Lesson 6.4


In Lesson 6.3, we used an inverse matrix to
solve systems represented by matrices. This
is a powerful strategy to use when the
number of variables is equal to the number of
equations.
In many applications, however, the number of
variables is not equal to the number of
equations. The row reduction method is such
a method. It requires only one matrix, and it
extends the method of elimination in a
systematic way.


In section 6.3 we replaced the system of equations
with an equivalent matrix equation and then used
the inverse matrix to solve the equation.
In this lesson we’ll write an augmented matrix,
which is a single matrix that contains columns for
the coefficients of each variable and a final column
for the constant terms.
2 1 5 


5
3
13






1 0 a 


0
1
b


The row reduction method transforms an
augmented matrix into a solution matrix.
Instead of combining equations and multiples
of equations until you are left with an equation
in one variable, you add multiples of rows to
other rows until you obtain the solution matrix.
The solution matrix for a consistent and
independent system contains the solutions in
the last column.
The rest of the matrix consists of 1’s along the
main diagonal and 0’s above and below it.

2x+y=5
Solve this system of equations. 
5x+3y=13
Because the equations are in standard form, you
can copy the coefficients and constants from each
equation into corresponding rows of the
augmented matrix.
2 1 5 


5
3
13


2 1 5 


5
3
13





1 0 a 


0
1
b


Using only the elementary row operations,
you can transform this matrix into the
solution matrix.
You need both m21 and m12 to be 0, and you
need both m11 and m22 to be 1.
.
2 1 5 


5
3
13


Add -2.5 times row 1 to
row 2 to get 0 for m21 .
Multiply row 2 by 2 to
change m22 to 1.
1 0 a 


0
1
b


Multiply equation 1 by
2.5 and add to equation
2 to eliminate x.
Multiply the equation by 2
to find y.
1 0 a 


0
1
b


2 1 5 


5
3
13



Add 1 times row 2 to
row 1 to get 0 for m12 .
Multiply row 1 by 0.5.

Multiply this new
equation by -1, and
add the result to
equation 1 to eliminate
y.
Multiply the equation by
0.5 to find x.
The last column of the solution matrix indicates that the solution
to the system is (2, 1).
You have learned how to interpret a
typical solution matrix.
For this matrix the solution is x = 12,
y = 30, and z = 8.
This matrix represents the equations
x =12 and y + 3z = 54. These
equations indicate that there was not
enough information to find a single
solution. Therefore, the system has
infinitely many solutions.
Can you find several ordered triples
that satisfy the equations?
possible answers (12, 0, 18), (12, 27, 9)
Or the final matrix may look like
The last row claims that 0 equals 5, which
is not true. Therefore, this matrix shows
that the system of equations has no
solution.