Suppose that $A(t)$ is a variable matrix with time... I'm writing to this group hoping to find help in the subsequent problem, please, I would appreciate a detailed explanation with which I can understand the solution of the exercise.

Suppose that $A(t)$ is a variable matrix with time, invertible for all $t$. Express $A(t)^{-1}$ in terms of $A(t)^{k}$ with $k=0,1,2 ... n-1$.
  Prove that if $\| A(t) \| ≤ \alpha <∞ $ for all $t$, then there exists a finite constant such that $\| A(t)^{-1} \| ≤ \beta$ for all $t$

Many thanks.
 A: Let $K\in \{\mathbb{R}, \mathbb{C} \}$. Let $A(t)\in GL(n, K)$. Recall that the characteristic polynomial associated to $A(t)$ is defined as
$$ P_{A(t)}(\lambda)= det(\lambda E_n - A(t)),$$
where $E_n\in GL(n, K)$ is the diagonal matrix with ones on the diagonal and zeroes else. When we write
$$ P_{A(t)}(\lambda)= \lambda^n + a_{n-1}(t) \lambda^{n-1} + a_{n-2}(t) \lambda^{n-2} + \dots + a_0(t),$$
then we have $a_0= (-1)^n det(A(t))$. Furthermore, there is the Cayley-Hamilton theorem telling you that
$$ P_{A(t)}(A(t)) = 0\in Mat(n\times n, K),$$
i.e. 
$$ A(t)^n + a_{n-1}(t) A(t)^{n-1} + a_{n-2}(t) A(t)^{n-2} + \dots + (-1)^n det(A(t)) E_n = 0.$$
From this we get
$$ A(t)(A(t)^{n-1}+a_{n-1}(t) A(t)^{n-2} + a_{n-2}(t) A(t)^{n-3} + \dots + a_1(t) E_n) = -(-1)^{n} det(A(t)) E_n. $$
As $A(t)\in GL(n,K)$ we have $det(A(t))\neq 0$. We set
$$ B(t)=  \frac{1}{-(-1)^{n} det(A(t))}(A(t)^{n-1}+a_{n-1}(t) A(t)^{n-2} + a_{n-2}(t) A(t)^{n-3} + \dots + a_1(t) E_n).$$
By the above computation we have
$$ A(t) B(t)= E_n $$
and hence (as we already know $A(t)\in GL(n,K)$ and that the leftinverse is equal to the rightinverse)
$$ A(t)^{-1}= B(t) = \frac{1}{-(-1)^{n} det(A(t))}(A(t)^{n-1}+a_{n-1}(t) A(t)^{n-2} + a_{n-2}(t) A(t)^{n-3} + \dots + a_1(t) E_n). $$
For your bound you need further assumptions. In general this will not work. Take for example  $A(t)= e^{-t^2} E_n$. Then you have
$$ \Vert A(t) \Vert = e^{-t^2} \leq 1.$$
However,
$$ \Vert A(t)^{-1} \Vert = \Vert e^{t^2} E_n \Vert = e^{t^2}$$
is not bounded from above.
