Exercise 2

According to the rules of matrix multiplication, Eq. 3.14 yields \[ \begin{flalign*} (\sigma_z)_{11} \cdot 1 + (\sigma_z)_{12} \cdot 0 &= 1 \Rightarrow (\sigma_z)_{11} = 1\\ (\sigma_z)_{21} \cdot 1 + (\sigma_z)_{22} \cdot 0 &= 0 \Rightarrow (\sigma_z)_{21} = 0\\ \end{flalign*} \]

And from Eq. 3.15 we get \[ \begin{flalign*} (\sigma_z)_{11} \cdot 0 + (\sigma_z)_{12} \cdot 1 &= 0 \Rightarrow (\sigma_z)_{12} = 0\\ (\sigma_z)_{21} \cdot 0 + (\sigma_z)_{22} \cdot 1 &= -1 \Rightarrow (\sigma_z)_{22} = -1\\ \end{flalign*} \]