Show that the $5\times 5$ matrix is not invertable 

Note that 
$$det\begin{pmatrix} 0 & a& 0 & 0 & 0 \\ b & 0 & c & 0 & 0 \\ 0 & d & 0 & e & 0 \\ 0 & 0 & f & 0 & g \\ 0 & 0 & 0 & h & 0 \end{pmatrix} $$
$$= (-1)^3b \times det\begin{pmatrix} a& 0 & 0 & 0  \\ d & 0 & e & 0 \\ 0 & f & 0 & g \\ 0 & 0 & h & 0 \end{pmatrix}        \text{(expanding on the first column})$$
$$= (-1)^3 b \times (-1)^{7}h  det\begin{pmatrix} a& 0 & 0 \\ d & 0 & 0 \\ 0 & f & g  \end{pmatrix}      \text{(expanding on the last row})$$
$$= b h \times det\begin{pmatrix} a& 0 & 0 \\ d & 0 & 0 \\ 0 & f & g \end{pmatrix}       \text{(using determinant formula for a lower diogonal matrix})$$
$$= b h  a \times 0 \times g=0.$$
Note that the determinant of a lower diogonal matrix is the product of the diogonal terms. We have used this fact in the last step above. 

Since the determinant is zero, the matrix is not invertible.