Recently came across to solve system of linear differential equation with the matrix method, I have

Make a variable substitution. If you have a homogeneous linear system of ODEs ${\displaystyle \dot{x}=Ax}$, where A has a basis of eigenvectors ${\displaystyle v_1,\dots ,v_n}$ and corresponding eigenvalues ${\displaystyle {\lambda }_1,\dots ,{\lambda }_n}$, then make a substitution in terms of a new vector variable y, the coordinate vector of x with respect to ${\displaystyle v_1,\dots ,v_n}$. Then,
${\displaystyle x=y_1{v}_1+\dots +y_n{v}_n\Rightarrow \dot{x}={\dot{y}}_1{v}_1+\dots +{\dot{y}}_n{v}_n.}$
Hence, the system becomes,
${\displaystyle \dot{x}=Ax}$
${\displaystyle {\dot{y}}_1{v}_1+\dots +{\dot{y}}_n{v}_n=A(y_1{v}_1+\dots +y_n{v}_n)}$
${\displaystyle =y_{1}A{v}_1+\dots +y_{n}A{v}_n}$
${\displaystyle ={\lambda }_1{y}_1{v}_1+\dots +{\lambda }_n{y}_n{v}_n.}$
By the linear independence of ${\displaystyle v_1,\dots ,v_n}$, we may equate the (non-constant, but still scalar) coefficients of ${\displaystyle v_i}$ to obtain
${\displaystyle {\dot{y}}_i={\lambda }_i{y}_i}$,
for each i. In other words, we have now decoupled the system into n totally independent first order ODEs, in terms of the coordinates of y. These have the usual solutions:
${\displaystyle y_i=A_i{e}^{{\lambda }_{i}t}}$,
for some constant ${\displaystyle A_i}$. Thus, changing back,
${\displaystyle x=y_1{v}_1+\dots +y_n{v}_n=A_1{e}^{{\lambda }_{1}t}{v}_1+\dots +A_n{e}^{{\lambda }_{n}t}{v}_n}$.