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Instruction: The Adjoint of a Matrix

Let A be a m n× matrix. Let aij represent the entry in the ith row and jth column of matrix A. Then, C is the transpose of A, denoted C= AT, if aij =cji. That is, the transpose of

A is the n m× matrix whose columns are formed by the rows of A. For example, if

1 2 3 4 0 6 B

= ⎢

,

then 1 3 0

2 4 6

BT

= ⎢

.

Let A be an n n× matrix. Let aij represent the entry in the ith row and jth column of matrix A. When the elements of the ith row and jth column of A are removed, the determinant of the remaining (n− × −1) (n 1) matrix is called the minor of aij, denoted mij. When this number is multiplied by ( )1 i j+ , it is called the cofactor of aij, denoted αij.

For example, consider

0 1 5 3 4 2 6 7 8 A

= ⎢

. Note that a23 =2. To find the minor of this entry, remove the second row and third column from the matrix.

0 1 6 7

The minor associated with a23 =2, is the determinant of this matrix. Thus, the minor of a23 =2 is 6. To find the cofactor associated with a23 =2, multiply the minor by ( )1 2 3+ = −( )15 = −1.

Thus, the cofactor of a23 =2, is 6.

The factor ( )1 i j+ is associated with a sign pattern that starts with + in the upper left corner moves left to right alternating from + to –. The sign pattern for 3 3× matrices is below.

+ − +

− + −

+ − +

The cofactor for any entry in a + position equals the associated minor while the cofactor for any entry in a – position equals the opposite of the associated minor.

0 1 5

3 4 2

6 7 8

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For example, consider

5 1 0

2 6 1

3 2 4

A

=

. To find the adjoint of A, first find all the minors.

The minor of 5 is 6 1

24 2 22

2 4

= − =

.

The minor of 1 is 2 1 8 ( )3 11

3 4

= − − = .

The minor of 0 is 2 6

4 18 22

3 2 = − − = −

.

The minor of 2 is 1 0

4 0 4

2 4 = − =

.

The minor of 6 is 5 0

20 0 20

3 4 = − = .

The minor of 1 is 5 1

10 3 13

3 2 = − − = −

.

The minor of 3 is 1 0

1 0 1

6 1= − − = −

.

The minor of 2 is 5 0

5 0 5

2 1= − − = −

.

The minor of 4 is 5 1

30 2 28

2 6 = − = .

Now construct the matrix of minors. Replace each entry in A with its associated minor.

22 11 22 4 20 13

1 5 28

− −

Next, multiply each i-jth minor by ( )1i j+ (or simply change the sign of any minor in a – position in the corresponding sign chart).

Let A be an n n× matrix whose entry in the ith row and jth column is denoted aij. Let αij be the cofactor of aij. The adjoint of A, denoted adj( )A , is the

transpose of the matrix of cofactors. That is, the adjoint of A is the matrix formed by replacing aij with αij for all i and j then transposing the result.

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22 11 22 4 20 13

1 5 28

Finally, transpose the result to obtain the adjoint of A.

( )

22 4 1

adj 11 20 5

22 13 28 A

= −

Instruction: The Inverse of a Matrix

Consider the matrix defined below.

In the set of real numbers, the number 1 is a special number called the multiplicative identity element because a⋅ = ⋅ =1 1 a a for any non-zero real number a. In the set of n n× matrices, In serves as the multiplicative identity element.

Since the set of n n× matrices has a multiplicative identity element, the question arises whether or not every n n× matrix has a multiplicative inverse. Recall that for any non-zero real number a there exist a real number called the multiplicative inverse of a, denoted a1, such that a a 1 =a1⋅ =a 1. An inverse matrix is defined similarly.

Given any arbitrary n n× non-zero matrixA, does there exist a multiplicative inverse, denoted A1, such that A A 1 =A1⋅ =A I? The answer is “No, but . . .”

An n n× matrix whose entries aij =0 whenever i j is called the n n× identity matrix, denoted In.

For example, 5

1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 I

=

.

n n

AI =I A=A for any non-zero n n× matrix A.

Let A be an n n× matrix. If det( )A =0, then A is a singular matrix. If det( )A 0, then A is a nonsingular matrix.

If A and B are n n× matrices such that AB=BA=In, then B is called the inverse of A, denoted B=A1, and A is called the inverse of B, denoted A=B1.

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In \, every non-zero real number has a multiplicative inverse. In the set of n n× matrices (with real entries), every nonsingular matrix has a multiplicative inverse.

For 2 2× matrices, calculating the inverse is simple.

For example, consider 1 5

2 6

W

= ⎢

.

( )

( )

1 1

1 6 2 5

1 6 10

1 16

6 5

16 16

2 1

16 16

3 5

8 16 1

1 1

8 16

6 5

2 1

6 5

2 1

6 5

2 1 W

W

⋅ − ⋅−

− −

=

=

= ⎢

= ⎢

= ⎢

Given the determinant and the adjoint of a nonsingular n n× matrix, the inverse matrix is a simple scalar product.

For example, consider

5 1 0

2 6 1

3 2 4

A

=

whose adjoint and determinant are

( )

22 4 1

adj 11 20 5

22 13 28 A

= −

and det( )A =99. The inverse of A is as follows.

If A is a nonsingular square matrix, then A is invertible, that is, A1 A A 1= A1⋅ =A I.

Let a b

A c d

= ⎢

. Then, 1 1 d b

A ad bc c a

= ⋅ ⎢ .

Let A be a nonsingular n n× matrix. Then,

( ) ( )

1 1

det adj

A A

A

= .

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1

22 4 1

99 99 99

20 5

11

99 99 99

13 28

22

99 99 99

2 4 1

9 99 99

1 1 20 5

9 99 99

13 28

2

9 99 99

22 4 1

1 11 20 5

99 22 13 28 A

A

=

= −

= −

Instruction: Matrix Equations

A matrix equation of the form AX =B can be solved using A1 if A is nonsingular.

Recall that matrix multiplication is not commutative, so solving the equation requires left- multiplication on both sides of the equation as follows.

1 1

1 1

AX B A AX A B

IX A B X A B

=

=

=

= For example, consider the equation below.

1 1 1 1

0 0 2 4

2 1 3 0

x y z

⎤ ⎛ ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟

⎜ ⎟ ⎜ ⎟=

⎜ ⎟ ⎜ ⎟

⎦ ⎝ ⎠ ⎝ ⎠

The inverse of

1 1 1 0 0 2 2 1 3

is

1 2 1

3 3 3

2 1 1

3 6 2

1

0 2 0

. Left multiply both sides of the equation to solve the equation as follows.

1 2 1 1 2 1

3 3 3 3 3 3

2 1 1 2 1 1

3 6 2 3 6 2

1 1

2 2

1 1 1 1

0 0 2 4

2 1 3 0

1 1 1 1

0 0 2 4

2 1 3 0

0 0 0 0

x y z x y z

⎤ ⎛ ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟

⎜ ⎟ ⎜ ⎟=

⎜ ⎟ ⎜ ⎟

⎦ ⎝ ⎠ ⎝ ⎠

⎤ ⎛ ⎞ ⎛ ⎞

⎜ ⎟= − ⎜ ⎟

⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟

⎦ ⎝ ⎠ ⎝ ⎠

Remember that A A1 =In and I Xn = X .

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1 2 1

3 3 3

2 1 1

3 6 2

1 2 8 1

3 3

2 4

3 6

1 3 2 3

1 0 0

0 1 0 1 4 0

0 0 1 0 0

0

0

0 2 0

x y z x y z x y z

⎛ ⎞

⎤ ⎛ ⎞

⎜ ⎟

⎜ ⎟⎜ ⎟= − + + ⎜ ⎟

⎜ ⎟ ⎜ ⎟

⎦ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞

⎜ ⎟= − + − +⎜ ⎟

⎜ ⎟ ⎜ ⎟

⎜ ⎟ ⎟ ⎜ ⎜ ⎟

⎝ ⎠ ⎠ ⎝ ⎝ ⎠

⎛ ⎞⎜ ⎟ = −

⎜ ⎟⎜ ⎟

⎝ ⎠

8 3 2 3

7 3 4 3

0 2

2 x

y z

⎟ + −

⎟ ⎜

⎠ ⎝

⎛ ⎞

⎜ ⎟ = −

⎜ ⎟⎜ ⎟ ⎜

⎝ ⎠ ⎝

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Instruction: Adjoint Matrices

Example

Finding the Adjoint of a 3×3 Matrix

Find the minors for each entry.

( ) ( )

1 2 11

1 2 12

13

1 0 0 0

0 0

2 0 2 2

4 0 2 1

0 4 4

4 0 m

m

m

= = − =

= = − − =

= = − − =

( )

21

22

23

4 0

0 0 0

0 0

1 0 0 0 0

4 0

1 4 0 16 16

4 0 m

m m

= = − =

= = − =

= = − − =

31 1

2

1 1

32 1 2 2

2

33

4 0

2 0 2 1

1 0 0

2

1 4 1 8 9

2 1 m

m

m

= = − =

= = − − = −

= = − − = −

Replace each entry in B with the corresponding minor.

1 2

0 2 4

0 0 16

2 9

Rewrite this matrix of minors as a matrix of cofactors by changing the signs of the appropriate entries.

1 2

0 2 4

0 0 16

2 9

Find the adjoint of B by transposing this matrix of cofactors. To transpose, rewrite the rows as columns.

( ) 12

0 0 2

2 0

4 16 9

adj B

= −

Given 12

1 4 0 2 1 4 0 0 B

= ⎢

. Find adj( )B .

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Instruction: Inverse Matrices

Example 1

Finding the Inverse of a 2×2 Matrix

The inverse of a 2×2 matrix is defined below.

11 12 1 22 12

21 22 11 22 21 12 21 11

If , c c then 1 c c

A A

c c c c c c c c

= = .

Accordingly, the inverse of a 2×2 matrix is defined if the determinant of the matrix does not equal zero.

( ) 1

5 3 0

2

15 0

2 2 15

30 h h h h

− + ≠

The inverse of B is defined for all values of h except 30.

Given

5 1 2 3 B

h

=

, find h such that B1 is defined.

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Example 2

Finding the Inverse of a 3×3 Matrix

Write the matrix of minors.

9 18 4

21 56 14

1 2 2

Change the matrix of minors into a matrix of cofactors. To do this, use the 3 3× sign chart to change the signs of entries appropriately.

9 18 4

21 56 14

1 2 2

Find the adjoint of the matrix of cofactors, that is, transpose the matrix of cofactors.

9 21 1

18 56 2

4 14 2

Divide every element in the adjoint of the matrix of cofactors by the determinant. The result is the inverse of C.

9 3 1

14 2 14

1 9 1

7 7

2 1

7 7

4 1 C

= −

Given

6 2 1

2 1 0

2 3 9

C

= ⎢

, find C1.

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Example 3 Verifying Inverses

Recall that if L2 2× and K2 2× are inverses of each other, then L2 2× K2 2× =I2 2× . Assume the elements of L2 2× K2 2× are c11, c12, c21, and c22, then

11 12

21 22

1 0 0 1 c c

c c

⎤ ⎡

⎥ ⎢=

.

Show that c11 =1 and c12 =0.

( ) ( )

11 11

sum of products of elements in row one of and corresponding elements in column one of 2 3 2 4 1 2 3 2 1

c L K

c

=

= − − + = − =

( ) ( )

12 12

sum of products of elements in row one of and corresponding elements in column two of 2 2 4 1 4 4 0

c L K

c

=

= − + = − + =

Find c21 and c22 in terms of h.

( ) ( )

21

21

sum of products of elements in row two of and corresponding elements in column one of

3 3

3 2 3 1 2

2 2

c L K

c h h

=

= + − = −

( ) ( )

22 22

sum of products of elements in row two of and corresponding elements in column two of

2 3 1 2 3

c L K

c h h

=

= + = +

Set c21 equal to zero and c22 equal to one. Solve the two equations. If h as the same solution for each equation, then the solution is the value of h such that L is the inverse of K.

3 3

2 2 0

3 3

2 2

1 h h h

− =

=

= −

2 3 1

2 2

1 h h h

+ =

= −

= − L and K are inverses when h = –1.

Given 2 4

L 3 h

= ⎢

& 3 2 2

1 2 1

K

= ⎢ , find h such that L and K are inverses.

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Instruction: Matrix Equations

Example

Solving a Matrix Equation

Left multiply both sides by the given inverse.

3

1 1

8 2 8 1

2

3 1 1

3

8 2 8

1 1 3

8 2 8 1

2

3 1 1

3

8 2 8

4

5 0 1 0 1

1 2 3 4 1 2 3

0 1 0 5 0 1 0 0 1 0

3 2 1 1 3 2 1

x x x

x x x

⎤ ⎛ ⎞

⎛ ⎞ ⎥ ⎜ ⎟

⎜ ⎟ = ⎥ ⎜ ⎟

⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

⎤ ⎛ ⎞

⎤ ⎛ ⎞ ⎡

⎥ ⎜ ⎟

⎜ ⎟⎜ ⎟= ⎥ ⎜ ⎟

⎜ ⎟ ⎜ ⎟

⎦ ⎝ ⎠ ⎣ ⎝ ⎠

Recall that A A1 =In and I Xn =X .

1 2 3 1

2 3 1

2 3

1 2 3 1 0 0

4 0 5 1 1 0 0 1 0

3 2 1 0 0 1

4 10 3

0 5 0

12 10 1

17 5

23

x x x x

x x x x x

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎡ ⎛ ⎞

⎜ ⎟

⎜ ⎟+ ⎜ ⎟+ ⎜ ⎟ ⎢= ⎜ ⎟

⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎢

⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎣ ⎦ ⎝ ⎠

⎛ ⎞

⎛ ⎞ ⎛ ⎞ ⎛ ⎞

⎜ ⎟

⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ + = ⎜ ⎟

⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞

⎛ ⎞ ⎜ ⎟

⎜ ⎟ =⎜ ⎟

⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ Solve

3

1 1

8 2 8 1

2

3 1 1

3

8 2 8

4

5 0 1 0

1

x x x

⎤ ⎛ ⎞

⎛ ⎞ ⎥ ⎜ ⎟

⎜ ⎟ = ⎥ ⎜ ⎟

⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

given that

3 1

1 1

8 2 8

3 1 1

8 2 8

1 2 3

0 1 0 0 1 0

3 2 1

=

.

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Suggested Homework from Blitzer Section 6.4: #1, #9, #13-17 odd.

Find the adjoint, determinant, and inverse of each matrix in #19-25 odd.

Solve the matrix equations in #33 and #35.

Application Exercise

Consider a horizontal elastic beam supported at each end. Imagine forces applied to the beam at points 1, 2, and 3 as shown in the figure below.

1

2

deflection at point 2

3

Let f 3 list the forces at these three points and let y 3 list the corresponding deflection quantities (the amount of movement) of the beam. Using Hooke’s law from physics, engineers derive the equation yDfwhere D is a flexibility matrix particular to a given beam. Suppose that for a particular beam the flexibility matrix is below.

0.005 0.002 0.001 0.002 0.004 0.002 0.001 0.002 0.005 D

 

Find the forces necessary to deflect the beam 0.02, 0.04, and 0.02 meters at points 1, 2, and 3 respectively.

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ANSWERS

#1 1 17 3 2 1 3 A

  #3

1 3 7

G 1 2

 

#5

1 4

1 1

3 1 2

0 0

0 0

0 0

R

#7 1 3 2 3 1 0 1 2 1 1 0 1

   

   

    #9

5 2

x 1 y

 

   

     Practice Problems

#1 If ,

3 1

2

3

A find A–1. #2 If

2

9 3

7 2 ,

B

  find B–1.

#3 If ,

3 1

7

2

G find G –1. #4 If ,

5 3

3

2

D find D –1.

#5 If

4 0 0

0 3 0 ,

0 0 2

R

find R –1. #6 If

0 0 3

1 2

2

1 0 1

Y , find Y –1.

#7 Use the product C∙D to verify that C and D are inverses of each other.

2 1

3

C 1 ,

1 1

3 D 2

#8 Use the product A∙B to verify that A and B are inverses of each other.

5 1 0

2 2 2

5 4 1

A ,

15

4 1

11 22 11

5 5 4

11 22 11

3

1 1

11 22 11

B

#9 Solve the matrix equation: 1 2 6

1 4 9

x y

    

   

    .

#10 Solve the matrix equation

1 2 3

1 0 0 10

0 2 0 20

0 0 3 30

x x x

 

   

   

 

 

given that

1

1 2

1 3

1 0 0

1 0 0

0 2 0 0 0

0 0 3 0 0

 

.

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