24. Let
\begin{equation*}
L^+ = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}, \quad L^- = \begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix},
\end{equation*}
\begin{equation*}
|-1\rangle = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}, \quad |0\rangle = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}, \quad |1\rangle = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \quad |\text{null}\rangle = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}.
\end{equation*}
Show that
\begin{align*}
L^+ |-1\rangle &= |0\rangle, & L^+ |0\rangle &= |1\rangle, & L^+ |1\rangle &= |\text{null}\rangle, \\
L^- |1\rangle &= |0\rangle, & L^- |0\rangle &= |-1\rangle, & L^- |-1\rangle &= |\text{null}\rangle.
\end{align*}
