## Number of Components in a Second Order System

We can write second order systems as a matrix. Any second order system can be written
$\mathbf{a}_{ij}$
or as a matrix.
$\left( \begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array} \right)$

In general the
$\mathbf{a}_{ij}$
are not related so there are 3{sup}3{/sup =9 components. If the i,j can run from 1 to
$n$
there are
$n^2$
components. For a symmetric system
$\mathbf{a}_{ij} = a_{ji}$

The matrix above becomes
$\left( \begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{12} & a_{22} & a_{23} \\ a_{13} & a_{23} & a_{33} \end{array} \right)$

There are
$\frac{3(3-1)}{2} +3=6$
components.
If the i,j can run from 1 to
$n$
there are
$\frac{n(n-1)}{2} +n=\frac{n(n+1)}{2}$
components. For a skew symmetric system
$\mathbf{a}_{ij} = -a_{ji}$

All the diagonal elements must be zero.
The matrix above becomes
$\left( \begin{array}{ccc} 0 & a_{12} & a_{13} \\ -a_{12} & 0 & a_{23} \\ -a_{13} & -a_{23} & 0 \end{array} \right)$

There are
$\frac{3(3-1)}{2}=3$
components.
If the i,j can run from 1 to
$n$
there are
$\frac{n(n-1)}{2}$
components.