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Section 8.1
Matrix Solutions to
Linear Systems
Solving Linear Systems
Using Matrices
This rectangular array of 24 numbers, arranged in rows and
columns and placed in red brackets, is an example of a matrix.
The numbers inside the brackets are called elements of the
matrix. Matrices are used to solve systems of linear equations.
They give us a shortened way of writing a system of equations.
The first step in solving a system of linear equations using matrices
is to write the augmented matrix. An augmented matrix has a
vertical bar separating the columns of the matrix into two groups.
The coefficients of each variable are placed to the left of the vertical
line and the constants are placed to the right. If any variable is
missing, its coefficient is 0.
Notice how the second matrix contains 1s down the
diagonal from upper left to lower right, called the main
diagonal, and 0s below the 1s. This arrangement makes
it easy to find the solution of the system of equations
with just a little back-substitution.
Main Diagonal
Example
Write the augmented matrix of the given system of equations.
3x+4y=7
4x-2y=5
A matrix with 1s down the main diagonal and 0s below the
1s is said to be in row-echelon form. We use row operations
on the augmented matrix. These row operations are just like
what you did when solving a linear system by the addition method.
Example
Write the system of equations corresponding to this
augmented matrix. Then perform the row operation
on the given augmented matrix. R 2  4r1  r2
 1 3 3 5



4

5

3

5


 3 2 4 6 
Gauss-Jordan Elimination
Using Gaussian elimination we obtain a matrix in row-echelon form,
with 1s down the main diagonal and 0s below the 1s. When you have
accomplished this you must back-substitute to get your answers.A second
method called Gauss-Jordan elimination continues the process until a matrix
with 1s down the main diagonal and 0s in every position above and below each
1 is found. Such a matrix is said to be in reduced row-echelon form. For a
system in three variables, x,y, and z, we must get the augmented matrix into
the form seen below.
Sometimes it is advantageous to write a matrix in
reduced row echelon form. In this form, row operations
are used to obtain entries that are 0 above as well as
below the leading 1 in a row. The advantage is that the
solution is readily found without needing to back-substitute.
There will be a second advantage in the future when we
discuss the inverse of a matrix.
Graphing Calculator-Matrices,
Reduced Row Echelon Form
To work with matrices we need to press 2nd x 1 keys
to get the Matrix Menu.
To type in a matrix, cursor to the right twice to getto EDIT,
press ENTER and type in the dimensions of matrix A. Press
ENTER after each number, then type in the numbers that
comprise the matrix, again pressing ENTER after each number.
To get out of the matrix menu press QUIT (2nd MODE).
To get the reduced row-echelon form bring up
the Matrix Menu (2nd x 1 ) and cursor to MATH.
Cursor down to B, rref (reduced row echelon form).
Press ENTER. (Just row echelon form is "A:ref.")
To type the name of the matrix again bring up
the Matrix Menu and under NAMES, choose
#1,  A press ENTER, press ENTER again any
you will have your new matrix.
Example
Solve this system of equations using matrices (row operations).
3x  5 y  3
15 x  5 y  21
Example
Solve this system of equations using matrices (row operations).
2 x  y  4
2 y  4 z  0
3x  2 z  11
Example
Solve this system of equations using matrices (row operations). If
the system has no solution, say that it is inconsistent.
x  y  z  1
4 x  3 y  2 z  16
2 x  2 y  3z  5
Write the system of equations as an augmented matrix.
Then perform this row operation on the given augmented
matrix. R 3  4r1  r3
x  3 y  z  5
4 x  5 y  3z  5
3 x  2 y  4 z  6
(a)
(c)
 1 3 3 5 



4

5

3

5


 1 14 16 14 
 4 12 12 20 


 4 5 3 5 
 1 14 16 14 
(b)
(d)
 1 3 3 5 



4

5

3

5


 4 12 12 20 
 1 14 16 14 


 4 5 3 5 
 1 14 16 14 
Find the solution to this system of equations using matrices.
x+2y=5
x+y=3
(a)  2,1
(b)  2, 2 
(c)  1,1
(d) 1, 2 