Now try Written Assignment, Question 4: 1 Find the inverse of the matrix A =2 -3 2 3 -2 5 -3 by performing row-reduction on the matrix [A | ). -41

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
icon
Related questions
icon
Concept explainers
Topic Video
Question
Written Assignment Question 4? Please solve in details.
3:23 0 &
A ll 11%I
2.2 Inverse of a Matrix.pdf
AB = : -6 9
(-a + 4c -b + 4d
12a + 3c 2b + 3al = l6 9
Looking at column 1, we have:
-a + 4c = 1
2a + 3c = 0
Looking at column 2, we have:
-b + 4d = 0
2b + 3d = 1
Which can be solved by reducing the matrices:
and
These can be combined into a single matrix (similar to the example above):
Then bring this matrix to RREF. If you obtain a matrix of the form
6 1:), then the right side of this matrix is the solution. That is, the right side is A-1.
Summary:
Let A be an n xn matrix. To find A-1:
Adjoin A to I, to obtain the matrix [A | J.
Bring the matrix to row reduced echelon form.
a. If the RREF looks like [ I | B), then A-1 = B.
b. If the left side of the reduced echelon form is not Ig, then A is not invertible.
1.
2.
Example 7:
[o 1 21
3
Find the inverse of the matrix A =
14
-3 81
Solution:
Set up the augmented matrix for [A 1], which is
so 1 2 1 0 o1
1
0 30 1 0
[4 ]
-3 8 0 0 1
Row reduce the above matrix until lz appears on the left hand side, which yields
[1 0 0 -9/2
7
-3/21
0 10
-2
-1
lo 0 1
3/2
-2
1/2
Link to Video Solution by Professor Matt Lewis:
https://www.educreations.com/lesson/view/example-2-2-4-1/32440060/
Now try Written Assignment, Question 4:
3 -21
Find the inverse of the matrix A = 2 5 -3 by performing row-reduction on the matrix [A | 13].
1
L-3 2 -4.
Theorem 2.2.3:
If A and B are invertible n xn matrices, then:
• (A-1)-1 = A
• (AB)-1 = B-A-1
Proof that (AB)-' = B-'A-:
Let C = (AB)-1
Since (AB)(AB)-1 = 1, then (AB)C = I.
Multiply both sides of the equation on the left by A-1:
Regrouping the terms on the left side using the associative property:
A-(AB)C = A-.
(A-'A)BC = A-
Then:
IBC = A-1.
BC = A-1
B-'BC = B-A-1
Multiplying both sides of the equation on the left by B-1:
IC = B-'A-1
so C = B-1A-1
Thus, C = (AB)-1 = B-'A-
Example 8:
Solve the equation below for X, assuming A, B, and C are n x n invertible matrices:
C(B + X)A-1 = In
Transcribed Image Text:3:23 0 & A ll 11%I 2.2 Inverse of a Matrix.pdf AB = : -6 9 (-a + 4c -b + 4d 12a + 3c 2b + 3al = l6 9 Looking at column 1, we have: -a + 4c = 1 2a + 3c = 0 Looking at column 2, we have: -b + 4d = 0 2b + 3d = 1 Which can be solved by reducing the matrices: and These can be combined into a single matrix (similar to the example above): Then bring this matrix to RREF. If you obtain a matrix of the form 6 1:), then the right side of this matrix is the solution. That is, the right side is A-1. Summary: Let A be an n xn matrix. To find A-1: Adjoin A to I, to obtain the matrix [A | J. Bring the matrix to row reduced echelon form. a. If the RREF looks like [ I | B), then A-1 = B. b. If the left side of the reduced echelon form is not Ig, then A is not invertible. 1. 2. Example 7: [o 1 21 3 Find the inverse of the matrix A = 14 -3 81 Solution: Set up the augmented matrix for [A 1], which is so 1 2 1 0 o1 1 0 30 1 0 [4 ] -3 8 0 0 1 Row reduce the above matrix until lz appears on the left hand side, which yields [1 0 0 -9/2 7 -3/21 0 10 -2 -1 lo 0 1 3/2 -2 1/2 Link to Video Solution by Professor Matt Lewis: https://www.educreations.com/lesson/view/example-2-2-4-1/32440060/ Now try Written Assignment, Question 4: 3 -21 Find the inverse of the matrix A = 2 5 -3 by performing row-reduction on the matrix [A | 13]. 1 L-3 2 -4. Theorem 2.2.3: If A and B are invertible n xn matrices, then: • (A-1)-1 = A • (AB)-1 = B-A-1 Proof that (AB)-' = B-'A-: Let C = (AB)-1 Since (AB)(AB)-1 = 1, then (AB)C = I. Multiply both sides of the equation on the left by A-1: Regrouping the terms on the left side using the associative property: A-(AB)C = A-. (A-'A)BC = A- Then: IBC = A-1. BC = A-1 B-'BC = B-A-1 Multiplying both sides of the equation on the left by B-1: IC = B-'A-1 so C = B-1A-1 Thus, C = (AB)-1 = B-'A- Example 8: Solve the equation below for X, assuming A, B, and C are n x n invertible matrices: C(B + X)A-1 = In
Expert Solution
steps

Step by step

Solved in 2 steps with 2 images

Blurred answer
Knowledge Booster
Sample space, Events, and Basic Rules of Probability
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
Basic Technical Mathematics
Basic Technical Mathematics
Advanced Math
ISBN:
9780134437705
Author:
Washington
Publisher:
PEARSON
Topology
Topology
Advanced Math
ISBN:
9780134689517
Author:
Munkres, James R.
Publisher:
Pearson,