After you have solved a few systems of equations using elimination, you may realize that you could write the solution faster if you could omit the letters x1, x2, and so on
as long as you could keep the coefficients lined up properly. To do this, we write out the coefficients in a rectangular array called a matrix. A general linear system of m equations in n unknowns will be represented by the matrix
a11 a12 alj a111 b1
a21 a22 a2j a211 b2
ail ai2 aij a;n b;
am! am2 amj a,,111 bm
where the coefficient aij appears in the i-th row and }-th column of the coefficient matrix. This is called the augmented matrix of the system; it is augmented because it includes as its last column the right-hand side of the equations. The matrix without this last column is called the coefficient matrix of the system:
au a12 alj a111
a21 a22 a2j a211
a;1 a;2 aij a;n
amt am2 amj amn
For convenience, we sometimes denote the augmented matrix of a system with coeffi-bi
cient matrix
A
and right-hand sideb
= by[A
Ib J.
In Chapter 3, we will develop bmanother way of representing a system of linear equations.
Write the coefficient matrix and augmented matrix for the following system:
3x1 + 8x2 - l 8x3 + X4 = 35 x1 + 2x2 - 4x3 = 11 x1+3x2 -7x3 + X4 =
l
0Solution: The coefficient matrix is formed by writing the coefficients of each equation as the rows of the matrix. Thus, we get the matrix
[
3 8 -=
1�
8�lj
A= � �
EXAMPLES
(continued)
EXAMPLE6
For the augmented matrix, we just add the right-hand side as the last column. We get
[ :
8 2 3 --4 0 -7 1 8 1 35111
10
W1ite the system of linear equations that has the augmented matrix
[
1 0 -1 1 0 20 0 1
Solution: The rows of the matrix tell us the coefficients and right-hand side of each equation. We get the system
Remark
x, + 2x3 = 3 -X2 + X3 = 1
X3 = -2
Another way to view the coefficient matrix is to see that the }-th column of the matrix is the vector containing all the coefficients of x1. This view will become very important in Chapter 3 and beyond.
Since each row in the augmented matrix corresponds to an equation in the system of linear equations, performing operations on the equations of the system corresponds to performing the same operations on the rows of the matrix. Thus, the steps in elimi
nation correspond to the following elementary row operations.
Types of Elementary Row Operations
(1) Multiply one row by a non-zero constant.
(2) Interchange two rows.
(3) Add a multiple of one row to another.
As with the steps in elimination, we do not combine operations of type (1) and type (3) into one operation.
The process of performing elementary row operations on a matrix to bring it into some simpler form is called row reduction.
Recall that if a system of equations is obtained from another system by one or more of the elimination steps, the systems are said to be equivalent. For matrices, if the matrix M is row reduced into a matrix N by a sequence of elementary row operations, then we say that Mis row equivalent to N. Just as elimination steps are reversible, so are elementary row operations. It follows that if M is row equivalent to N, then N is row equivalent to M, so we may say that Mand N are row equivalent. It also follows that if A is row equivalent to B and Bis row equivalent to C, then A is row equivalent to C.
Let us see how the elimination in Example 3 appears in matrix notation. To do this, we introduce notation to indicate this elementary row operation. We write Ri + cR J
EXAMPLE 7
EXAMPLES
EXERCISE 2
Section 2.1 Systems of Linear Equations and Elimination 71
to indicate adding c times row j to row i,
R;
!R
J to indicate interchanging row i and row j. We writecR;
to indicate multiplying row i by a non-zero scalar c. Additionally, at each step we will use � to indicate that the matrices are row equivalent. Note that it would be incorrect to use = or ::::} instead of �. As one becomes confident with elementary row operations, one may omfr these indicators of which elementary row operations were used. However, including them can make checking the steps easier, and instructors may require them in work submitted for grading.The augmented matrix for the system in Example
3
is[ 1 1 -2 4 1 3 -1
7l
2 1 -5
7The first step in the elimination was to add
(-1)
times the first equation to the second.Here we add
(-1)
multiplied by the first row to the second. We write[ ; 1
3-2 -1 -5 n R2+(-l)R1
-[ � 1 2 -2 -5 1 n
The remaining steps are
-[ �
[ 1 1 -2
02 1
2 1 -5 -1 2 -2 -1 1 � l R3 - 2 R1 -! l (-l)R2
-[ � -[ �
2 -2 -1 -2 -1 1 1
2 1
j] R2
!R3
4 1 l [
�1
03 R3 - 2R2
01 -2 1 1
0-1 n
All the elementary row operations corresponding to the elimination in Example
3
havebeen performed. Observe that the final matrix is the augmented matrix for the final system of linear equations that we obtained in Example
3.
The matrix representation of the elimination in Example
4
is[ 1 1
0 02 1 114 J 3 3 19 R2
+(-l)R1
�[ 1
0 0 02 21 1154 J 1
Write out the matrix representation of the elimination used in Exercise
1.
In the next example, we will solve a system of linear equations using Gaussian elimination with back-substitution entirely in matrix form.
EXAMPLE9 Find the general solution of the system
3x1 + 8x2 -l 8x3 + x4 = 35 Xi+ 2x2 - 4x3 = 11 Xi + 3x2 -7 X3 + X4 = 10
Solution: W rite the augmented matrix of the system and row reduce:
[ :
8 2 3 -18 -4 -7To find the general solution, we now interpret the final matrix as the augmented matrix of the equivalent system. We get the system
Check this solution by substituting these values for x1, x2, x3, x4 into the original equations.
Observe that there are many different ways that we could choose to row reduce the augmented matrix in any of these examples. For instance, in Example 9 we could interchange row 1 and row 3 instead of interchanging row 1 and row 2. Alternatively, we could use the elementary row operations