Example 3. Solve
the homogeneous linear system of equations
Solution 3.
Enter the equations into Mathematica.
![[Graphics:../Images/HomogenLinSystemMod_gr_201.gif]](../Images/HomogenLinSystemMod_gr_201.gif)
Identify the matrix of coefficients A and column vector B for the matrix problem AX = B.
![]()
Form the augmented matrix M = [A, B] and find its reduced row echelon form.
The equation form for this matrix is
There are two free variables which we choose to
be
and
. They
is used in computing
.
Solve the previous equations for
.
Make the substitutions
and
.
![]()
Get
![]()
The solution vector
is
We are done.
Aside. We can
verify that this is the solution by direct multiplication A
X. This is just for fun !
Aside. We can let Mathematica find the reduced row echelon matrix. This is just for fun !
Notice. Since the
last two rows are entirely zero, the system has reduced to one
equations and three unknowns.
We can add the equations
and
.
to those in the reduced row echelon form and then row reduce one more
time to get the solution.
The 3×3 identity matrix appears in the left 3 columns
of M, and the given
linear system is equivalent to:
![[Graphics:../Images/HomogenLinSystemMod_gr_247.gif]](../Images/HomogenLinSystemMod_gr_247.gif)
The solution vector is the fourth column of M.
We can verify that this is the solution by direct multiplication
A X. This is just for fun !
We are really done.
Aside. We can
graph the situation. This is just for fun.
![[Graphics:../Images/HomogenLinSystemMod_gr_259.gif]](../Images/HomogenLinSystemMod_gr_259.gif)
Aside. We check
out
.
Looking at the above calculations we see
that
,
and the theorem guarantees that the system has an infinite number of
solution. However, it might be easier to just
check out the determinant.
(c) John H. Mathews 2004