Solving Linear Systems, Continued and The Inverse of a Matrix

Solving Linear

Systems

Math 240

Solving Linear

Systems

Gauss-Jordan

elimination

Rank

Inverse

matrices

Definition

Computing

inverses

Properties of

inverses

Using inverse

matrices

Conclusion

Solving Linear Systems, Continued

and

The Inverse of a Matrix

Math 240 ¡ª Calculus III

Summer 2015, Session II

Tuesday, July 7, 2015

Solving Linear

Systems

Agenda

Math 240

Solving Linear

Systems

Gauss-Jordan

elimination

Rank

Inverse

matrices

Definition

Computing

inverses

Properties of

inverses

Using inverse

matrices

Conclusion

1. Solving Linear Systems

Gauss-Jordan elimination

The rank of a matrix

2. The inverse of a square matrix

Definition

Computing inverses

Properties of inverses

Using inverse matrices

Conclusion

Solving Linear

Systems

Gaussian elimination

Math 240

Solving Linear

Systems

Gauss-Jordan

elimination

Rank

Inverse

matrices

Definition

Computing

inverses

Properties of

inverses

Using inverse

matrices

Conclusion

Gaussian elimination solves a linear system by reducing to

REF via elementary row ops and then using back substitution.

Example

3x1 ? 2x2 +

x1 ? 2x2 +

2x1 ? x2 ?

?

1

¡ú ?0

0

2x3 = 9

x3 = 5

2x3 = ?1

?2

1

0

1

3

1

?

?

3 ?2

2

9

?1 ?2

1

5?

2 ?1 ?2 ?1

?

5

x1 ? 2x2 + x3 = 5

5?

x2 + 3x3 = 5

2

x3 = 2

Steps

1. P12

3. A13 (?2)

5. A23 (?3)




2. A12 (?3)

4. A32 (?1)

6. M3 ?1

13

Back substitution gives the solution (1, ?1, 2).

Solving Linear

Systems

Gauss-Jordan elimination

Math 240

Solving Linear

Systems

Gauss-Jordan

elimination

Rank

Inverse

matrices

Definition

Computing

inverses

Properties of

inverses

Using inverse

matrices

Conclusion

Reducing the augmented matrix to RREF makes the system

even easier to solve.

Example

?

?

?

1 ?2 1 5

1

?0

1 3 5? ¡ú ? 0

0

0 1 2

0

0

1

0

?

0 1

0 ?1?

1 2

x1

x2

= 1

= ?1

x3 = 2

Steps

1. A32 (?3)

2. A31 (?1)

3. A21 (2)

Now, without any back substitution, we can see that the

solution is (1, ?1, 2).

The method of solving a linear system by reducing its

augmented matrix to RREF is called Gauss-Jordan

elimination.

Solving Linear

Systems

The rank of a matrix

Math 240

Solving Linear

Systems

Gauss-Jordan

elimination

Rank

Inverse

matrices

Definition

Computing

inverses

Properties of

inverses

Using inverse

matrices

Conclusion

Definition

The rank of a matrix, A, is the number of nonzero rows it has

after reduction to REF. It is denoted by rank(A).

If A is the coefficient matrix of an m ¡Á n linear system and

rank(A# ) = rank(A) = n then the REF looks like

?

?

1 ? ? ¡¤¡¤¡¤ ?

x1 = ?

? 1 ? . . . ??

?

?

x2 = ?

?

.. ?

..

?

?

..

.

.

?

?

.

?

1 ??

xn = ?

0 ........... 0

0

Lemma

Suppose Ax = b is an m ¡Á n linear system with augmented

matrix A# . If rank(A# ) = rank(A) = n then the system has a

unique solution.

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