Example 3. Solve
the linear system
Solution 3.
Enter the equations into Mathematica.
![[Graphics:../Images/EchelonFormMod_gr_134.gif]](../Images/EchelonFormMod_gr_134.gif)
Identify the matrix of coefficients A and column vector B for the matrix problem AX = B.
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Form the augmented matrix M = [A, B] and perform Gauss-Jordan elimination with row interchanges.
Again, Gauss-Jordan elimination encountered division by error and
could not find a solution.
We might have suspected a problem because the determinant
of A is zero.
Let us investigate further and the reduced row echelon form of the augmented matrix M = [A, B].
The equation form for this matrix is
There is one free variable which we choose to
be
. It
is used in computing
.
Solve the previous equations for
.
Make the substitution
.
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 row is entirely zero, the system has reduced to two equations
and three unknowns.
We can add the equation
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:
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 might wonder if Gauss-Jordan
elimination could be use to find some solution, after all there is an
infinite number of them.
Form the augmented matrix M = [A, B] and perform Gauss-Jordan elimination with row interchanges.
So we have found one of the
infinite number of solutions to
be
.
Notice that this is merely substituting t=1
in the general solution . ![]()
Aside. We can graph general solution. This is just for fun.
![[Graphics:../Images/EchelonFormMod_gr_223.gif]](../Images/EchelonFormMod_gr_223.gif)
Aside. We check
out
.
Looking at the above calculations we see
that
and
.
Since
the
theorem guarantees that there will be an infinite number of
solutions.
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 !
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(c) John H. Mathews 2004