MATH 105: Finite Mathematics 2-6: The Inverse of a...

Solving a Matrix Equation The Inverse of a Matrix Solving Systems of Equations Conclusion

MATH 105: Finite Mathematics2-6: The Inverse of a Matrix

Prof. Jonathan Duncan

Walla Walla College

Winter Quarter, 2006

Outline

1 Solving a Matrix Equation

2 The Inverse of a Matrix

3 Solving Systems of Equations

4 Conclusion

Outline

4 Conclusion

Solving Equations

Recall that last time we saw that a system of equations can berepresented as a matrix equation as shown below.

Example

Write the following system of equations in matrix form.

Solving Equations

Example

{2x + 3y = 73x − 4y = 2

Solving Equations

Example

{2x + 3y = 73x − 4y = 2

[2 33 −4

Solving Equations

Example

{2x + 3y = 73x − 4y = 2

[2 33 −4

]AX = B

Solving Equations

Example

{2x + 3y = 73x − 4y = 2

[2 33 −4

]AX = B

If we wish to use the matrix equation on the right to solve a systemof equations, then we need to review how we solve basic equationsinvolving numbers.

Solving a Simple Equation

The most basic algebra equation, ax = b, is solved using themultiplicative inverse of a.

Example

Solve the equation 3x = 6 for x .

Step 1: Multiply by 13

13 · (3x) = 1

3 · (6)Step 2: Simplify the Right

(13 · 3

)x = 2

Step 3: Simplify the Left 1 · x = 2Step 4: Solution x = 2

Example

13 · (3x) = 1

(13 · 3

)x = 2

Example

13 · (3x) = 1

(13 · 3

)x = 2

Example

13 · (3x) = 1

(13 · 3

)x = 2

Example

13 · (3x) = 1

(13 · 3

)x = 2

Example

13 · (3x) = 1

(13 · 3

)x = 2

Example

13 · (3x) = 1

(13 · 3

)x = 2

This solution process worked because 13 is the inverse of 3, so that

13 · 3 = 1, the identity for multiplication.

Outline

4 Conclusion

Matrix Inverse

To solve the matrix equation AX = B we need to find a matrixwhich we can multiply by A to get the identity In.

Matrix Inverse

Let A be an n× n matrix. Then a matrix A−1 is the inverse of A ifAA−1 = A−1A = In.

Caution:

Just as with numbers, not every matrix will have an inverse!

Matrix Inverse

Caution:

Matrix Inverse

Caution:

Verifying Matrix Inverses

Example

Show that

[−1 −23 4

[2 1−3

2 −12

]are inverses.

[−1 −23 4

] [2 1−3

2 −12

[1 00 1

[2 1−3

2 −12

] [−1 −23 4

[1 00 1

Example

Show that

[−1 −23 4

[2 1−3

2 −12

]are inverses.

[−1 −23 4

] [2 1−3

2 −12

[1 00 1

[2 1−3

2 −12

] [−1 −23 4

[1 00 1

Example

Show that

[−1 −23 4

[2 1−3

2 −12

]are inverses.

[−1 −23 4

] [2 1−3

2 −12

[1 00 1

[2 1−3

2 −12

] [−1 −23 4

[1 00 1

Example

Show that

[−1 −23 4

[2 1−3

2 −12

]are inverses.

[−1 −23 4

] [2 1−3

2 −12

[1 00 1

[2 1−3

2 −12

] [−1 −23 4

[1 00 1

While it is relatively easy to verify that matrices are inverses, wereally need to be able to find the inverse of a given matrix.

Finding a Matrix Inverse

To find the inverse of a matrix A we will use the fact thatAA−1 = In.

Find A−1

Let A =

[3 2−1 4

]and find A−1 =

[x1 x2

AA−1 = I2 ⇒[

3 2−1 4

] [x1 x2

[1 00 1

3x1 + 2x3 3x2 + 2x4

−x1 + 4x3 −x2 + 4x4

[1 00 1

]This gives two systems of equations:{

3x1 + 2x3 = 1−x1 + 4x3 = 0

{3x2 + 2x4 = 0−x2 + 4x4 = 1

Find A−1

Let A =

[3 2−1 4

]and find A−1 =

[x1 x2

AA−1 = I2 ⇒[

3 2−1 4

] [x1 x2

[1 00 1

3x1 + 2x3 3x2 + 2x4

−x1 + 4x3 −x2 + 4x4

[1 00 1

3x1 + 2x3 = 1−x1 + 4x3 = 0

{3x2 + 2x4 = 0−x2 + 4x4 = 1

Find A−1

Let A =

[3 2−1 4

]and find A−1 =

[x1 x2

AA−1 = I2 ⇒[

3 2−1 4

] [x1 x2

[1 00 1

3x1 + 2x3 3x2 + 2x4

−x1 + 4x3 −x2 + 4x4

[1 00 1

3x1 + 2x3 = 1−x1 + 4x3 = 0

{3x2 + 2x4 = 0−x2 + 4x4 = 1

Find A−1

Let A =

[3 2−1 4

]and find A−1 =

[x1 x2

AA−1 = I2 ⇒[

3 2−1 4

] [x1 x2

[1 00 1

3x1 + 2x3 3x2 + 2x4

−x1 + 4x3 −x2 + 4x4

[1 00 1

3x1 + 2x3 = 1−x1 + 4x3 = 0

{3x2 + 2x4 = 0−x2 + 4x4 = 1

Find A−1

Let A =

[3 2−1 4

]and find A−1 =

[x1 x2

AA−1 = I2 ⇒[

3 2−1 4

] [x1 x2

[1 00 1

3x1 + 2x3 3x2 + 2x4

−x1 + 4x3 −x2 + 4x4

[1 00 1

3x1 + 2x3 = 1−x1 + 4x3 = 0

{3x2 + 2x4 = 0−x2 + 4x4 = 1

Finding the Matrix Inverse, Cont. . .

Finding a Matrix Inverse, Continued

Solve the systems of equations:{3x1 + 2x3 = 1−x1 + 4x3 = 0

{3x2 + 2x4 = 0−x2 + 4x4 = 1[

3 2 1 0−1 4 0 1

][1 0 4

14 − 214

0 1 114

A−1 =

[414 − 2

{3x2 + 2x4 = 0−x2 + 4x4 = 1[

3 2 1 0−1 4 0 1

][1 0 4

14 − 214

0 1 114

A−1 =

[414 − 2

{3x2 + 2x4 = 0−x2 + 4x4 = 1[

3 2 1 0−1 4 0 1

][1 0 4

14 − 214

0 1 114

A−1 =

[414 − 2

{3x2 + 2x4 = 0−x2 + 4x4 = 1[

3 2 1 0−1 4 0 1

][1 0 4

14 − 214

0 1 114

A−1 =

[414 − 2

General Rule for Finding An Inverse

Applying the lessons of the previous example yields a generalprocedure for finding the inverse of a matrix.

To find the inverse of a n× n matrix A, form the augmented matrix[A | In] and use row reduction to transform it into

[In | A−1

Caution:

It may not be possible to get In on the left side of the matrix. If itis not possible, then the matrix A has no inverse.

[In | A−1

Caution:

[In | A−1

Caution:

Finding the Inverse of a 3× 3 Matrix

Example

Find the inverse of the following matrix, if it exists.1 1 13 −4 20 0 0

Finding the Inverse of a 3× 3 Matrix

Example

Find the inverse of the following matrix, if it exists.1 1 13 −4 20 0 0

Row operations yield:1 0 6

0 1 17

37 −1

7 00 0 0 0 0 1

And therefore there is no inverse.

Outline

4 Conclusion

Using A−1 to Solve a System of Equations

The main reason we are interested in matrix inverses is to solve asystem of equations written in matrix form.

Solving a System of Equations

If A is the matrix of coefficients for a system of equations, X is thecolumn vector containing the system variables, and B is thecolumn vector containing the constants, then:

AX = B ⇒ A−1AX = A−1B

⇒ InX = A−1B

⇒ X = A−1B

Caution:

A system of equations can only be solved in this way if it has aunique solution.

AX = B ⇒ A−1AX = A−1B

⇒ InX = A−1B

⇒ X = A−1B

Caution:

AX = B ⇒ A−1AX = A−1B

⇒ InX = A−1B

⇒ X = A−1B

Caution:

An Example

Example

Use a matrix equation to set-up and solve each system ofequations given below.

{−x − 2y = 13x + 4y = 3

x + y − z = 63x − y = 82x − 3y + 4z = −3

An Example

Example

{−x − 2y = 13x + 4y = 3

x + y − z = 63x − y = 82x − 3y + 4z = −3

An Example

Example

{−x − 2y = 13x + 4y = 3

x + y − z = 63x − y = 82x − 3y + 4z = −3

Reusing Results

One major advantage of solving a system of equations using amatrix equation is that if your matrix of coefficients A stays thesame, but your matrix B changes, you can reuse most of your work.

Example

Solve each of the following systems of equations using the resultsfrom the last part of the previous question.

x + y − z = 23x − y = 12x − 3y + 4z = 0

x + y − z = 03x − y = −142x − 3y + 4z = −13

Reusing Results

Example

x + y − z = 23x − y = 12x − 3y + 4z = 0

x + y − z = 03x − y = −142x − 3y + 4z = −13

Reusing Results

Example

x + y − z = 23x − y = 12x − 3y + 4z = 0

x + y − z = 03x − y = −142x − 3y + 4z = −13

Reusing Results

Example

x + y − z = 23x − y = 12x − 3y + 4z = 0

x + y − z = 03x − y = −142x − 3y + 4z = −13

Outline

4 Conclusion

Important Concepts

Things to Remember from Section 2-6

1 A−1A = AA−1 = In for an n × n matrix A

2 Find A−1 by reducing [A | In] to [In | A]

3 Solving systems of equations with matrix equations.

Important Concepts

Next Time. . .

Next time we will review for our third and final in class exam. Itwill be over the sections we covered from chapters 1 and 2.

For next time

Review for Exam

Next Time. . .

Next time we will review for our third and final in class exam. Itwill be over the sections we covered from chapters 1 and 2.

For next time

Review for Exam

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