Download Math 313 Lecture #1 §1.1: Systems of Linear Equations Consistency

Math 313 Lecture #1
§1.1: Systems of Linear Equations
Consistency and Inconsistency of Systems. An m × n system of linear equations (or simply system) is m linear equations in n unknowns:
a11 x1
a21 x1
..
.
+ · · · + a1n xn = b1 ,
+ · · · + a2n xn = b2 ,
..
..
...
.
.
+ · · · + amn xn = bm ,
+ a12 x2
+ a22 x2
..
.
am1 x1 + am2 x2
where the aij ’s and the bi ’s are all real numbers, and the xi ’s are the variables.
A solution of an m × n system is an ordered n-tuple of numbers (x1 , x2 , . . . , xn ) that
satisfy all m equations simultaneously.
Example. Consider the three 2 × 2 systems:
2x1 + 4x2 = 6,
x1 + 2x2 = 5,
x1 − 2x2 = 3,
2x1 − 4x2 = 6.
2x1 + x2 = 1,
x1 + x2 = −1,
Each linear equation here can be represented as a line in the plane:
5
!5
!4
!3
!2
4
4
3
3
3
2
2
2
1
1
1
0
0
!1
0
!1
x1
5
5
4
1
2
3
4
5
!5
!4
!2
!3
x1
!1
0
0
1
2
3
4
!5
!4
!1
!2
!3
0
!1
!2
!2
!3
!3
!3
!4
!4
!4
x2
2
3
4
5
!2
x2
!5
1
!1
x1
!5
x2
!5
The ordered pair (2, −3) is a solution for the second system. [Check it X.] And there
are no other solutions.
The ordered pair (3, 0) is a solution for the third system. Actually, an ordered pair of
the form (3 + 2t, t) for any real t is a solution for the third system. [Check them X.]
However, the first systems has no solution. Why? Because if the ordered pair (x1 , x2 ) is
a solution, then x1 = 5 − 2x2 and 2x1 = 6 − 4x2 ; but the second of these can be written
as x1 = 3 − 2x2 , so that 5 − 2x2 = x1 = 3 − 2x2 which implies that 5 = 3.
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The solution set of an m × n system is the collection of all of its solutions.
The three systems in the above example illustrate what is possible for the solution set:
no solutions, exactly one solution, or infinitely many solutions.
An m×n system is consistent if it has at least one solution; otherwise it is inconsistent.
To solve a consistent linear system is to find all of its solutions.
Equivalent Systems. How is the solution set of a consistent system found?
Consider the two 3 × 3 systems:
x1 +
x2 + 2x3 = 1,
2x2
= 4,
x3 = −1,
x1 + x2 + 2x3 = 1,
2x1 + 4x2 + 4x3 = 6,
−x1 − x2 − x3 = −2.
The first system is easily solved: x3 = −1, x2 = 2, and x1 = 1 − x2 − 2x3 = 1 − 2 + 2 = 1;
so the (one and only) solution of this system is (1, 2, −1).
A solution (x1 , x2 , x3 ) of the second system is a solution of each of the equations in the
second system; thus it is a solution of any new equation formed
I. by switching the order of the equations in the system,
II. by a nonzero multiple of any of the equations in the system, or
III. by adding a multiple of one of the equations in the system to another.
Adding the first and the third equations in the second system gives x3 = −1.
Adding minus two times the first equation to the second gives 2x2 = 4, so that x2 = 2.
The first equation in the second system then gives x1 = 1.
The solution of the second system is (1, 2, −1). [Check that (1, 2, −1) is a solution of the
second system X.]
The solution of the second system is the same as the first system.
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Two systems involving the same variables are said to be equivalent if they have the
same solution set.
The above example illustrates that operations I, II, and III listed above can be applied
to a system to obtain an equivalent system that is easier to solve.
n × n Systems and Strict Triangular Form. The first of the 3 × 3 matrices above
was easy to solve because of its form.
An n × n system is said to be in strict triangular form if for each k = 1, . . . , n, the
coefficients of the first k − 1 variables in the k th equation are all zero and the coefficient
of xk in the k th equation is nonzero.
An n × n system in strict triangular form is solved by back substitution:
· the value of xn is found by solving the nth equation,
· the value of xn−1 is found by substituting the value of xn into the (n − 1)th equation
and solving,
· the value of xn−2 is found by substituting the values of xn and xn−1 into the (n−2)th
equation and solving,
· continuing until the values of xn , . . . , x2 are substituted in the first equation to solve
for the value of x1 .
When the operations I, II, and III can be used on an n×n system to obtain an equivalent
strictly triangular system, then back substitution can be used to find the unique solution.
Example. Solve the strict triangular system:
2x1 + 3x2 +
x2 −
x3 − x4
x3 + 2x4
2x3 + 3x4
3x4
= 1,
= 0,
= −1,
= 9.
By back substitution,
x4
x3
x2
x1
= 3,
= (−1 − 3x4 )/2 = (−1 − 9)/2 = −5,
= x3 − 2x4 = −5 − 2(3) = −11,
= (1 − 3x2 − x3 + x4 )/2 = (1 + 33 + 5 + 3)/2 = 21.
Check that (21, −11, −5, 3) is the unique solution X.
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Augmented Matrices and Row Operations. When applying the operations I, II,
and III to an m × n system, the numbers aij ’s and bi ’s change, but the placement of the
variables xi ’s and the right hand side do not.
Only the numbers in the system need to be tracked while obtaining an equivalent system.
The numbers in an m × n system

a11
 a21

 ..
 .
am1
can be encoded by the augmented matrix,

a12 . . . an1 | b1
a22 . . . an2 | b2 

..
..
.. ..  .
..
.
.
.
. . 
am2 . . . amn | bm
The aij part of this is called the coefficient matrix of the m × n system.
Operations I, II, and III for a system correspond to the elementary row operations
on its augmented matrix:
I. interchange two rows,
II. multiply a row through by a nonzero real number,
III. replace a row by its sum with a multiple of another row.
With Ri representing the ith row of the augmented matrix, the elementary row operations
are symbolically I. Ri ↔ Rj , II. αRi → Ri for nonzero α, and III. Ri + αRj → Ri .
Applying these elementary row operations to the augmented matrix of a system may lead
to an equivalent system that is easier to solve than the original.
Example. Find, if possible, the augmented matrix of a strictly triangular system that
is equivalent to the 3 × 3 system,
2x1
2x1
4x2 − 3x3 = 8,
+ x3 = 2,
+ 2x2
= 5.
The augmented matrix for this system is


0 4 −3 | 8
2 0 1 | 2 .
2 2 0 | 5
The goal is to apply a sequence of elementary row operations to this to obtain an augmented matrix in which the “diagonal” entries of the coefficient matrix part are nonzero
while the entries below this diagonal are all zero.
The algorithm to do this is illustrated for this augmented matrix.
The first nonzero entry in the first column is called the pivot element; it occurs in the
second row here, which row is called the pivotal row.
Interchange the first

2 0
0 4
2 2
and second rows so that the new first row is the pivotal row:

1 | 2 ← pivotal row with pivot element a11 = 2.
−3 | 8
0 | 5
Now eliminate all of the nonzero entries of the first column below the pivot element by
elementary row operation III to obtain


2 0 1 | 2
0 4 −3 | 8 ← new pivotal row with pivot element a22 = 4.
0 2 −1 | 3
The first nonzero entry in the second column below the first row is the new pivot element,
and the row it is in the new pivotal row. (An interchange of rows may be needed to make
the second row the new pivotal row.)
Now eliminate all of the nonzero entries of the second column below the pivot element
by elementary row operation III to obtain the desired augmented matrix:


2 0 1 | 2
0 4 −3 | 8  .
0 0 1/2 | −1
The whole of

0
2
2
the computation is


4 −3 | 8
2
R ↔ R2
0 1 | 2 1
⇔ 0
2 0 | 5
2

2

⇔ 0
0

2

⇔ 0
0

0 1 | 2
4 −3 | 8
2 0 | 5 R3 − R1 → R3

0 1 | 2
4 −3 | 8
2 −1 | 3 R3 − (1/2)R2 → R2

0 1 | 2
4 −3 | 8  .
0 1/2 | −1