Question

The system matrix of a linear time-invariant continuous-time system is

Given by \(A\left[ {\begin{array}{*{20}{c}} 0&1\\ { - 4}&{ - 5} \end{array}} \right]\). What are the roots of the characteristic equation?

1. -1, -4
2. -1, -5
3. -4, -5
4. 0, -1

Answer 1

Correct Answer - Option 1 : -1, -4

Concept:

For the root of the characteristic equation:

[sI - A] = 0

Where

I = Unit matrix

\({\rm{I}} = \left[ {\;\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]\)

A = Given matrix

Explanation:

Given

\(A=\left[ {\begin{array}{*{20}{c}} 0&1\\ { - 4}&{ - 5} \end{array}} \right]\)

For the root of the characteristic equation:

[sI - A] = 0

\({\left[ {\left[ {\begin{array}{*{20}{c}} s&0\\ 0&s \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 0&1\\ -4&-5 \end{array}} \right]} \right]^{ }}=0\)

\(\left[ {\begin{array}{*{20}{c}} s&-1\\ { 4}&{ s+ 5} \end{array}} \right]=0\)

s(s + 5) + 4 = 0

s² + 5s + 4 = 0

(s+4)(s+1) = 0

s = -4, -1