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5. Gauss Jordan Elimination - Massachusetts … - augmented matrix calculator

5. Gauss Jordan Elimination - Massachusetts …-augmented matrix calculator

5. Gauss Jordan Elimination
Gauss Jordan elimination is very similar to Gaussian elimination,
except that one "keeps going". To apply Gauss Jordan elimination,
first apply Gaussian elimination until A is in echelon form. Then pick
the pivot furthest to the right (which is the last pivot created). If there
is a non-zero entry lying above the pivot (after all, by definition of
echelon form there are no non-zero entries below the pivot) then apply
elimination to create a zero in that spot, in the usual way. Repeat this
until there are no more non-zero entries in that column. Once we are
finished with this pivot then move left to the next closest pivot and
keep going. An example will hopefully make this clear. Let us start
with a matrix in echelon form:

1 -1 2 3
0 0 1 -2 .
The pivot furthest to the right is the (3, 4) entry. So we take the third
row multiply by 2 and by -3 and add it to the second and first rows,
to get

1 -1 2 0
0 0 1 0 .
0 0 01
So much for the pivot furthest to the right. The next pivot (reading
right to left) is at position (2, 3). Multiply the second row by -2 and
add it to the first row, to get

1 -1 0 0
0 0 1 0 .
0 0 01
This completes Gauss Jordan elimination.
Definition 5.1. Let A be an m ? n matrix.
We say that A is in reduced row echelon form if A in echelon
form and in addition
? every other entry of a column which contains a pivot is zero.
The end product of Gauss Jordan elimination is a matrix in reduced
row echelon form. Note that if one has a matrix in reduced row echelon
form, then it is very easy to solve equations. One can read off the
solutions with almost no work.
We can exploit this fact to come up with a very pretty way to com-
pute the inverse of a matrix. First a couple of easy (but important):
Lemma 5.2. Let A be a square matrix.
If A is invertible then every equation Ax = b has a unique solution.
Proof. Let B be the inverse of A. Suppose that v is a solution of the
equation Ax = b. Then
b = Av.
Multiply both sides by B, to get
Bb = B(Av)
= (BA)v
= Inv = v.
Thus v = Bb. It follows that there is at most one solution to the
equation Ax = b. On the other hand let v = Bb. Then
Av = A(Bb) = (AB)b = Inb = b,
so that v = Bb is a solution to the equation.
Suppose that we are given a square n ? n invertible matrix A and
we want to compute its inverse matrix B. We can view B as be-
ing composed of n column vectors, say v1, v2, . . . , vn Rn. Suppose
that we also view the identity matrix as being composed of columns
e1, e2, . . . , en. Then ei is the column vector whose ith entry is 1 and
every other entry is zero. The way that matrix multiplication works is
Avi = ei.
Since we are trying to determine vi, we can think of vi as being the
solution to the linear equation
Ax = ei.
How can we solve this equation? Well create the augmented matrix
If we apply Gaussian elimination then we get to a matrix U in echelon
(U |fi).
Here fi is the result of applying the steps of Gaussian elimination to ei.
Now if we continue we get a matrix, call it R, in reduced row echelon
form and another column vector gi,
Now let us make three clever observations. The first is that the ma-
trices U and R do not depend on i. That is to say the steps of both
Gaussian elimination and Gauss Jordan elimination only depend on
the coefficient matrix A and not on ei. The second is that the matrix
R must be the identity matrix. Indeed we cannot get a row of zeroes
when we apply Gaussian elimination, since we know that every equa-
tion has a solution. It follows that every row contains a pivot and so
every column contains a pivot (R is a square matrix). Since R is a
square matrix in reduced row echelon form it follows that R = In. But
now if we apply back substitution the equations immediately tell us
that vi = gi.
So in principle we could compute the inverse matrix B, column by
column, by solving n sets of linear equations. But this sounds like a lot
of hard work. The final observation is that for each one of these n sets
of linear equations the coefficient matrix is the same, namely A, and
the steps of Gauss Jordan are the same. So why not put all n sets of
equations into one super augmented matrix? If one puts the n column
vectors e1, e2, . . . , en together side by side, then one will automatically
get the identity matrix In,
Let us see an example. Suppose that

1 2 -3
A = 2 5 -8 .
-3 -5 8
First we make the super augmented matrix

1 2 -3 | 1 0 0
2 5 -8 | 0 1 0 .
-3 -5 8 | 0 0 1
Now apply Gaussian elimination. Multiply the first row by -2 and
3 and add them to the second and third row.

1 2 -3 | 1 0 0
0 1 -2 | -2 1 0 .
0 1 -1 | 3 0 1
Now multiply the second row by -1 and add it to the third row to get

1 2 -3 | 1 0 0
0 1 -2 | -2 1 0 .
0 0 1 | 5 -1 1
This completes the Gaussian elimination. Now we continue Gauss Jor-
dan elimination Multiply the third row by 2 and 3 and add it to the

How do I solve an augmented matrix? An augmented matrix contains the coefficients of the unknowns and the "pure" coefficients. You can manipulate the rows of this matrix (elementary row operations) to transform the coefficients and to "read", at the end, the solutions of your system. The two row operations allowed are: 1) swap rows;