Example 4. Solve
the linear system
Solution 4.
Enter the equations into Mathematica.
Identify the matrix of coefficients A and column vector B for the matrix problem AX = B.
![[Graphics:../Images/HomogenLinSystemMod_gr_280.gif]](../Images/HomogenLinSystemMod_gr_280.gif)
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 three
equations and five 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 5×5 identity matrix appears in the left 5 columns
of M, and the given
linear system is equivalent to:
![[Graphics:../Images/HomogenLinSystemMod_gr_326.gif]](../Images/HomogenLinSystemMod_gr_326.gif)
The solution vector is the sixth 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 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.
We are really really
done.
Aside. The
following Mathematica steps will also solve the problem
automatically. It starts with the equations, creates the
matrices, and ends up with the vector form of the
solution. This is just for fun !
![[Graphics:../Images/HomogenLinSystemMod_gr_351.gif]](../Images/HomogenLinSystemMod_gr_351.gif)
(c) John H. Mathews 2004