\[\begin{array}{l} {\rm{We \,\,have, }}\,\,A = \left[ {\begin{array}{*{20}{c}} 0&{ - 1}\\ 1&0 \end{array}} \right]\\ {\rm{Now, }}\,\,{A^2} = A.A = \left( {\begin{array}{*{20}{c}} 0&{ - 1}\\ 1&0 \end{array}} \right)\left( {\begin{array}{*{20}{c}} 0&{ - 1}\\ 1&0 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { - 1}&0\\ 0&{ - 1} \end{array}} \right) = - I\\ {\rm{where }}\,\,I = \left( {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right){\rm{ is\,\, identity\,\, matrix}} \end{array}\]
$$\eqalign{ & {\left( {{A^2}} \right)^8} = {\left( { - I} \right)^8} = I. \cr & {\text{Hence, }}{A^{16}} = I \cr} $$

The given equations are
$$\eqalign{ & - x + cy + bz = 0 \cr & cx - y + az = 0 \cr & bx + ay - z = 0 \cr} $$
$$\because \,\,x,y,z$$   are not all zero
∴ The above system should not have unique (zero) solution
$$ \Rightarrow \,\,\Delta = 0$$
\[ \Rightarrow \,\left| \begin{array}{l} - 1\,\,\,\,\,\,\,\,c\,\,\,\,\,\,\,\,\,b\\ \,\,c\,\,\,\,\,\, - 1\,\,\,\,\,\,\,\,a\\ \,\,b\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\, - 1 \end{array} \right| = 0\]
$$\eqalign{ & \Rightarrow \,\, - 1\left( {1 - {a^2}} \right) - c\left( { - c - ab} \right) + b\left( {ac + b} \right) = 0 \cr & \Rightarrow \,\, - 1 + {a^2} + {b^2} + {c^2} + 2abc = 0 \cr & \Rightarrow \,\,{a^2} + {b^2} + {c^2} + 2abc = 1 \cr} $$