{pmatrix}-\sigma_i\sigma_j&0\\0&-\sigma_i\sigma_j\end{pmatrix}.
$$
Thus the mixed anticommutators vanish and the spatial ones are $-2\delta_{ij}I_4$. With the <Minkowski metric> convention $\eta=\operatorname{diag}(1,-1,-1,-1)$,
$$
\boxed{\{\gamma^\mu,\gamma^\nu\}=2\eta^{\mu\nu}I_4.}
$$
These matrices therefore represent the required <Clifford algebra>.
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