Uploaded bymath260
2,034 views

6.4 inverse matrices

The document discusses inverse matrices. It begins by explaining that the inverse of a nonzero number a is its reciprocal 1/a. It then states that the inverse of a matrix A, written as A-1, is a matrix such that AA-1 = A-1A = I, where I is the identity matrix. However, not all matrices have inverses. There are two types of matrices regarding inverses: invertible matrices that have inverses, and noninvertible matrices that do not. The document provides an example of finding the inverse of a matrix and discusses when a matrix may not have an inverse.

Embed presentation

Inverse Matrices
Inverse Matrices For numbers, the multiplicative inverse of a nonzero number a is the reciprocal 1/a.
Inverse Matrices For numbers, the multiplicative inverse of a nonzero number a is the reciprocal 1/a. The most important property of 1/a is that 1 a * a = a * 1 a = 1
Inverse Matrices For numbers, the multiplicative inverse of a nonzero number a is the reciprocal 1/a. The most important property of 1/a is that 1 a * a = a * 1 a = 1 The most important usage of the inverse 1/a is to solve the equation ax = b.
Inverse Matrices For numbers, the multiplicative inverse of a nonzero number a is the reciprocal 1/a. The most important property of 1/a is that 1 a * a = a * 1 a = 1 The most important usage of the inverse 1/a is to solve the equation ax = b. For example, to solve 2x = 3
Inverse Matrices For numbers, the multiplicative inverse of a nonzero number a is the reciprocal 1/a. The most important property of 1/a is that 1 a * a = a * 1 a = 1 The most important usage of the inverse 1/a is to solve the equation ax = b. For example, to solve 2x = Multiply both sides by ½½ * ½ * 3 x = 3/2
Inverse Matrices For numbers, the multiplicative inverse of a nonzero number a is the reciprocal 1/a. The most important property of 1/a is that 1 a * a = a * 1 a = 1 The most important usage of the inverse 1/a is to solve the equation ax = b. For example, to solve 2x = Multiply both sides by ½½ * x = 3/2 Let A be an n x n matrix. ½ * 3
Inverse Matrices For numbers, the multiplicative inverse of a nonzero number a is the reciprocal 1/a. The most important property of 1/a is that 1 a * a = a * 1 a = 1 The most important usage of the inverse 1/a is to solve the equation ax = b. For example, to solve 2x = Multiply both sides by ½½ * x = 3/2 Let A be an n x n matrix. Since In plays the role of 1 in multiplication, we say B is the inverse of A if AB = BA = In and we write B as A-1. ½ * 3
Inverse Matrices For numbers, the multiplicative inverse of a nonzero number a is the reciprocal 1/a. The most important property of 1/a is that 1 a * a = a * 1 a = 1 The most important usage of the inverse 1/a is to solve the equation ax = b. For example, to solve 2x = Multiply both sides by ½½ * x = 3/2 We can use A-1 to solve matrix equations AX = C, i.e. to find a matrix X where A and C are matrices. Let A be an n x n matrix. Since In plays the role of 1 in multiplication, we say B is the inverse of A if AB = BA = In and we write B as A-1. ½ * 3
1 0 0 . . 0 0 1 0 . . . 0 0 1 . . . 0 0 . . . 1 . . . . . . Inverse Matrices AA-1 = A-1A = = In The matrices A, A-1 is a pair of inverse matrices if
1 0 0 . . 0 0 1 0 . . . 0 0 1 . . . 0 0 . . . 1 . . . . . . Inverse Matrices AA-1 = A-1A = = In Where as the inverse of nonzero numbers always exist, this is not the case for matrices. The matrices A, A-1 is a pair of inverse matrices if
1 0 0 . . 0 0 1 0 . . . 0 0 1 . . . 0 0 . . . 1 . . . . . . Inverse Matrices AA-1 = A-1A = = In Where as the inverse of nonzero numbers always exist, this is not the case for matrices. There're two types of matrices with regard to inverses. The matrices A, A-1 is a pair of inverse matrices if
1 0 0 . . 0 0 1 0 . . . 0 0 1 . . . 0 0 . . . 1 . . . . . . Inverse Matrices AA-1 = A-1A = = In I. The ones that have inverses, these are called invertible (or nonsingular) matrices The matrices A, A-1 is a pair of inverse matrices if Where as the inverse of nonzero numbers always exist, this is not the case for matrices. There're two types of matrices with regard to inverses.
1 0 0 . . 0 0 1 0 . . . 0 0 1 . . . 0 0 . . . 1 . . . . . . Inverse Matrices AA-1 = A-1A = = In I. The ones that have inverses, these are called invertible (or nonsingular) matrices II. The ones that don‘t have an inverse are called noninvertible (or singular) matrices The matrices A, A-1 is a pair of inverse matrices if Where as the inverse of nonzero numbers always exist, this is not the case for matrices. There're two types of matrices with regard to inverses.
An example of a noninvertible matrix is Inverse Matrices 0 -2 0 3
Inverse Matrices because for any matrix , 0 -2 0 3 a b c d = 0 # 0 # a b c d An example of a noninvertible matrix is 0 -2 0 3
Inverse Matrices because for any matrix , 0 -2 0 3 a b c d = 0 # 0 # = 1 0 0 1 a b c d An example of a noninvertible matrix is 0 -2 0 3
Inverse Matrices because for any matrix , 0 -2 0 3 a b c d = 0 # 0 # = 1 0 0 1 a b c d Method for Finding Inverse Matrices: An example of a noninvertible matrix is 0 -2 0 3
Inverse Matrices To find the inverse, A-1 of a matrix A, extend the matrix A by the identity matrix I on the right as shown, because for any matrix , 0 -2 0 3 a b c d = 0 # 0 # = 1 0 0 1 a b c d Method for Finding Inverse Matrices: A A I An example of a noninvertible matrix is 0 -2 0 3
Inverse Matrices To find the inverse, A-1 of a matrix A, extend the matrix A by the identity matrix I on the right as shown, because for any matrix , 0 -2 0 3 a b c d = 0 # 0 # = 1 0 0 1 a b c d Method for Finding Inverse Matrices: A A I Apply row operations to tranform it so the identity matrix I is on the left side as , then B = A-1.I B An example of a noninvertible matrix is 0 -2 0 3
Example A: Find the inverse matrix A-1 if Inverse Matrices a. A = 1 2 1 3
Example A: Find the inverse matrix A-1 if Inverse Matrices a. A = 1 2 1 3 Extend 1 2 1 3 to 1 2 1 0 1 3 0 1
Example A: Find the inverse matrix A-1 if Inverse Matrices a. A = 1 2 1 3 Extend 1 2 1 3 to 1 2 1 0 1 3 0 1 Apply row operations to tranform it so the identity matrix I is on the left side:
Example A: Find the inverse matrix A-1 if Inverse Matrices a. A = 1 2 1 3 Extend 1 2 1 3 to 1 2 1 0 1 3 0 1 Apply row operations to tranform it so the identity matrix I is on the left side: 1 2 1 0 1 3 0 1 (-1)R1 Add to R2
Example A: Find the inverse matrix A-1 if Inverse Matrices a. A = 1 2 1 3 Extend 1 2 1 3 to 1 2 1 0 1 3 0 1 Apply row operations to tranform it so the identity matrix I is on the left side: 1 2 1 0 1 3 0 1 (-1)R1 Add to R2 -1 -2 -1 0
Example A: Find the inverse matrix A-1 if Inverse Matrices a. A = 1 2 1 3 Extend 1 2 1 3 to 1 2 1 0 1 3 0 1 Apply row operations to tranform it so the identity matrix I is on the left side: 1 2 1 0 1 3 0 1 (-1)R1 Add to R2 -1 -2 -1 0 1 2 1 0 0 1 -1 1
Example A: Find the inverse matrix A-1 if Inverse Matrices a. A = 1 2 1 3 Extend 1 2 1 3 to 1 2 1 0 1 3 0 1 Apply row operations to tranform it so the identity matrix I is on the left side: 1 2 1 0 1 3 0 1 (-1)R1 Add to R2 -1 -2 -1 0 1 2 1 0 0 1 -1 1 (-2)R2 Add to R1
Example A: Find the inverse matrix A-1 if Inverse Matrices a. A = 1 2 1 3 Extend 1 2 1 3 to 1 2 1 0 1 3 0 1 Apply row operations to tranform it so the identity matrix I is on the left side: 1 2 1 0 1 3 0 1 (-1)R1 Add to R2 -1 -2 -1 0 1 2 1 0 0 1 -1 1 0 -2 2 -2 (-2)R2 Add to R1
Example A: Find the inverse matrix A-1 if Inverse Matrices a. A = 1 2 1 3 Extend 1 2 1 3 to 1 2 1 0 1 3 0 1 Apply row operations to tranform it so the identity matrix I is on the left side: 1 2 1 0 1 3 0 1 (-1)R1 Add to R2 -1 -2 -1 0 1 2 1 0 0 1 -1 1 0 -2 2 -2 1 0 3 -2 0 1 -1 1 (-2)R2 Add to R1
Example A: Find the inverse matrix A-1 if Inverse Matrices a. A = 1 2 1 3 Extend 1 2 1 3 to 1 2 1 0 1 3 0 1 Apply row operations to tranform it so the identity matrix I is on the left side: 1 2 1 0 1 3 0 1 (-1)R1 Add to R2 -1 -2 -1 0 1 2 1 0 0 1 -1 1 0 -2 2 -2 1 0 3 -2 0 1 -1 1 (-2)R2 Add to R1 = identity matrix
Example A: Find the inverse matrix A-1 if Inverse Matrices a. A = 1 2 1 3 Extend 1 2 1 3 to 1 2 1 0 1 3 0 1 Apply row operations to tranform it so the identity matrix I is on the left side: 1 2 1 0 1 3 0 1 (-1)R1 Add to R2 -1 -2 -1 0 1 2 1 0 0 1 -1 1 0 -2 2 -2 1 0 3 -2 0 1 -1 1 (-2)R2 Add to R1 = identity matrix Hence A-1 = 3 -2 -1 1 .
Example A: Find the inverse matrix A-1 if Inverse Matrices a. A = 1 2 1 3 Extend 1 2 1 3 to 1 2 1 0 1 3 0 1 Apply row operations to tranform it so the identity matrix I is on the left side: 1 2 1 0 1 3 0 1 (-1)R1 Add to R2 -1 -2 -1 0 1 2 1 0 0 1 -1 1 0 -2 2 -2 1 0 3 -2 0 1 -1 1 (-2)R2 Add to R1 = identity matrix Hence A-1 = 3 -2 -1 1 . One checks easily that AA-1 = A-1A = 1 0 0 1 .
Inverse Matrices b. A = 0 2 0 3
Inverse Matrices b. A = 0 2 0 3 Extend 0 2 0 3 to 0 2 1 0 0 3 0 1
Inverse Matrices b. A = 0 2 0 3 Extend 0 2 0 3 to 0 2 1 0 0 3 0 1 Since the first column consists of 0's only, there is no way we may introduce a 1 at the a11 position by row operations.
Inverse Matrices b. A = 0 2 0 3 Extend 0 2 0 3 to 0 2 1 0 0 3 0 1 Since the first column consists of 0's only, there is no way we may introduce a 1 at the a11 position by row operations. Hence A-1 doesn't exist.
Inverse Matrices b. A = 0 2 0 3 Extend 0 2 0 3 to 0 2 1 0 0 3 0 1 Since the first column consists of 0's only, there is no way we may introduce a 1 at the a11 position by row operations. Hence A-1 doesn't exist. c. A = 0 1 -1 1 -1 2 1 1 0
Inverse Matrices b. A = 0 2 0 3 Extend 0 2 0 3 to 0 2 1 0 0 3 0 1 Since the first column consists of 0's only, there is no way we may introduce a 1 at the a11 position by row operations. Hence A-1 doesn't exist. c. A = 0 1 -1 1 -1 2 1 1 0 Extend 0 1 -1 1 -1 2 1 1 0 0 1 -1 1 0 0 1 -1 2 0 1 0 1 1 0 0 0 1
Inverse Matrices Apply row operations: 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 R1 R2
Inverse Matrices Apply row operations: 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 R1 R2
Inverse Matrices Apply row operations: 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 R1 R2 (-1)R1 add to R3
Inverse Matrices Apply row operations: 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 R1 R2 -1 -1 -2 0 -1 0 (-1)R1 add to R3
Inverse Matrices Apply row operations: 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 R1 R2 -1 -1 -2 0 -1 0 0 1 1 1 0 0 1 1 2 0 1 0 0 0 -2 0 -1 1 (-1)R1 add to R3
Inverse Matrices Apply row operations: 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 (-1)R2 add to R1 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 R1 R2 -1 -1 -2 0 -1 0 0 1 1 1 0 0 1 1 2 0 1 0 0 0 -2 0 -1 1 (-1)R1 add to R3
Inverse Matrices Apply row operations: 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 R1 R2 -1 -1 -2 0 -1 0 0 1 1 1 0 0 1 1 2 0 1 0 0 0 -2 0 -1 1 0 -1 -1 -1 0 0 (-1)R1 add to R3 (-1)R2 add to R1
Inverse Matrices Apply row operations: 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 (-1)R2 add to R1 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 R1 R2 -1 -1 -2 0 -1 0 0 1 1 1 0 0 1 1 2 0 1 0 0 0 -2 0 -1 1 0 -1 -1 -1 0 0 0 1 -1 1 0 0 1 0 1 -1 1 0 0 0 -2 0 -1 1 (-1)R1 add to R3
Inverse Matrices Apply row operations: 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 (-1)R2 add to R1 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 R1 R2 -1 -1 -2 0 -1 0 0 1 1 1 0 0 1 1 2 0 1 0 0 0 -2 0 -1 1 0 -1 -1 -1 0 0 0 1 -1 1 0 0 1 0 1 -1 1 0 0 0 -2 0 -1 1 (-1)R1 add to R3 (-1/2)R3
Inverse Matrices Apply row operations: 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 (-1)R2 add to R1 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 R1 R2 -1 -1 -2 0 -1 0 0 1 1 1 0 0 1 1 2 0 1 0 0 0 -2 0 -1 1 0 -1 -1 -1 0 0 0 1 -1 1 0 0 1 0 1 -1 1 0 0 0 -2 0 -1 1 (-1)R1 add to R3 0 1 -1 1 0 0 1 0 1 -1 1 0 0 0 1 0 1/2 -1/2 (-1/2)R3
Inverse Matrices Apply row operations: 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 (-1)R2 add to R1 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 R1 R2 -1 -1 -2 0 -1 0 0 1 1 1 0 0 1 1 2 0 1 0 0 0 -2 0 -1 1 0 -1 -1 -1 0 0 0 1 -1 1 0 0 1 0 1 -1 1 0 0 0 -2 0 -1 1 (-1)R1 add to R3 0 1 -1 1 0 0 1 0 1 -1 1 0 0 0 1 0 1/2 -1/2 (-1)R3 + R1 (-1/2)R3 R3 + R2
Inverse Matrices Apply row operations: 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 (-1)R2 add to R1 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 R1 R2 -1 -1 -2 0 -1 0 0 1 1 1 0 0 1 1 2 0 1 0 0 0 -2 0 -1 1 0 -1 -1 -1 0 0 0 1 -1 1 0 0 1 0 1 -1 1 0 0 0 -2 0 -1 1 (-1)R1 add to R3 0 1 -1 1 0 0 1 0 1 -1 1 0 0 0 1 0 1/2 -1/2 0 0 -1 0 -1/2 1/2 (-1)R3 + R1 (-1/2)R3 R3 + R2
Inverse Matrices Apply row operations: 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 (-1)R2 add to R1 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 R1 R2 -1 -1 -2 0 -1 0 0 1 1 1 0 0 1 1 2 0 1 0 0 0 -2 0 -1 1 0 -1 -1 -1 0 0 0 1 -1 1 0 0 1 0 1 -1 1 0 0 0 -2 0 -1 1 (-1)R1 add to R3 0 1 0 1 -1/2 1/2 1 0 0 -1 1/2 1/2 0 1 -1 1 0 0 1 0 1 -1 1 0 0 0 1 0 1/2 -1/2 0 0 -1 0 -1/2 1/2 (-1)R3 + R1 (-1/2)R3 R3 + R2 0 0 1 0 1/2 -1/2
Inverse Matrices Apply row operations: 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 (-1)R2 add to R1 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 R1 R2 -1 -1 -2 0 -1 0 0 1 1 1 0 0 1 1 2 0 1 0 0 0 -2 0 -1 1 0 -1 -1 -1 0 0 0 1 -1 1 0 0 1 0 1 -1 1 0 0 0 -2 0 -1 1 (-1)R1 add to R3 0 1 0 1 -1/2 1/2 1 0 0 -1 1/2 1/2 0 1 -1 1 0 0 1 0 1 -1 1 0 0 0 1 0 1/2 -1/2 0 0 -1 0 -1/2 1/2 (-1)R3 + R1 (-1/2)R3 R3 + R2 0 0 1 0 1/2 -1/2 = A-1
Inverse Matrices Apply row operations: 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 (-1)R2 add to R1 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 R1 R2 -1 -1 -2 0 -1 0 0 1 1 1 0 0 1 1 2 0 1 0 0 0 -2 0 -1 1 0 -1 -1 -1 0 0 0 1 -1 1 0 0 1 0 1 -1 1 0 0 0 -2 0 -1 1 (-1)R1 add to R3 0 1 0 1 -1/2 1/2 1 0 0 -1 1/2 1/2 0 1 -1 1 0 0 1 0 1 -1 1 0 0 0 1 0 1/2 -1/2 0 0 -1 0 -1/2 1/2 (-1)R3 + R1 (-1/2)R3 R3 + R2 0 0 1 0 1/2 -1/2 = A-1We check that A-1*A = A*A-1 = I.
Inverse Matrices Apply row operations: 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 (-1)R2 add to R1 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 R1 R2 -1 -1 -2 0 -1 0 0 1 1 1 0 0 1 1 2 0 1 0 0 0 -2 0 -1 1 0 -1 -1 -1 0 0 0 1 -1 1 0 0 1 0 1 -1 1 0 0 0 -2 0 -1 1 (-1)R1 add to R3 0 1 0 1 -1/2 1/2 1 0 0 -1 1/2 1/2 0 1 -1 1 0 0 1 0 1 -1 1 0 0 0 1 0 1/2 -1/2 0 0 -1 0 -1/2 1/2 (-1)R3 + R1 (-1/2)R3 R3 + R2 0 0 1 0 1/2 -1/2 = A-1We check that A-1*A = A*A-1 = I. Let’s use an inverse matrix to solve a matrix equation.
Example B: a. Write the following system as a matrix equation x + 2y = 4 x + 3y = -1 An Application of Inverse Matrices
Example B: a. Write the following system as a matrix equation x + 2y = 4 x + 3y = -1 The system may be expressed as 1 2 1 3 x y = 4 -1 An Application of Inverse Matrices
Example B: a. Write the following system as a matrix equation If we let A = 1 2 1 3 x + 2y = 4 x + 3y = -1 , X = x y , The system may be expressed as 1 2 1 3 x y = 4 -1 then the matrix equation is simply AX = B. and B = 4 -1 . An Application of Inverse Matrices
Example B: a. Write the following system as a matrix equation If we let A = 1 2 1 3 x + 2y = 4 x + 3y = -1 , X = x y , The system may be expressed as 1 2 1 3 x y = 4 -1 then the matrix equation is simply AX = B. and B = 4 -1 . An Application of Inverse Matrices To solve for the matrix X, we use the following inverse matrix method.
To solve the equation AX = B for the matrix X, we multiply to the left of both sides by A-1 (if it exists) An Application of Inverse Matrices
To solve the equation AX = B for the matrix X, we multiply to the left of both sides by A-1 (if it exists) AX = B An Application of Inverse Matrices
To solve the equation AX = B for the matrix X, we multiply to the left of both sides by A-1 (if it exists) AX = B A-1AX = A-1B An Application of Inverse Matrices
To solve the equation AX = B for the matrix X, we multiply to the left of both sides by A-1 (if it exists) AX = B A-1AX = A-1B IX = A-1B or that X = A-1B An Application of Inverse Matrices
To solve the equation AX = B for the matrix X, we multiply to the left of both sides by A-1 (if it exists) AX = B A-1AX = A-1B IX = A-1B or that X = A-1B b. Solve the matrix equation AX = B from part a. by the inverse matrix method. An Application of Inverse Matrices
To solve the equation AX = B for the matrix X, we multiply to the left of both sides by A-1 (if it exists) AX = B A-1AX = A-1B IX = A-1B or that X = A-1B b. Solve the matrix equation AX = B from part a. by the inverse matrix method. from the previous example we've that Given that A = 1 2 1 3 3 -2 -1 1 . A-1 = An Application of Inverse Matrices
Multiply A-1 to the left on both sides of the equation Inverse Matrices 1 2 1 3 x y = 4 -1
Multiply A-1 to the left on both sides of the equation Inverse Matrices 1 2 1 3 x y = 4 -1 we get 3 -2 -1 1 1 2 1 3 x y = 3 -2 -1 1 4 -1
Multiply A-1 to the left on both sides of the equation Inverse Matrices 1 2 1 3 x y = 4 -1 we get 3 -2 -1 1 1 2 1 3 x y = 3 -2 -1 1 4 -1 0 1 1 0 x y = 14 -5
Multiply A-1 to the left on both sides of the equation Inverse Matrices 1 2 1 3 x y = 4 -1 we get 3 -2 -1 1 1 2 1 3 x y = 3 -2 -1 1 4 -1 0 1 1 0 x y = 14 -5 Hence x y = 14 -5 or that x = 14, and y = -5.
Multiply A-1 to the left on both sides of the equation Inverse Matrices 1 2 1 3 x y = 4 -1 we get 3 -2 -1 1 1 2 1 3 x y = 3 -2 -1 1 4 -1 0 1 1 0 x y = 14 -5 Hence x y = 14 -5 or that x = 14, and y = -5. With inverse matrices, we may do matrix algebra that is similar to the “traditional” algebra.
Inverse Matrices Exercise A. Find the inverse matrix of each of the following matrices, if it exits 1 0 0 1 1) 2) 3) 4)1 5 0 1 a 5 0 𝑏 0 𝑏 𝑎 5 3 5 –2 1 5) 6) 7) 8) 1 –2 7 –1 𝑎 𝑏 𝑏 𝑎 𝑎 𝑏 𝑎 𝑏 9) 10) 11) 12) 1 0 0 0 1 0 0 0 1 1 2 3 0 1 4 0 0 1 a 0 0 0 b 0 0 0 c a 2 3 0 b 4 0 0 c 13) 14) 1 0 2 0 1 0 −1 0 1 1 2 3 0 1 4 −1 2 1
2. Verify the following inverse formula for 2x2 matrices. 𝑎 𝑏 𝑐 𝑑 –1 = 1 ad – bc 𝑑 –𝑏 –𝑐 𝑎 (Hint: you don’t have to compute the inverse, just check that their product is I so it must be the one and only inverse.) 3. Verify the that (AB) –1 = B –1 A–1. So if A and B are invertible then AB is invertible. Ex B. A matrix that has an inverse is said to be invertible. 1. (A–1 is unique) Verify that if A is invertible, then there is only one A–1. This means to show that if there are matrices B and C that invert A so that AB = BA = I and AC = CA = I, then B = C. 4. Verify that if AB is invertible, then A and B must be invertible. (Hint: Let C is (AB) –1, then A–1 or B –1 can be “solved”.) So if A or B is not invertible then AB is not invertible. Inverse Matrices
(Answers to the odd problems) Exercise A. 𝐴−1 = 1 0 0 1 1) 𝐴−1 = 1/𝑎 −5/𝑎𝑏 0 1/𝑏 𝐴−1 = 1 13 1 −5 2 3 5) 𝐴−1 = 1 𝑎2 − 𝑏2 𝑎 −𝑏 −𝑏 𝑎 9) 𝐴−1 = 1 0 0 0 1 0 0 0 1 𝐴−1 = 1/a 0 0 0 1/b 0 0 0 1/c 13) 𝐴−1 = 1 3 1 0 −2 0 3 0 1 0 1 3) 7) 11) Inverse Matrices

More Related Content

PPTX
6.3 matrix algebra
bymath260
 
PPTX
20 methods of division x
bymath260
 
PPTX
28 more on log and exponential equations x
bymath260
 
PPTX
18Ellipses-x.pptx
bymath260
 
PPTX
17 conic sections circles-x
bymath260
 
PPTX
6.1 system of linear equations and matrices
bymath260
 
PPTX
6.2 special cases system of linear equations
bymath260
 
PPT
Systems of linear equations
bygandhinagar
 
6.3 matrix algebra
bymath260
 
20 methods of division x
bymath260
 
28 more on log and exponential equations x
bymath260
 
18Ellipses-x.pptx
bymath260
 
17 conic sections circles-x
bymath260
 
6.1 system of linear equations and matrices
bymath260
 
6.2 special cases system of linear equations
bymath260
 
Systems of linear equations
bygandhinagar
 

What's hot

PPTX
13 graphs of factorable polynomials x
bymath260
 
PPTX
22 the fundamental theorem of algebra x
bymath260
 
PPTX
25 continuous compound interests perta x
bymath260
 
PPTX
12 graphs of second degree functions x
bymath260
 
PPTX
8 inequalities and sign charts x
bymath260
 
PPTX
4 5 fractional exponents
bymath123b
 
PPT
DISCRIMINANT.ppt
byNelsonNelson56
 
PPTX
9 the basic language of functions x
bymath260
 
PPTX
5.2 arithmetic sequences and sums
bymath260
 
PPTX
3.1 Characteristics of Polynomial Functions.pptx
byAlaa480924
 
PPT
8.2 Exploring exponential models
byswartzje
 
PPT
7.5 proportions in triangles
byRana Kkkk
 
PPT
Systemsof3 equations
bygandhinagar
 
PPTX
7.2 Alg 2 exploring exponential models
byswartzje
 
PPTX
Types of function
bySanaullahMemon10
 
PPT
L4 one sided limits limits at infinity
byJames Tagara
 
PPTX
10 3 double and half-angle formulas
byhisema01
 
PPT
Lu decomposition
bygilandio
 
PPTX
Linear Equations
byAdjex Academy
 
PPT
5 1 quadratic transformations
bylothomas
 
13 graphs of factorable polynomials x
bymath260
 
22 the fundamental theorem of algebra x
bymath260
 
25 continuous compound interests perta x
bymath260
 
12 graphs of second degree functions x
bymath260
 
8 inequalities and sign charts x
bymath260
 
4 5 fractional exponents
bymath123b
 
DISCRIMINANT.ppt
byNelsonNelson56
 
9 the basic language of functions x
bymath260
 
5.2 arithmetic sequences and sums
bymath260
 
3.1 Characteristics of Polynomial Functions.pptx
byAlaa480924
 
8.2 Exploring exponential models
byswartzje
 
7.5 proportions in triangles
byRana Kkkk
 
Systemsof3 equations
bygandhinagar
 
7.2 Alg 2 exploring exponential models
byswartzje
 
Types of function
bySanaullahMemon10
 
L4 one sided limits limits at infinity
byJames Tagara
 
10 3 double and half-angle formulas
byhisema01
 
Lu decomposition
bygilandio
 
Linear Equations
byAdjex Academy
 
5 1 quadratic transformations
bylothomas
 

Similar to 6.4 inverse matrices

PPTX
6.4 inverse matrices t
bymath260
 
PPT
Matrices - Mathematics
byDrishti Bhalla
 
PDF
Lecture 3 Inverse matrices(hotom).pdf
bySakith1
 
PPT
systems of linear equations & matrices
byStudent
 
PPT
determinants.ppt
byrushikumar17
 
PDF
23 Matrix Algorithms
byAndres Mendez-Vazquez
 
PPTX
matrix algebra
bykganu
 
PPT
Inverse of Matrices Matrix Algebra maths.ppt
byvsignacademy
 
PPT
Inverse of Matrices, for two by two and three by three
byCalvinskkvv
 
PPTX
linear algebra (malak,).pptx
bymalakahmadwsu
 
PDF
ppt power point presentation physics.pdf
bymdsaifshiddique123
 
PDF
3.-Matrix.pdf
byJohnCarlocatilo
 
PPT
determinants and matrices in mathematics.ppt
byNurulAbrar5
 
PPT
determinants.ppt
byTGBSmile
 
PDF
R.Ganesh Kumar
byGaneshKumar1103
 
PPTX
matrix mathmatics information technology .pptx
byadasaram276
 
PPT
Matrix inverses
byST ZULAIHA NURHAJARURAHMAH
 
PDF
7 4
byELIMENG
 
PDF
Ma3bfet par 10.5 31 julie 2014
byCelumusa Godfrey Nkosi
 
PDF
Matrixprop
byFrank Lucas
 
6.4 inverse matrices t
bymath260
 
Matrices - Mathematics
byDrishti Bhalla
 
Lecture 3 Inverse matrices(hotom).pdf
bySakith1
 
systems of linear equations & matrices
byStudent
 
determinants.ppt
byrushikumar17
 
23 Matrix Algorithms
byAndres Mendez-Vazquez
 
matrix algebra
bykganu
 
Inverse of Matrices Matrix Algebra maths.ppt
byvsignacademy
 
Inverse of Matrices, for two by two and three by three
byCalvinskkvv
 
linear algebra (malak,).pptx
bymalakahmadwsu
 
ppt power point presentation physics.pdf
bymdsaifshiddique123
 
3.-Matrix.pdf
byJohnCarlocatilo
 
determinants and matrices in mathematics.ppt
byNurulAbrar5
 
determinants.ppt
byTGBSmile
 
R.Ganesh Kumar
byGaneshKumar1103
 
matrix mathmatics information technology .pptx
byadasaram276
 
Matrix inverses
byST ZULAIHA NURHAJARURAHMAH
 
7 4
byELIMENG
 
Ma3bfet par 10.5 31 julie 2014
byCelumusa Godfrey Nkosi
 
Matrixprop
byFrank Lucas
 

More from math260

PPTX
14 graphs of factorable rational functions x
bymath260
 
PPTX
11 graphs of first degree functions x
bymath260
 
PPTX
10 rectangular coordinate system x
bymath260
 
PPTX
36 Matrix Algebra-x.pptx
bymath260
 
PPTX
35 Special Cases System of Linear Equations-x.pptx
bymath260
 
PPTX
18 ellipses x
bymath260
 
PPTX
19 more parabolas a& hyperbolas (optional) x
bymath260
 
PPTX
24 exponential functions and periodic compound interests pina x
bymath260
 
PPTX
7 sign charts of factorable formulas y
bymath260
 
PPTX
1 exponents yz
bymath260
 
PPTX
11 graphs of first degree functions x
bymath260
 
PPTX
16 slopes and difference quotient x
bymath260
 
PPTX
21 properties of division and roots x
bymath260
 
PPTX
29 inverse functions x
bymath260
 
PPTX
15 translations of graphs x
bymath260
 
PPTX
27 calculation with log and exp x
bymath260
 
PPTX
23 looking for real roots of real polynomials x
bymath260
 
PPTX
9 the basic language of functions x
bymath260
 
PPTX
10.5 more on language of functions x
bymath260
 
PPTX
26 the logarithm functions x
bymath260
 
14 graphs of factorable rational functions x
bymath260
 
11 graphs of first degree functions x
bymath260
 
10 rectangular coordinate system x
bymath260
 
36 Matrix Algebra-x.pptx
bymath260
 
35 Special Cases System of Linear Equations-x.pptx
bymath260
 
18 ellipses x
bymath260
 
19 more parabolas a& hyperbolas (optional) x
bymath260
 
24 exponential functions and periodic compound interests pina x
bymath260
 
7 sign charts of factorable formulas y
bymath260
 
1 exponents yz
bymath260
 
11 graphs of first degree functions x
bymath260
 
16 slopes and difference quotient x
bymath260
 
21 properties of division and roots x
bymath260
 
29 inverse functions x
bymath260
 
15 translations of graphs x
bymath260
 
27 calculation with log and exp x
bymath260
 
23 looking for real roots of real polynomials x
bymath260
 
9 the basic language of functions x
bymath260
 
10.5 more on language of functions x
bymath260
 
26 the logarithm functions x
bymath260
 

Recently uploaded

PDF
The Accel 2025 Globalscape: Race for compute
byPhilippe Botteri
 
PPTX
Enhancing Web Security: Key Concepts & Strategies.pptx
byParham Abolghasemi
 
PDF
Database Performance at Scale: The TL;DR
byScyllaDB
 
PDF
From Pixels to Insights: Getting Started with Imagery in FME
bySafe Software
 
PDF
MathML Core in WebKit
byIgalia
 
PDF
Automating ArcGIS Enterprise Compliance for Operational Excellence
bySafe Software
 
PPTX
Bulwark Pokemon League Top 8 graphic archive
byGc Howard
 
PPTX
Expanding Your Toolbox: Beginners Guide to Controlling Kubernetes Logs
byEric D. Schabell
 
PDF
Revolutionizing SQL Database Design for the Modern Era.pdf
bySQL DBM
 
PPSX
AppSec Role Based Training OWASP Global AppSec USA 2025-11-06
byRobert Grupe, CSSLP CISSP PE PMP
 
PDF
Install and Update Software - RHCSA (RH124).pdf
byLinuxCert Guru
 
PDF
Configure and Secure SSH - RHCSA (RH124).pdf
byLinuxCert Guru
 
PDF
Manage Networking in RHEL - RHCSA (RH124).pdf
byLinuxCert Guru
 
PPTX
Getting Started with MSP360 Backup for M365/Google
byMSP360
 
PPTX
Odoo_by_Infinys_All_in_One_Business_Solution_for_Small_Companies.pptx
byAgusto Sipahutar
 
PDF
Supercharge Your JVM with Project Leyden
byAna-Maria Mihalceanu
 
PDF
Inclusive India_Samir_Dash_10 NOV_FINAL 2025.pdf
bySamir Dash
 
PDF
Q3'25 Financial Results and Earnings Presentation
byDropbox
 
PDF
ENTSO-E's Response to the European Commission Call for Evidence on the Strate...
byReynir Orn Bachmann Gudmundsson
 
PDF
Do more with the HP Z2 Mini G1a
byPrincipled Technologies
 
The Accel 2025 Globalscape: Race for compute
byPhilippe Botteri
 
Enhancing Web Security: Key Concepts & Strategies.pptx
byParham Abolghasemi
 
Database Performance at Scale: The TL;DR
byScyllaDB
 
From Pixels to Insights: Getting Started with Imagery in FME
bySafe Software
 
MathML Core in WebKit
byIgalia
 
Automating ArcGIS Enterprise Compliance for Operational Excellence
bySafe Software
 
Bulwark Pokemon League Top 8 graphic archive
byGc Howard
 
Expanding Your Toolbox: Beginners Guide to Controlling Kubernetes Logs
byEric D. Schabell
 
Revolutionizing SQL Database Design for the Modern Era.pdf
bySQL DBM
 
AppSec Role Based Training OWASP Global AppSec USA 2025-11-06
byRobert Grupe, CSSLP CISSP PE PMP
 
Install and Update Software - RHCSA (RH124).pdf
byLinuxCert Guru
 
Configure and Secure SSH - RHCSA (RH124).pdf
byLinuxCert Guru
 
Manage Networking in RHEL - RHCSA (RH124).pdf
byLinuxCert Guru
 
Getting Started with MSP360 Backup for M365/Google
byMSP360
 
Odoo_by_Infinys_All_in_One_Business_Solution_for_Small_Companies.pptx
byAgusto Sipahutar
 
Supercharge Your JVM with Project Leyden
byAna-Maria Mihalceanu
 
Inclusive India_Samir_Dash_10 NOV_FINAL 2025.pdf
bySamir Dash
 
Q3'25 Financial Results and Earnings Presentation
byDropbox
 
ENTSO-E's Response to the European Commission Call for Evidence on the Strate...
byReynir Orn Bachmann Gudmundsson
 
Do more with the HP Z2 Mini G1a
byPrincipled Technologies
 
In this document
Powered by AI

Introduction to inverse matrices and their properties. Discusses multiplicative inverse and its role in solving equations.

Explanation of the condition AB = BA = In for inverses of matrices. Introduces notation for inverse matrices.

Discusses types of matrices related to inverses: invertible and noninvertible (singular) matrices.

Presents a specific example of a noninvertible matrix and discusses methods for finding inverses.

Method for finding the inverse of a matrix using the identity matrix and row operations.

Step-by-step example on how to find the inverse of a 2x2 matrix, concluding with the result.

Presents a system of equations as a matrix equation and discusses the conditions for inverses.

Demonstrates how to solve linear equations using the inverse matrix method, with a specific example.Exercises on finding inverses of given matrices and checking properties of inverse matrices.

6.4 inverse matrices

  • 1.
  • 2.
    Inverse Matrices For numbers,the multiplicative inverse of a nonzero number a is the reciprocal 1/a.
  • 3.
    Inverse Matrices For numbers,the multiplicative inverse of a nonzero number a is the reciprocal 1/a. The most important property of 1/a is that 1 a * a = a * 1 a = 1
  • 4.
    Inverse Matrices For numbers,the multiplicative inverse of a nonzero number a is the reciprocal 1/a. The most important property of 1/a is that 1 a * a = a * 1 a = 1 The most important usage of the inverse 1/a is to solve the equation ax = b.
  • 5.
    Inverse Matrices For numbers,the multiplicative inverse of a nonzero number a is the reciprocal 1/a. The most important property of 1/a is that 1 a * a = a * 1 a = 1 The most important usage of the inverse 1/a is to solve the equation ax = b. For example, to solve 2x = 3
  • 6.
    Inverse Matrices For numbers,the multiplicative inverse of a nonzero number a is the reciprocal 1/a. The most important property of 1/a is that 1 a * a = a * 1 a = 1 The most important usage of the inverse 1/a is to solve the equation ax = b. For example, to solve 2x = Multiply both sides by ½½ * ½ * 3 x = 3/2
  • 7.
    Inverse Matrices For numbers,the multiplicative inverse of a nonzero number a is the reciprocal 1/a. The most important property of 1/a is that 1 a * a = a * 1 a = 1 The most important usage of the inverse 1/a is to solve the equation ax = b. For example, to solve 2x = Multiply both sides by ½½ * x = 3/2 Let A be an n x n matrix. ½ * 3
  • 8.
    Inverse Matrices For numbers,the multiplicative inverse of a nonzero number a is the reciprocal 1/a. The most important property of 1/a is that 1 a * a = a * 1 a = 1 The most important usage of the inverse 1/a is to solve the equation ax = b. For example, to solve 2x = Multiply both sides by ½½ * x = 3/2 Let A be an n x n matrix. Since In plays the role of 1 in multiplication, we say B is the inverse of A if AB = BA = In and we write B as A-1. ½ * 3
  • 9.
    Inverse Matrices For numbers,the multiplicative inverse of a nonzero number a is the reciprocal 1/a. The most important property of 1/a is that 1 a * a = a * 1 a = 1 The most important usage of the inverse 1/a is to solve the equation ax = b. For example, to solve 2x = Multiply both sides by ½½ * x = 3/2 We can use A-1 to solve matrix equations AX = C, i.e. to find a matrix X where A and C are matrices. Let A be an n x n matrix. Since In plays the role of 1 in multiplication, we say B is the inverse of A if AB = BA = In and we write B as A-1. ½ * 3
  • 10.
    1 0 0. . 0 0 1 0 . . . 0 0 1 . . . 0 0 . . . 1 . . . . . . Inverse Matrices AA-1 = A-1A = = In The matrices A, A-1 is a pair of inverse matrices if
  • 11.
    1 0 0. . 0 0 1 0 . . . 0 0 1 . . . 0 0 . . . 1 . . . . . . Inverse Matrices AA-1 = A-1A = = In Where as the inverse of nonzero numbers always exist, this is not the case for matrices. The matrices A, A-1 is a pair of inverse matrices if
  • 12.
    1 0 0. . 0 0 1 0 . . . 0 0 1 . . . 0 0 . . . 1 . . . . . . Inverse Matrices AA-1 = A-1A = = In Where as the inverse of nonzero numbers always exist, this is not the case for matrices. There're two types of matrices with regard to inverses. The matrices A, A-1 is a pair of inverse matrices if
  • 13.
    1 0 0. . 0 0 1 0 . . . 0 0 1 . . . 0 0 . . . 1 . . . . . . Inverse Matrices AA-1 = A-1A = = In I. The ones that have inverses, these are called invertible (or nonsingular) matrices The matrices A, A-1 is a pair of inverse matrices if Where as the inverse of nonzero numbers always exist, this is not the case for matrices. There're two types of matrices with regard to inverses.
  • 14.
    1 0 0. . 0 0 1 0 . . . 0 0 1 . . . 0 0 . . . 1 . . . . . . Inverse Matrices AA-1 = A-1A = = In I. The ones that have inverses, these are called invertible (or nonsingular) matrices II. The ones that don‘t have an inverse are called noninvertible (or singular) matrices The matrices A, A-1 is a pair of inverse matrices if Where as the inverse of nonzero numbers always exist, this is not the case for matrices. There're two types of matrices with regard to inverses.
  • 15.
    An example ofa noninvertible matrix is Inverse Matrices 0 -2 0 3
  • 16.
    Inverse Matrices because forany matrix , 0 -2 0 3 a b c d = 0 # 0 # a b c d An example of a noninvertible matrix is 0 -2 0 3
  • 17.
    Inverse Matrices because forany matrix , 0 -2 0 3 a b c d = 0 # 0 # = 1 0 0 1 a b c d An example of a noninvertible matrix is 0 -2 0 3
  • 18.
    Inverse Matrices because forany matrix , 0 -2 0 3 a b c d = 0 # 0 # = 1 0 0 1 a b c d Method for Finding Inverse Matrices: An example of a noninvertible matrix is 0 -2 0 3
  • 19.
    Inverse Matrices To findthe inverse, A-1 of a matrix A, extend the matrix A by the identity matrix I on the right as shown, because for any matrix , 0 -2 0 3 a b c d = 0 # 0 # = 1 0 0 1 a b c d Method for Finding Inverse Matrices: A A I An example of a noninvertible matrix is 0 -2 0 3
  • 20.
    Inverse Matrices To findthe inverse, A-1 of a matrix A, extend the matrix A by the identity matrix I on the right as shown, because for any matrix , 0 -2 0 3 a b c d = 0 # 0 # = 1 0 0 1 a b c d Method for Finding Inverse Matrices: A A I Apply row operations to tranform it so the identity matrix I is on the left side as , then B = A-1.I B An example of a noninvertible matrix is 0 -2 0 3
  • 21.
    Example A: Findthe inverse matrix A-1 if Inverse Matrices a. A = 1 2 1 3
  • 22.
    Example A: Findthe inverse matrix A-1 if Inverse Matrices a. A = 1 2 1 3 Extend 1 2 1 3 to 1 2 1 0 1 3 0 1
  • 23.
    Example A: Findthe inverse matrix A-1 if Inverse Matrices a. A = 1 2 1 3 Extend 1 2 1 3 to 1 2 1 0 1 3 0 1 Apply row operations to tranform it so the identity matrix I is on the left side:
  • 24.
    Example A: Findthe inverse matrix A-1 if Inverse Matrices a. A = 1 2 1 3 Extend 1 2 1 3 to 1 2 1 0 1 3 0 1 Apply row operations to tranform it so the identity matrix I is on the left side: 1 2 1 0 1 3 0 1 (-1)R1 Add to R2
  • 25.
    Example A: Findthe inverse matrix A-1 if Inverse Matrices a. A = 1 2 1 3 Extend 1 2 1 3 to 1 2 1 0 1 3 0 1 Apply row operations to tranform it so the identity matrix I is on the left side: 1 2 1 0 1 3 0 1 (-1)R1 Add to R2 -1 -2 -1 0
  • 26.
    Example A: Findthe inverse matrix A-1 if Inverse Matrices a. A = 1 2 1 3 Extend 1 2 1 3 to 1 2 1 0 1 3 0 1 Apply row operations to tranform it so the identity matrix I is on the left side: 1 2 1 0 1 3 0 1 (-1)R1 Add to R2 -1 -2 -1 0 1 2 1 0 0 1 -1 1
  • 27.
    Example A: Findthe inverse matrix A-1 if Inverse Matrices a. A = 1 2 1 3 Extend 1 2 1 3 to 1 2 1 0 1 3 0 1 Apply row operations to tranform it so the identity matrix I is on the left side: 1 2 1 0 1 3 0 1 (-1)R1 Add to R2 -1 -2 -1 0 1 2 1 0 0 1 -1 1 (-2)R2 Add to R1
  • 28.
    Example A: Findthe inverse matrix A-1 if Inverse Matrices a. A = 1 2 1 3 Extend 1 2 1 3 to 1 2 1 0 1 3 0 1 Apply row operations to tranform it so the identity matrix I is on the left side: 1 2 1 0 1 3 0 1 (-1)R1 Add to R2 -1 -2 -1 0 1 2 1 0 0 1 -1 1 0 -2 2 -2 (-2)R2 Add to R1
  • 29.
    Example A: Findthe inverse matrix A-1 if Inverse Matrices a. A = 1 2 1 3 Extend 1 2 1 3 to 1 2 1 0 1 3 0 1 Apply row operations to tranform it so the identity matrix I is on the left side: 1 2 1 0 1 3 0 1 (-1)R1 Add to R2 -1 -2 -1 0 1 2 1 0 0 1 -1 1 0 -2 2 -2 1 0 3 -2 0 1 -1 1 (-2)R2 Add to R1
  • 30.
    Example A: Findthe inverse matrix A-1 if Inverse Matrices a. A = 1 2 1 3 Extend 1 2 1 3 to 1 2 1 0 1 3 0 1 Apply row operations to tranform it so the identity matrix I is on the left side: 1 2 1 0 1 3 0 1 (-1)R1 Add to R2 -1 -2 -1 0 1 2 1 0 0 1 -1 1 0 -2 2 -2 1 0 3 -2 0 1 -1 1 (-2)R2 Add to R1 = identity matrix
  • 31.
    Example A: Findthe inverse matrix A-1 if Inverse Matrices a. A = 1 2 1 3 Extend 1 2 1 3 to 1 2 1 0 1 3 0 1 Apply row operations to tranform it so the identity matrix I is on the left side: 1 2 1 0 1 3 0 1 (-1)R1 Add to R2 -1 -2 -1 0 1 2 1 0 0 1 -1 1 0 -2 2 -2 1 0 3 -2 0 1 -1 1 (-2)R2 Add to R1 = identity matrix Hence A-1 = 3 -2 -1 1 .
  • 32.
    Example A: Findthe inverse matrix A-1 if Inverse Matrices a. A = 1 2 1 3 Extend 1 2 1 3 to 1 2 1 0 1 3 0 1 Apply row operations to tranform it so the identity matrix I is on the left side: 1 2 1 0 1 3 0 1 (-1)R1 Add to R2 -1 -2 -1 0 1 2 1 0 0 1 -1 1 0 -2 2 -2 1 0 3 -2 0 1 -1 1 (-2)R2 Add to R1 = identity matrix Hence A-1 = 3 -2 -1 1 . One checks easily that AA-1 = A-1A = 1 0 0 1 .
  • 33.
  • 34.
    Inverse Matrices b. A= 0 2 0 3 Extend 0 2 0 3 to 0 2 1 0 0 3 0 1
  • 35.
    Inverse Matrices b. A= 0 2 0 3 Extend 0 2 0 3 to 0 2 1 0 0 3 0 1 Since the first column consists of 0's only, there is no way we may introduce a 1 at the a11 position by row operations.
  • 36.
    Inverse Matrices b. A= 0 2 0 3 Extend 0 2 0 3 to 0 2 1 0 0 3 0 1 Since the first column consists of 0's only, there is no way we may introduce a 1 at the a11 position by row operations. Hence A-1 doesn't exist.
  • 37.
    Inverse Matrices b. A= 0 2 0 3 Extend 0 2 0 3 to 0 2 1 0 0 3 0 1 Since the first column consists of 0's only, there is no way we may introduce a 1 at the a11 position by row operations. Hence A-1 doesn't exist. c. A = 0 1 -1 1 -1 2 1 1 0
  • 38.
    Inverse Matrices b. A= 0 2 0 3 Extend 0 2 0 3 to 0 2 1 0 0 3 0 1 Since the first column consists of 0's only, there is no way we may introduce a 1 at the a11 position by row operations. Hence A-1 doesn't exist. c. A = 0 1 -1 1 -1 2 1 1 0 Extend 0 1 -1 1 -1 2 1 1 0 0 1 -1 1 0 0 1 -1 2 0 1 0 1 1 0 0 0 1
  • 39.
    Inverse Matrices Apply rowoperations: 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 R1 R2
  • 40.
    Inverse Matrices Apply rowoperations: 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 R1 R2
  • 41.
    Inverse Matrices Apply rowoperations: 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 R1 R2 (-1)R1 add to R3
  • 42.
    Inverse Matrices Apply rowoperations: 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 R1 R2 -1 -1 -2 0 -1 0 (-1)R1 add to R3
  • 43.
    Inverse Matrices Apply rowoperations: 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 R1 R2 -1 -1 -2 0 -1 0 0 1 1 1 0 0 1 1 2 0 1 0 0 0 -2 0 -1 1 (-1)R1 add to R3
  • 44.
    Inverse Matrices Apply rowoperations: 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 (-1)R2 add to R1 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 R1 R2 -1 -1 -2 0 -1 0 0 1 1 1 0 0 1 1 2 0 1 0 0 0 -2 0 -1 1 (-1)R1 add to R3
  • 45.
    Inverse Matrices Apply rowoperations: 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 R1 R2 -1 -1 -2 0 -1 0 0 1 1 1 0 0 1 1 2 0 1 0 0 0 -2 0 -1 1 0 -1 -1 -1 0 0 (-1)R1 add to R3 (-1)R2 add to R1
  • 46.
    Inverse Matrices Apply rowoperations: 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 (-1)R2 add to R1 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 R1 R2 -1 -1 -2 0 -1 0 0 1 1 1 0 0 1 1 2 0 1 0 0 0 -2 0 -1 1 0 -1 -1 -1 0 0 0 1 -1 1 0 0 1 0 1 -1 1 0 0 0 -2 0 -1 1 (-1)R1 add to R3
  • 47.
    Inverse Matrices Apply rowoperations: 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 (-1)R2 add to R1 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 R1 R2 -1 -1 -2 0 -1 0 0 1 1 1 0 0 1 1 2 0 1 0 0 0 -2 0 -1 1 0 -1 -1 -1 0 0 0 1 -1 1 0 0 1 0 1 -1 1 0 0 0 -2 0 -1 1 (-1)R1 add to R3 (-1/2)R3
  • 48.
    Inverse Matrices Apply rowoperations: 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 (-1)R2 add to R1 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 R1 R2 -1 -1 -2 0 -1 0 0 1 1 1 0 0 1 1 2 0 1 0 0 0 -2 0 -1 1 0 -1 -1 -1 0 0 0 1 -1 1 0 0 1 0 1 -1 1 0 0 0 -2 0 -1 1 (-1)R1 add to R3 0 1 -1 1 0 0 1 0 1 -1 1 0 0 0 1 0 1/2 -1/2 (-1/2)R3
  • 49.
    Inverse Matrices Apply rowoperations: 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 (-1)R2 add to R1 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 R1 R2 -1 -1 -2 0 -1 0 0 1 1 1 0 0 1 1 2 0 1 0 0 0 -2 0 -1 1 0 -1 -1 -1 0 0 0 1 -1 1 0 0 1 0 1 -1 1 0 0 0 -2 0 -1 1 (-1)R1 add to R3 0 1 -1 1 0 0 1 0 1 -1 1 0 0 0 1 0 1/2 -1/2 (-1)R3 + R1 (-1/2)R3 R3 + R2
  • 50.
    Inverse Matrices Apply rowoperations: 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 (-1)R2 add to R1 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 R1 R2 -1 -1 -2 0 -1 0 0 1 1 1 0 0 1 1 2 0 1 0 0 0 -2 0 -1 1 0 -1 -1 -1 0 0 0 1 -1 1 0 0 1 0 1 -1 1 0 0 0 -2 0 -1 1 (-1)R1 add to R3 0 1 -1 1 0 0 1 0 1 -1 1 0 0 0 1 0 1/2 -1/2 0 0 -1 0 -1/2 1/2 (-1)R3 + R1 (-1/2)R3 R3 + R2
  • 51.
    Inverse Matrices Apply rowoperations: 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 (-1)R2 add to R1 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 R1 R2 -1 -1 -2 0 -1 0 0 1 1 1 0 0 1 1 2 0 1 0 0 0 -2 0 -1 1 0 -1 -1 -1 0 0 0 1 -1 1 0 0 1 0 1 -1 1 0 0 0 -2 0 -1 1 (-1)R1 add to R3 0 1 0 1 -1/2 1/2 1 0 0 -1 1/2 1/2 0 1 -1 1 0 0 1 0 1 -1 1 0 0 0 1 0 1/2 -1/2 0 0 -1 0 -1/2 1/2 (-1)R3 + R1 (-1/2)R3 R3 + R2 0 0 1 0 1/2 -1/2
  • 52.
    Inverse Matrices Apply rowoperations: 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 (-1)R2 add to R1 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 R1 R2 -1 -1 -2 0 -1 0 0 1 1 1 0 0 1 1 2 0 1 0 0 0 -2 0 -1 1 0 -1 -1 -1 0 0 0 1 -1 1 0 0 1 0 1 -1 1 0 0 0 -2 0 -1 1 (-1)R1 add to R3 0 1 0 1 -1/2 1/2 1 0 0 -1 1/2 1/2 0 1 -1 1 0 0 1 0 1 -1 1 0 0 0 1 0 1/2 -1/2 0 0 -1 0 -1/2 1/2 (-1)R3 + R1 (-1/2)R3 R3 + R2 0 0 1 0 1/2 -1/2 = A-1
  • 53.
    Inverse Matrices Apply rowoperations: 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 (-1)R2 add to R1 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 R1 R2 -1 -1 -2 0 -1 0 0 1 1 1 0 0 1 1 2 0 1 0 0 0 -2 0 -1 1 0 -1 -1 -1 0 0 0 1 -1 1 0 0 1 0 1 -1 1 0 0 0 -2 0 -1 1 (-1)R1 add to R3 0 1 0 1 -1/2 1/2 1 0 0 -1 1/2 1/2 0 1 -1 1 0 0 1 0 1 -1 1 0 0 0 1 0 1/2 -1/2 0 0 -1 0 -1/2 1/2 (-1)R3 + R1 (-1/2)R3 R3 + R2 0 0 1 0 1/2 -1/2 = A-1We check that A-1*A = A*A-1 = I.
  • 54.
    Inverse Matrices Apply rowoperations: 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 (-1)R2 add to R1 0 1 1 1 0 0 1 1 2 0 1 0 1 1 0 0 0 1 R1 R2 -1 -1 -2 0 -1 0 0 1 1 1 0 0 1 1 2 0 1 0 0 0 -2 0 -1 1 0 -1 -1 -1 0 0 0 1 -1 1 0 0 1 0 1 -1 1 0 0 0 -2 0 -1 1 (-1)R1 add to R3 0 1 0 1 -1/2 1/2 1 0 0 -1 1/2 1/2 0 1 -1 1 0 0 1 0 1 -1 1 0 0 0 1 0 1/2 -1/2 0 0 -1 0 -1/2 1/2 (-1)R3 + R1 (-1/2)R3 R3 + R2 0 0 1 0 1/2 -1/2 = A-1We check that A-1*A = A*A-1 = I. Let’s use an inverse matrix to solve a matrix equation.
  • 55.
    Example B: a. Writethe following system as a matrix equation x + 2y = 4 x + 3y = -1 An Application of Inverse Matrices
  • 56.
    Example B: a. Writethe following system as a matrix equation x + 2y = 4 x + 3y = -1 The system may be expressed as 1 2 1 3 x y = 4 -1 An Application of Inverse Matrices
  • 57.
    Example B: a. Writethe following system as a matrix equation If we let A = 1 2 1 3 x + 2y = 4 x + 3y = -1 , X = x y , The system may be expressed as 1 2 1 3 x y = 4 -1 then the matrix equation is simply AX = B. and B = 4 -1 . An Application of Inverse Matrices
  • 58.
    Example B: a. Writethe following system as a matrix equation If we let A = 1 2 1 3 x + 2y = 4 x + 3y = -1 , X = x y , The system may be expressed as 1 2 1 3 x y = 4 -1 then the matrix equation is simply AX = B. and B = 4 -1 . An Application of Inverse Matrices To solve for the matrix X, we use the following inverse matrix method.
  • 59.
    To solve theequation AX = B for the matrix X, we multiply to the left of both sides by A-1 (if it exists) An Application of Inverse Matrices
  • 60.
    To solve theequation AX = B for the matrix X, we multiply to the left of both sides by A-1 (if it exists) AX = B An Application of Inverse Matrices
  • 61.
    To solve theequation AX = B for the matrix X, we multiply to the left of both sides by A-1 (if it exists) AX = B A-1AX = A-1B An Application of Inverse Matrices
  • 62.
    To solve theequation AX = B for the matrix X, we multiply to the left of both sides by A-1 (if it exists) AX = B A-1AX = A-1B IX = A-1B or that X = A-1B An Application of Inverse Matrices
  • 63.
    To solve theequation AX = B for the matrix X, we multiply to the left of both sides by A-1 (if it exists) AX = B A-1AX = A-1B IX = A-1B or that X = A-1B b. Solve the matrix equation AX = B from part a. by the inverse matrix method. An Application of Inverse Matrices
  • 64.
    To solve theequation AX = B for the matrix X, we multiply to the left of both sides by A-1 (if it exists) AX = B A-1AX = A-1B IX = A-1B or that X = A-1B b. Solve the matrix equation AX = B from part a. by the inverse matrix method. from the previous example we've that Given that A = 1 2 1 3 3 -2 -1 1 . A-1 = An Application of Inverse Matrices
  • 65.
    Multiply A-1 tothe left on both sides of the equation Inverse Matrices 1 2 1 3 x y = 4 -1
  • 66.
    Multiply A-1 tothe left on both sides of the equation Inverse Matrices 1 2 1 3 x y = 4 -1 we get 3 -2 -1 1 1 2 1 3 x y = 3 -2 -1 1 4 -1
  • 67.
    Multiply A-1 tothe left on both sides of the equation Inverse Matrices 1 2 1 3 x y = 4 -1 we get 3 -2 -1 1 1 2 1 3 x y = 3 -2 -1 1 4 -1 0 1 1 0 x y = 14 -5
  • 68.
    Multiply A-1 tothe left on both sides of the equation Inverse Matrices 1 2 1 3 x y = 4 -1 we get 3 -2 -1 1 1 2 1 3 x y = 3 -2 -1 1 4 -1 0 1 1 0 x y = 14 -5 Hence x y = 14 -5 or that x = 14, and y = -5.
  • 69.
    Multiply A-1 tothe left on both sides of the equation Inverse Matrices 1 2 1 3 x y = 4 -1 we get 3 -2 -1 1 1 2 1 3 x y = 3 -2 -1 1 4 -1 0 1 1 0 x y = 14 -5 Hence x y = 14 -5 or that x = 14, and y = -5. With inverse matrices, we may do matrix algebra that is similar to the “traditional” algebra.
  • 70.
    Inverse Matrices Exercise A.Find the inverse matrix of each of the following matrices, if it exits 1 0 0 1 1) 2) 3) 4)1 5 0 1 a 5 0 𝑏 0 𝑏 𝑎 5 3 5 –2 1 5) 6) 7) 8) 1 –2 7 –1 𝑎 𝑏 𝑏 𝑎 𝑎 𝑏 𝑎 𝑏 9) 10) 11) 12) 1 0 0 0 1 0 0 0 1 1 2 3 0 1 4 0 0 1 a 0 0 0 b 0 0 0 c a 2 3 0 b 4 0 0 c 13) 14) 1 0 2 0 1 0 −1 0 1 1 2 3 0 1 4 −1 2 1
  • 71.
    2. Verify thefollowing inverse formula for 2x2 matrices. 𝑎 𝑏 𝑐 𝑑 –1 = 1 ad – bc 𝑑 –𝑏 –𝑐 𝑎 (Hint: you don’t have to compute the inverse, just check that their product is I so it must be the one and only inverse.) 3. Verify the that (AB) –1 = B –1 A–1. So if A and B are invertible then AB is invertible. Ex B. A matrix that has an inverse is said to be invertible. 1. (A–1 is unique) Verify that if A is invertible, then there is only one A–1. This means to show that if there are matrices B and C that invert A so that AB = BA = I and AC = CA = I, then B = C. 4. Verify that if AB is invertible, then A and B must be invertible. (Hint: Let C is (AB) –1, then A–1 or B –1 can be “solved”.) So if A or B is not invertible then AB is not invertible. Inverse Matrices
  • 72.
    (Answers to theodd problems) Exercise A. 𝐴−1 = 1 0 0 1 1) 𝐴−1 = 1/𝑎 −5/𝑎𝑏 0 1/𝑏 𝐴−1 = 1 13 1 −5 2 3 5) 𝐴−1 = 1 𝑎2 − 𝑏2 𝑎 −𝑏 −𝑏 𝑎 9) 𝐴−1 = 1 0 0 0 1 0 0 0 1 𝐴−1 = 1/a 0 0 0 1/b 0 0 0 1/c 13) 𝐴−1 = 1 3 1 0 −2 0 3 0 1 0 1 3) 7) 11) Inverse Matrices