Find determinant of matrix dependent on one variable given specific conditions. The conditions given are:
$$\frac{d}{dt} A(t) = \begin{pmatrix}
2 & 5 \\
1 & -2 
\end{pmatrix}A(t)$$
\begin{equation}
{A(0)= I}
\end{equation}
And using this I am supposed to find $\det(A(t))$ for $t=1,10,100$
I tried taking determinants of both sides, component wise approach but don't seem to be getting anywhere. Would appreciate any push in the right direction.
 A: If $B=\left(\begin{smallmatrix}2&5\\1&-2\end{smallmatrix}\right)$, then the solution of that differential equation is $A(t)=\exp(tB)$. Now, $(5,1)$ and $(-1,1)$ are eigenvectors of $tB$; the eigenvalues are $3$ and $-3$ respectively. If $M=\left(\begin{smallmatrix}5&-1\\1&1\end{smallmatrix}\right)$ (the columns of $M$ are the eigenvectors of $tB$), then$$M^{-1}.(tB).M=\begin{pmatrix}3t&0\\0&-3t\end{pmatrix}.$$Therefore,$$M^{-1}.e^{tB}.M=\begin{pmatrix}e^{3t}&0\\0&e^{-3t}\end{pmatrix},$$from which it follows that\begin{align}A(t)&=e^{tB}\\&=M.\begin{pmatrix}e^{3t}&0\\0&e^{-3t}\end{pmatrix}.M^{-1}\\&=\frac16\begin{pmatrix}5e^{3t}+e^{-3t}&5e^{3t}-5e^{-3t}\\e^{3t}-e^{-3t}&e^{3t}+5e^{-3t}\end{pmatrix}.\end{align}
A: You don’t need to solve the differential equation explicitly in order to solve this problem if you know two key properties of matrices: First, the determinant of a matrix is equal to the product of its eigenvalues, taking multiplicity into account. Second, if $f$ is an analytic function and $\lambda$ an eigenvalue of the matrix $M$, then $f(\lambda)$ is an eigenvalue of $f(M)$. Letting $M$ be the coefficient matrix on the right-hand side of the differential equation, you know from the initial conditions that $A(t)=e^{tM}$, so if $\lambda_1$ and $\lambda_2$ are the eigenvalues of $M$, then the eigenvalues of $A(t)$ are $e^{\lambda_1 t}$ and $e^{\lambda_2 t}$, and so $$\det{A(t)}=e^{\lambda_1 t}e^{\lambda_2 t}=e^{(\lambda_1+\lambda_2)t} = e^{t \operatorname{tr}M}.$$ (Here I used the additional fact that the trace of a matrix is equal to the sum of its eigenvalues.) 
In this case, $M$ is traceless, so $\det{A(t)}=1$ for all values of $t$.
