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Matrix Inverse, IMT
- 1. Announcements Quiz 2 on Wednesday Jan 27 on sections 1.4, 1.5, 1.7 and 1.8 Test 1 will be on Feb 1, Monday in class. More details later.
- 2. Last Class a b Let A = . If ad − bc = 0 then A is invertible and c d −1 1 d −b A = ad − bc −c a .
- 3. Last Class a b Let A = . If ad − bc = 0 then A is invertible and c d −1 1 d −b A = ad − bc −c a . So if the determinant of A (or det A) is equal to 0, A−1 does not exist.
- 4. Last Class Steps for a 2 × 2 matrix A 1. First check whether det A=0. If so, stop. A is not invertible.
- 5. Last Class Steps for a 2 × 2 matrix A 1. First check whether det A=0. If so, stop. A is not invertible. 2. If det A = 0, swap the main diagonal elements of A.
- 6. Last Class Steps for a 2 × 2 matrix A 1. First check whether det A=0. If so, stop. A is not invertible. 2. If det A = 0, swap the main diagonal elements of A. 3. Then change the sign of both o diagonal elements (don't swap these)
- 7. Last Class Steps for a 2 × 2 matrix A 1. First check whether det A=0. If so, stop. A is not invertible. 2. If det A = 0, swap the main diagonal elements of A. 3. Then change the sign of both o diagonal elements (don't swap these) 4. Divide this matrix (after steps 2 and 3) by detA to give A−1 . (This divides each element of the resultant matrix.) 5. If you want to check your answer, you can see whether 1 0 AA−1 = 0 1 6. This method will not work for 3 × 3 or bigger matrices.
- 8. Example 7(a) section 2.2, One Case 1 2 Let A = . Find A−1 and use it to solve the equation 5 12 2 Ax = b3 where b3 = 6 Solution: Here detA = (1)(12) − (5)(2) = 2 = 0. So we can nd A−1 .
- 9. Example 7(a) section 2.2, One Case 1 2 Let A = . Find A−1 and use it to solve the equation 5 12 2 Ax = b3 where b3 = 6 Solution: Here detA = (1)(12) − (5)(2) = 2 = 0. So we can nd A−1 . Interchange the positions of 1 and 12. Change the signs of 2 and 5. Then we get 12 −2 −5 1 .
- 10. Example 7(a) section 2.2, One Case 1 2 Let A = . Find A−1 and use it to solve the equation 5 12 2 Ax = b3 where b3 = 6 Solution: Here detA = (1)(12) − (5)(2) = 2 = 0. So we can nd A−1 . Interchange the positions of 1 and 12. Change the signs of 2 and 5. Then we get 12 −2 −5 1 . Divide each element of the matrix by detA which is 2. This gives −1 6 −1 A = −5/2 1/2 x1 6 −1 2 6 x= = = x2 −5/2 1 /2 6 −2 A−1 b
- 11. Algorithm to nd A−1 To nd the inverse of a square matrix A of any size, 1. Write the augmented matrix with A and I (the identity matrix of proper size) written side by side.
- 12. Algorithm to nd A−1 To nd the inverse of a square matrix A of any size, 1. Write the augmented matrix with A and I (the identity matrix of proper size) written side by side. 2. Do proper row reductions on both A and I till A is reduced to I.
- 13. Algorithm to nd A−1 To nd the inverse of a square matrix A of any size, 1. Write the augmented matrix with A and I (the identity matrix of proper size) written side by side. 2. Do proper row reductions on both A and I till A is reduced to I. 3. This changes I to a new matrix which is A−1
- 14. Algorithm to nd A−1 To nd the inverse of a square matrix A of any size, 1. Write the augmented matrix with A and I (the identity matrix of proper size) written side by side. 2. Do proper row reductions on both A and I till A is reduced to I. 3. This changes I to a new matrix which is A−1 4. If you cannot reduce A to I (if you get a row full of zeros for example), A−1 does not exist
- 15. Example 30, section 2.2 5 10 Let A = . Find A−1 using the algorithm. 4 7 Solution: Start with the augmented matrix 5 10 1 0 4 7 0 1
- 16. Example 30, section 2.2 5 10 Let A = . Find A−1 using the algorithm. 4 7 Solution: Start with the augmented matrix 5 10 1 0 4 7 0 1 Divide R1 by 5 1 2 1/5 0 4 7 0 1
- 17. Example 30, section 2.2 1 2 1/5 0 R2-4R1 4 7 0 1
- 18. Example 30, section 2.2 1 2 1/5 0 R2-4R1 4 7 0 1 1 2 1/5 0 R1+2R2 0 −1 −4/5 1 1 0 −7/5 2 =⇒ 0 −1 −4/5 1
- 19. Example 30, section 2.2 1 2 1/5 0 R2-4R1 4 7 0 1 1 2 1/5 0 R1+2R2 0 −1 −4/5 1 1 0 −7/5 2 =⇒ 0 −1 −4/5 1 1 0 −7/5 2 Divide R2 by -1 =⇒ = I A−1 0 1 4/5 −1
- 20. Example of a 3 × 3 Matrix 1 0 −2 Let A = 3 −1 −4 . Find A−1 using the algorithm. −2 3 −4 Solution: Start with the augmented matrix
- 21. Example of a 3 × 3 Matrix 1 0 −2 Let A = 3 −1 −4 . Find A−1 using the algorithm. −2 3 −4 Solution: Start with the augmented matrix 1 0 −2 1 0 0 R2-3R1 3 −1 −4 0 1 0 −2 3 −4 0 0 1 1 0 −2 1 0 0 0 −1 2 −3 1 0 R3+2R1 −2 3 −4 0 0 1
- 22. Example of a 3 × 3 Matrix 1 0 −2 1 0 0 0 −1 2 −3 1 0 R3+3R2 0 3 −8 2 0 1
- 23. Example of a 3 × 3 Matrix 1 0 −2 1 0 0 0 −1 2 −3 1 0 R3+3R2 0 3 −8 2 0 1 1 0 −2 1 0 0 =⇒ 0 −1 2 −3 1 0 0 0 −2 −7 3 1 Divide R3 by -2 1 0 −2 1 0 0 =⇒ 0 −1 2 −3 1 0 0 0 1 7/2 −3/2 −1/2
- 24. Example of a 3 × 3 Matrix 1 0 −2 1 0 0 0 −1 2 −3 1 0 R2-2R3 0 0 1 7 /2 −3/2 −1/2 1 0 −2 1 0 0 0 −1 0 −10 4 1 0 0 1 7 /2 −3/2 −1/2 Divide R2 by -1 1 0 −2 1 0 0 0 1 0 10 −4 −1 R1+2R2 0 0 1 7/2 −3/2 −1/2
- 25. Example of a 3 × 3 Matrix
- 26. Example of a 3 × 3 Matrix 1 0 0 8 −3 −1 0 1 0 10 −4 −1 0 0 1 7 /2 −3/2 −1/2 = I A−1
- 27. Example 32, section 2.2 1 −2 1 Let A = 4 −7 3 . Find A−1 using the algorithm. −2 6 −4 Solution: Start with the augmented matrix 1 −2 1 1 0 0 R2-4R1 4 −7 3 0 1 0 −2 6 −4 0 0 1 1 −2 1 1 0 0 0 1 −1 −4 1 0 R3+2R1 −2 6 −4 0 0 1
- 28. Example 32, section 2.2 1 −2 1 1 0 0 0 1 −1 −4 1 0 R3-2R2 0 2 −2 2 0 1
- 29. Example 32, section 2.2 1 −2 1 1 0 0 0 1 −1 −4 1 0 R3-2R2 0 2 −2 2 0 1 1 −2 1 1 0 0 0 1 −1 −4 1 0 R3-2R2 0 0 0 10 −2 1
- 30. Very Important Conclusion: You cannot reduce A to I because of the zero row and so A−1 does not exist 1. Here A does not have a pivot in every row.
- 31. Very Important Conclusion: You cannot reduce A to I because of the zero row and so A−1 does not exist 1. Here A does not have a pivot in every row. 2. Here A does not have pivot in every column (there is a free variable)
- 32. Very Important Conclusion: You cannot reduce A to I because of the zero row and so A−1 does not exist 1. Here A does not have a pivot in every row. 2. Here A does not have pivot in every column (there is a free variable) 3. The equation Ax = 0 has Non-trivial solution
- 33. Very Important Conclusion: You cannot reduce A to I because of the zero row and so A−1 does not exist 1. Here A does not have a pivot in every row. 2. Here A does not have pivot in every column (there is a free variable) 3. The equation Ax = 0 has Non-trivial solution 4. The columns of A are linearly
- 34. Very Important Conclusion: You cannot reduce A to I because of the zero row and so A−1 does not exist 1. Here A does not have a pivot in every row. 2. Here A does not have pivot in every column (there is a free variable) 3. The equation Ax = 0 has Non-trivial solution 4. The columns of A are linearly Dependent
- 35. Very Important Conclusion: You cannot reduce A to I because of the zero row and so A−1 does not exist 1. Here A does not have a pivot in every row. 2. Here A does not have pivot in every column (there is a free variable) 3. The equation Ax = 0 has Non-trivial solution 4. The columns of A are linearly Dependent 5. A cannot be row-reduced to the 3 × 3 identity matrix
- 36. Section 2.3 Invertible Matrices If a square matrix A of size n × n is invertible the following is true
- 37. Section 2.3 Invertible Matrices If a square matrix A of size n × n is invertible the following is true Theorem The Invertible Matrix Theorem (IMT) 1. A has n pivot positions (n pivot rows and n pivot columns)
- 38. Section 2.3 Invertible Matrices If a square matrix A of size n × n is invertible the following is true Theorem The Invertible Matrix Theorem (IMT) 1. A has n pivot positions (n pivot rows and n pivot columns) 2. A can be row-reduced to the n × n identity matrix
- 39. Section 2.3 Invertible Matrices If a square matrix A of size n × n is invertible the following is true Theorem The Invertible Matrix Theorem (IMT) 1. A has n pivot positions (n pivot rows and n pivot columns) 2. A can be row-reduced to the n × n identity matrix 3. The equation Ax = 0 has only the trivial solution (no free variables)
- 40. Section 2.3 Invertible Matrices If a square matrix A of size n × n is invertible the following is true Theorem The Invertible Matrix Theorem (IMT) 1. A has n pivot positions (n pivot rows and n pivot columns) 2. A can be row-reduced to the n × n identity matrix 3. The equation Ax = 0 has only the trivial solution (no free variables) 4. The columns of A are linearly
- 41. Section 2.3 Invertible Matrices If a square matrix A of size n × n is invertible the following is true Theorem The Invertible Matrix Theorem (IMT) 1. A has n pivot positions (n pivot rows and n pivot columns) 2. A can be row-reduced to the n × n identity matrix 3. The equation Ax = 0 has only the trivial solution (no free variables) 4. The columns of A are linearly independent
- 42. Section 2.3 Invertible Matrices If a square matrix A of size n × n is invertible the following is true Theorem The Invertible Matrix Theorem (IMT) 1. A has n pivot positions (n pivot rows and n pivot columns) 2. A can be row-reduced to the n × n identity matrix 3. The equation Ax = 0 has only the trivial solution (no free variables) 4. The columns of A are linearly independent 5. The equation Ax = b is consistent for every b in Rn
- 43. Section 2.3 Invertible Matrices If a square matrix A of size n × n is invertible the following is true Theorem The Invertible Matrix Theorem (IMT) 1. A has n pivot positions (n pivot rows and n pivot columns) 2. A can be row-reduced to the n × n identity matrix 3. The equation Ax = 0 has only the trivial solution (no free variables) 4. The columns of A are linearly independent 5. The equation Ax = b is consistent for every b in Rn 6. The columns of A span Rn (because every row has a pivot and recall theorem 4, sec 1.4)
- 44. Example 2, Section 2.3 −4 6 Determine whether A = is invertible. Why(not)? 6 −9
- 45. Example 2, Section 2.3 −4 6 Determine whether A = is invertible. Why(not)? 6 −9 Solution: Clearly the determinant is (-4)(-9)-(6)(6)=36-36=0. So not invertible. Checking invertibility of 2 × 2 matrices are thus easy. Not so obvious fact: The second column is -1.5 times the rst column. Since we have linearly dependent columns, by IMT, A is not invertible.
- 46. Example 4, Section 2.3 −7 0 4 Determine whether A = 3 0 −1 is invertible. Why(not)? 2 0 9 Solution: Remember what happens to a set of vectors if the zero vector is present in that set?
- 47. Example 4, Section 2.3 −7 0 4 Determine whether A = 3 0 −1 is invertible. Why(not)? 2 0 9 Solution: Remember what happens to a set of vectors if the zero vector is present in that set? The vectors are linearly dependent. Since the second column of A is full of zeros, we have linearly dependent columns, and so A is not invertible.
- 48. Example 6, Section 2.3 1 −5 −4 Determine whether A = 0 3 4 is invertible. Why(not)? −3 6 0 1 −5 −4 0 3 4 R3+3R1 −3 6 0 1 −5 −4 0 3 4 R3+3R2 0 −9 −12
- 49. Example 6, Section 2.3 1 −5 −4 0 3 4 0 0 0
- 50. Example 6, Section 2.3 1 −5 −4 0 3 4 0 0 0 Since the third row (and the third column) does not have a pivot, we have linearly dependent columns and by the IMT A is not invertible.
- 51. Example 8, Section 2.3 1 3 7 4 0 5 9 6 Determine whether A = is invertible. Why(not)? 0 0 2 8 0 0 0 10
- 52. Example 8, Section 2.3 1 3 7 4 0 5 9 6 Determine whether A = is invertible. Why(not)? 0 0 2 8 0 0 0 10 Solution: This matrix is already in the echelon form. There are 4 pivot rows and 4 pivot columns. So A is invertible.