Given that equilibrum point $x_e=0$ and a dynamical system:
$\dot x(t)=\begin{bmatrix} a(t) & 1\\ 0 & -1\\\end{bmatrix}x(t)$
for each $a(t)$ below, determine if it is uniformly exponentially stable
a) $a(t)=0$
b) $a(t)=-1$
Please help me with this question, I know the theorem behind this which is that if there exist constants $C$, $\lambda >0$ and $r\gt 0$ such that $||x(t)||\le Ce^{-\lambda(t-t_o)}||x(t_o)|| $ for all $t\ge t_o$ and $||x(t_o)||\le r$. But I don't how to use this in a real problem. Also this the $\lambda$ denote eigenvalue of the A matrix?
First you should know that the answer of the linear dynamic system: $$ x'=Ax , x ∈R^n $$ can be achieved by calculating the following value: $$x=e^(At)x_0=L^-1{(s*I-A)^-1}x_0$$ In which L denotes the Laplace transform.If you do that for $$a(t)=0$$ you have: $$x(t)=\begin{matrix} [1 & 1-e^(-t)\\ 0 & e^(-t)]\\ \end{matrix}*x_0 $$ As you see the answer is not asymptotic stable so it can not be exponentionally stable!!!( Take care about $$ A(1,1)$$ which is constant).For the second part calculate the answer in this way.Try to find the lambda and C constant according to definition. Lambda is not the matrix eigenvalue. For finding them use the following inequality: $$ ||[a\ b]^T||< 2 max(|a|,|b|)$$