Linear Algebra with Differential Equations/Heterogeneous Linear Differential Equations/Diagonalization

First of all (and kind of obvious suggested by the title), $$\mathbf{A}$$ must be diagonalizable. Second, the eigenvalues and eigenvectors of $$\mathbf{A}$$ are found, and form the matrix $$\mathbf{T}$$ which is an augemented matrix of eigenvectors, and $$\mathbf{D}$$ which is a matrix consisting of the corresponding eigenvalues on the main diagonal in the same column as their corresponding eigenvectors. Then with our central problem:

$$\mathbf{X}' = \mathbf{AX} + \mathbf{G}(t)$$

We substitute:

$$\mathbf{TY}' = \mathbf{ATY} + \mathbf{G}(t)$$

Then left multiply by $$\mathbf{T}^{-1}$$

$$\mathbf{Y}' = \mathbf{T}^{-1}\mathbf{ATY} + \mathbf{T}^{-1}\mathbf{G}(t)$$

As a consequence of Linear Algebra we take the following identity:

$$\mathbf{D} = \mathbf{T}^{-1}\mathbf{AT}$$

Thus:

$$\mathbf{Y}' = \mathbf{DY} + \mathbf{T}^{-1}\mathbf{G}(t)$$

And because of the nature of the diagonal the problem is a series of one-dimensional normal differential equations which can be solved for $$\mathbf{Y}$$ and used to find out $$\mathbf{X}$$.