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Homogeneous and non-homogeneous equations

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

$\displaystyle A(t_1X_1 + t_2X_2 + \cdots + t_nX_n) = $

$\displaystyle t_1A(X_1) + t_2A(X_2) + \cdots + t_nA(X_n) =
0+0+\cdots+0 = 0.$

This is the same reason that the general solution to a homogeneous linear differential equation is a linear combination of particular solutions, such as

$\displaystyle y = A\cos x + B \sin x.$

In the case of differential equations, the number of different particular solutions, or the number of constants in the general solution, depends on the order of the differential equation; one solution for a first order equation, two different solutions for a second order equation, etc. In the case of matrix equations, the number of particular solutions is the number of paramters in the general or complete solution, the dimension of the solution space.

We can also see property (3) in action by solving a matrix equation. Here's the equation:

$\displaystyle \left(\begin{array}{cccc} 2 & 4 & 0 & 8 \\  1 & 2 & 1 & 10
\end{...
... y\\  z
\end{array}\right) = \left(\begin{array}{c} 0\\  0
\end{array}\right).
$

The augmented matrix of this equation has the row echelon form

$\displaystyle \left(\begin{array}{cccccc} 1 & 2 & 0 & 4 &\vdots & 0 \\  0 & 0 & 1 & 6 & \vdots &
0
\end{array}\right)$

so we can write down the complete general solution

$\displaystyle \left(\begin{array}{c} w \\  x \\  y \\  z
\end{array}\right) =
\left(\begin{array}{c} -2s -4t \\  s \\  -6t \\  t
\end{array}\right).
$

We can rewrite this as

$\displaystyle \left(\begin{matrix}w\cr x \cr y\cr z\cr\end{matrix}\right) =
s ...
...trix}\right) +
t \left(\begin{matrix}-4\cr 0\cr -6\cr 1\cr\end{matrix}\right).$

The particular solutions from which we can put together this complete solution are

$\displaystyle \left(\begin{matrix}-2\cr 1\cr 0\cr 0\cr\end{matrix}\right) \qquad\hbox{and}\qquad
\left(\begin{matrix}-4\cr 0\cr -6\cr 1\cr\end{matrix}\right).$

The really nice thing we get out of this is a method for finding solutions to non-homogeneous systems of linear equations (or non-homogeneous matrix equations.) It works exactly the same way as solutions for linear differential equations:

If the matrix equation

$\displaystyle AX = B$

has one particular solution $ X_p$, and the associated homogeneous equation

$\displaystyle AX = 0$

has the complete solution $ X_h$, then the complete solution to the original non-homogeneous equation is

$\displaystyle X = X_p + X_h.$

Example:

$\displaystyle \left(\begin{array}{cccc} 2 & 4 & 0 & 8 \\  1 & 2 & 1 & 10
\end{...
...  y\\  z
\end{array}\right) = \left(\begin{array}{c} 4\\  3
\end{array}\right)
$

has the complete solution (which we computed earlier)

$\displaystyle \left(\begin{array}{c} w \\  x \\  y \\  z
\end{array}\right) =
\left(\begin{array}{c} -2s -4t +2\\  s \\  -6t +1\\  t
\end{array}\right),
$

which we can rewrite as

$\displaystyle \left(\begin{matrix}w\cr x \cr y\cr z\cr\end{matrix}\right) =
s ...
...atrix}\right) +
\left(\begin{matrix}2 \cr 0 \cr 1 \cr 0 \cr\end{matrix}\right).$

This is the sum of the solution to the associated homogeneous system, which we wrote down in the previous example,

$\displaystyle X_h = s \left(\begin{matrix}-2\cr 1\cr 0\cr 0\cr\end{matrix}\right) +
t \left(\begin{matrix}-4\cr 0\cr -6\cr 1\cr\end{matrix}\right),$

and a particular solution to this inhomogeneous system

$\displaystyle X_p = \left(\begin{matrix}2 \cr 0 \cr 1 \cr 0 \cr\end{matrix}\right).$

Example: The homogeneous system of linear equations

$\displaystyle x + y + z = 0$

$\displaystyle x - y + 2z = 0$

$\displaystyle 3x - y + 5z = 0$

has the complete solution

$\displaystyle x = -\frac{3}{2}t$

$\displaystyle y = \frac{1}{2}t$

$\displaystyle z=t.$

The non-homongeous system

$\displaystyle x + y + z = 3$

$\displaystyle x - y + 2z = 2$

$\displaystyle 3x - y + 5z = 7$

has one particular solution

$\displaystyle x=1$

$\displaystyle y=1$

$\displaystyle z=1.$

To get the complete solution to the non-homongeneous system

$\displaystyle x + y + z = 3$

$\displaystyle x - y + 2z = 2$

$\displaystyle 3x - y + 5z = 7$

we add these together:

$\displaystyle x = -\frac{3}{2}t + 1$

$\displaystyle y = \frac{1}{2}t + 1$

$\displaystyle z=t + 1.$

Exercise 1   Rewrite the systems of linear equations of Exercise 4 (in the last handout) as matrix equations.

Exercise 2   Rewrite the following matrix equations as systems of linear equations.

$\displaystyle \left(\begin{array}{ccc} 1 & 1 & 1 \\  3 & 1 & -1
\end{array}\rig...
... y \\  z
\end{array}\right) =
\left(\begin{array}{c} 4 \\  2
\end{array}\right)$

$\displaystyle \left(\begin{array}{cc} 3 & 1 \\  0 & 1 \\  9 & 8 \\  -1 & 1
\en...
...{array}\right) =
\left(\begin{array}{c} -1 \\  1 \\  3 \\  2
\end{array}\right)$

$\displaystyle \left(\begin{array}{ccc} 1 & 0 & 0 \\  0 & 1 & 0 \\  0 & 0 & 1
\e...
...
z
\end{array}\right) =
\left(\begin{array}{c} 3 \\  1 \\  4
\end{array}\right)$

Exercise 3   Carry out the following matrix multiplications, or explain why they cannot be carried out.

$\displaystyle \left(\begin{array}{cc} 3 & 2 \\  2 & 1
\end{array}\right)
\left(\begin{array}{c} 1 \\  -1
\end{array}\right)
$

$\displaystyle \left(\begin{array}{ccc} 1 & 2 & 4 \\  0 & 1 & 3
\end{array}\right)
\left(\begin{array}{c} -1 \\  2 \\  1
\end{array}\right)
$

$\displaystyle \left(\begin{array}{ccc} 3 & 1 & 2 \\  1 & -1 & 4
\end{array}\right)
\left(\begin{array}{c} 2 \\  9
\end{array}\right)
$

$\displaystyle \left(\begin{array}{cc} 0 & 1 \\  1 & 0
\end{array}\right)
\left(\begin{array}{c} x \\  y
\end{array}\right)
$

Exercise 4   Solve the following matrix equations using row-reduction.

$\displaystyle \left(\begin{array}{cc} 1 & 3 \\  -1 & 4
\end{array}\right)
\lef...
...x \\  y
\end{array}\right) =
\left(\begin{array}{c} 4 \\  3
\end{array}\right)
$

$\displaystyle \left(\begin{array}{ccc} 2 & 1 & -1 \\  4 & 2 & 8
\end{array}\rig...
...w \\  x
\end{array}\right) =
\left(\begin{array}{c} 6 \\  2
\end{array}\right)
$

Exercise 5   What does it say about the set of solutions to a system of linear equations if its augmented matrix, when put into row-reduced form:

(a.) Has a row whose leading entry is in the last column (the column corresponding to the constant terms)?

(b.) Has all zeroes in the last column?

(c.) Has a column, other than the last column, in which no row has a leading entry?

(d.) Has rows with leading entries in every column except the last one?

Exercise 6   Write down the associated homogeneous matrix equations for the matrix equations in exercise 4. Now write down the complete solution to each of these homogeneous matrix equations.

Exercise 7   The matrix equation

$\displaystyle \left(\begin{array}{ccc} 2 & 1 & -1 \\  4 & 2 & 8
\end{array}\rig...
...w \\  x
\end{array}\right) =
\left(\begin{array}{c} 1\\  26
\end{array}\right)
$

has one solution given by

$\displaystyle \left(\begin{array}{c} v \\  w \\  x
\end{array}\right) = \left(\begin{matrix}2 \cr 1 \cr 2 \cr \end{matrix}\right).$

Give the complete solution to this matrix equation.


next up previous
Next: About this document ... Up: Solutions To Matrix Equations Previous: Linear functions
Peter Kostelec
2000-05-05