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

Given by $A [\begin{matrix} 0 & 1 \\ - 4 & - 5 \end{matrix}]$ . What are the roots of the characteristic equation?

Question

Question

This question was previously asked in

ESE Electrical 2013 Paper 1: Official Paper

Answer 1

Concept:

For the root of the characteristic equation:

[sI - A] = 0

Where

I = Unit matrix

$I = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$

A = Given matrix

Explanation:

Given

$A = [\begin{matrix} 0 & 1 \\ - 4 & - 5 \end{matrix}]$

For the root of the characteristic equation:

[sI - A] = 0

${[[\begin{matrix} s & 0 \\ 0 & s \end{matrix}] - [\begin{matrix} 0 & 1 \\ - 4 & - 5 \end{matrix}]]}^{} = 0$

$[\begin{matrix} s & - 1 \\ 4 & s + 5 \end{matrix}] = 0$

s(s + 5) + 4 = 0

s²+ 5s + 4 = 0

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

s = -4, -1

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