Differentiating the matrix with exponential function? I need to show that
$$ae^{\lambda t}(1,0)+be^{\lambda t}(t,1)$$ is a general solution of 
$$
X'=\begin{bmatrix} \lambda & 1 \\ 0 & \lambda \end{bmatrix}X
$$
I guess that the derivative of the first part is just $a \lambda e^{\lambda t } (1.0)$, but I am not sure how to differentiate the second part. 
 A: HINT: Write the solutions as
\begin{eqnarray}
x(t) &=& ae^{\lambda t} + b t e^{\lambda t} \\
y(t) &=& be^{\lambda t}
\end{eqnarray}
And show they solve the problem
\begin{eqnarray}
\frac{dx}{dt} &=& \lambda x(t) + y(t) \\
\frac{dy}{dt} &=& \lambda y(t)
\end{eqnarray}
A: Well, for general solution of 
$$ 
X'=MX
$$
($M$ a constant matrix), the technique is always the same. 


*

* consider an invertible solution (i.e. not zero !) of 
$$ 
C'=MC
$$
(with $C$ matrices with variable coefficients) like $C=e^{tM}$

* The general solution is of the form $X=e^{tM}X_0$ where $X_0$ is the "initial vector" (i.e. at $t=0$). 


Here your "multiplier" is 
$$
M=\begin{bmatrix} \lambda & 1 \\ 0 & \lambda \end{bmatrix}=
\begin{bmatrix} \lambda & 0 \\ 0 & \lambda\end{bmatrix}+
\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}
$$
and, taking into account that scalar matrices ($\lambda I$) commute with everybody, you can apply $e^{A+B}=e^{A}e^{B}$ when $A,B$ commute. Then
$$
e^{tM}=e^{\begin{bmatrix} t\lambda & 0 \\ 0 & t\lambda\end{bmatrix}+
\begin{bmatrix} 0 & t \\ 0 & 0 \end{bmatrix}}=
e^{\begin{bmatrix} t\lambda & 0 \\ 0 & t\lambda\end{bmatrix}}
e^{\begin{bmatrix} 0 & t \\ 0 & 0 \end{bmatrix}}=
\begin{bmatrix} e^{t\lambda} & 0 \\ 0 & e^{t\lambda}\end{bmatrix}\begin{bmatrix} 1 & t \\ 0 & 1 \end{bmatrix}=
\begin{bmatrix} e^{t\lambda} & te^{t\lambda} \\ 0 & e^{t\lambda}\end{bmatrix}
$$
Hope this helps.
A: 
Note that 
$$be^{\lambda t}\pmatrix {t \\ 1}=be^{\lambda t}(t \pmatrix{ 1 \\0}+ \pmatrix {0 \\ 1})=be^{\lambda t}t \pmatrix{ 1 \\0}+be^{\lambda t} \pmatrix {0 \\ 1}$$
You know how to differentiate that ? 
