Question

How to calculate particular solution of linear iquation in matrix form

Answer 1

You recall that a linear differential equation

$\displaystyle f_n(x) y^{(n)} + \cdots + f_1(x) y' + f_0(x) y = g(x)$

was called homogeneous if $ g(x) = 0$, and non-homogeneous or inhomogeneous otherwise. We use the same terminology for systems of linear equations and for matrix equations:

A matrix equation

$\displaystyle AX = B$

is called homogeneous if $ B$ is the zero vector (all entries are zero). A system of linear equations is called homogeneous if the equivalent matrix equation is homogeneous.

Homogeneous matrix equations have some special properties:

1.

The matrix equation

$\displaystyle AX = 0$

always has at least one solution, the zero solution

$\displaystyle X=0.$

(Here 0 stands for a column vector all of whose entries are zero.)

2.

If the column vectors $ X_1$ and $ X_2$ are two solutions to the matrix equation

$\displaystyle AX = 0$

then so is any linear combination of them, $ aX_1 + bX_2$.

3.

The complete solution to a matrix equation

$\displaystyle AX = 0$

is always given in the form

$\displaystyle X = t_1X_1 + t_2X_2 + \cdots + t_nX_n,$

where $ X_1$, $ X_2$, ...$ X_n$ are solutions and $ t_1$, $ t_2$, ..., $ t_n$ are parameters. The number of parameters depends on the dimension of the ``solution space.''

You can see why property (1) holds; a system of linear equations like

$\displaystyle a_1x_1 + a_2x_2 + \cdots + a_nx_n = 0$

$\displaystyle b_1x_1 + b_2x_2 + \cdots

+ b_nx_n = 0$

$\displaystyle c_1x_1 + c_2x_2 + \cdots + c_nx_n = 0$

$\displaystyle \cdot \hspace{1.65in} \cdot$

$\displaystyle \cdot \hspace{1.65in} \cdot$

$\displaystyle \cdot \hspace{1.65in} \cdot$

will always be satisfied by setting all the variables equal to zero. (This is the same reason a homogeneous linear differential equation can always be satisfied by setting $ y=0$.)

Property (2) depends on the linearity of multiplication by $ A$. If

$\displaystyle AX_1 = 0 \qquad \hbox{and} \qquad AX_2 = 0$

then we have that

$\displaystyle A(aX_1+bX_2)= A(aX_1) + A(bX_2) = aAX_1 + bAX_2 = 0+0 = 0.$

Property (3) also really comes from the linearity, since if we have

$\displaystyle AX_1 = 0 \qquad \qquad AX_2 = 0 \qquad \cdots \qquad AX_n=0$

then we have that