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
⎡ ⎤
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎣ ⎦
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.
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 a−1, such that a a⋅ −1 =a−1⋅ =a 1. An inverse matrix is defined similarly.
Given any arbitrary n n× non-zero matrixA, does there exist a multiplicative inverse, denoted A−1, such that A A⋅ −1 =A−1⋅ =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=A−1, and A is called the inverse of B, denoted A=B−1.
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, ∃ A−1 ∋ A A⋅ −1= A−1⋅ =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
− = ⋅ .
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 A−1 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 A−1 =In and I Xn = X .
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
⎛− ⎞
⎛ ⎞
⎜ ⎟
⎜ ⎟ + −⎜ ⎟
⎜ ⎟
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
⎛− ⎞
⎛ ⎞ ⎜ ⎟
⎜ ⎟ = −⎜ ⎟
⎜ ⎟⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
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 .
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 B−1 is defined.
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 C−1.
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.
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 A−1 =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
− − −
⎡ ⎤ ⎡ ⎤
⎢ ⎥ =⎢ ⎥
⎢ ⎥ ⎢ ⎥
⎢ − − ⎥ ⎢⎣ ⎥⎦
⎣ ⎦
.
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.
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
.