Question

True or False? If the trace and the determinant of a $\mathbf{3}\mathbf{\times }\mathbf{3}$matrix A are both negative, then the origin is a stable equilibrium solution of the system$\frac{\mathbf{d}\overrightarrow{\mathbf{x}}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{A}\overrightarrow{\mathbf{x}}$. Justify your answer.

Short answer

The statement is false.

Step-by-step solution

Given

$\frac{d\overrightarrow{x}}{dt}=A\overrightarrow{x}$

Explanation of the solution

Consider the given statement as follows.

If the trace and the determinant of a $3\times 3$matrix$A$ are both negative, then the origin is stable equilibrium solution of the system as follows.

$\frac{d\overrightarrow{x}}{dt}=A\overrightarrow{x}$

It is given that the system is$\frac{d\overrightarrow{x}}{dt}=A\overrightarrow{x}$ stable if and only if the trace is less than 0 and the determinant is greater than zero.

Since, the system is stable and as follows.

$trA<0$and $dtA>0$.

Hence, the given statement is false.