Linear Algebra Chapter 4: Determinants Section 4.3. Computation of - PowerPoint PPT Presentation

Linear Algebra Chapter 4: Determinants Section 4.3. Computation of Determinants and Cramer’s Rule—Proofs of Theorems November 8, 2018 () Linear Algebra November 8, 2018 1 / 19

Table of contents Page 271 Number 6 1 Theorem 4.5. Cramer’s Rule 2 Page 272 Number 26 3 Page 272 Number 18 (find adjoint) 4 Theorem 4.6. Property of the Adjoint 5 Page 272 Number 18 (find inverse) 6 Page 272 Number 22 7 Page 273 Number 36 8 Page 273 Number 38 9 () Linear Algebra November 8, 2018 2 / 19

Page 271 Number 6 Page 271 Number 6  3 2 0 0 0  − 1 4 1 0 0     Page 271 Number 6. Find det( A ) where A = 0 − 3 5 2 0 .     0 0 0 1 4   0 0 0 − 1 2 Solution. We state the elementary row operations and keep track of how they affect the determinant based on Theorem 4.2.A, “Properties of the Determinant.” We have: � � 3 2 0 0 0 � � � � − 1 4 1 0 0 � � � � det( A ) = 0 − 3 5 2 0 � � � � 0 0 0 1 4 � � � � 0 0 0 − 1 2 � � () Linear Algebra November 8, 2018 3 / 19

Page 271 Number 6 Page 271 Number 6  3 2 0 0 0  − 1 4 1 0 0     Page 271 Number 6. Find det( A ) where A = 0 − 3 5 2 0 .     0 0 0 1 4   0 0 0 − 1 2 Solution. We state the elementary row operations and keep track of how they affect the determinant based on Theorem 4.2.A, “Properties of the Determinant.” We have: � � 3 2 0 0 0 � � � � − 1 4 1 0 0 � � � � det( A ) = 0 − 3 5 2 0 � � � � 0 0 0 1 4 � � � � 0 0 0 − 1 2 � � () Linear Algebra November 8, 2018 3 / 19

Page 271 Number 6 Page 271 Number 6 (continued 1) Solution (continued). � 3 2 0 0 0 � � − 1 4 1 0 0 � � � � � � � � � − 1 4 1 0 0 3 2 0 0 0 Row Exchange: � � � � � � � � 0 − 3 5 2 0 = − 0 − 3 5 2 0 R 1 ↔ R 2 � � � � � � � � 0 0 0 1 4 0 0 0 1 4 � � � � � � � � 0 0 0 − 1 2 0 0 0 − 1 2 � � � � � � − 1 4 1 0 0 � � � � 0 14 3 0 0 Row Addition: � � � � = − 0 − 3 5 2 0 R 2 → R 2 + 3 R 1 and R 5 → R 5 + R 4 � � � � 0 0 0 1 4 � � � � 0 0 0 0 6 � � () Linear Algebra November 8, 2018 4 / 19

Page 271 Number 6 Page 271 Number 6 (continued 2) Solution (continued). . . . � � � � − 1 4 1 0 0 − 1 4 1 0 0 � � � � � � � � 0 14 3 0 0 0 14 3 0 0 Row Addition: � � � � � � � � − 0 − 3 5 2 0 = − 0 0 79 / 14 2 0 R 3 → R 3 + (3 / 14) R 2 � � � � � � � � 0 0 0 1 4 0 0 0 1 4 � � � � � � � � 0 0 0 0 6 0 0 0 0 6 � � � � = − ( − 1)(14)(79 / 14)(1)(6) = 474 by Example 4.2.4 . � () Linear Algebra November 8, 2018 5 / 19

Theorem 4.5. Cramer’s Rule Theorem 4.5 Theorem 4.5. Cramer’s Rule. x = � Consider the linear system A � b , where A = [ a ij ] is an n × n invertible matrix,     x 1 b 1 x 2 b 2      and � x = � b =  . . .     . .     . .   x n b n The system has a unique solution given by x k = det( B k ) det( A ) for k = 1 , 2 , . . . , n , where B k is the matrix obtained from A by replacing the k th column vector of A by the column vector � b . x = � Proof. Since A is invertible, we know that the linear system A � b has a unique solution by Theorem 1.16. Let � x be this unique solution. () Linear Algebra November 8, 2018 6 / 19

Theorem 4.5. Cramer’s Rule Theorem 4.5 Theorem 4.5. Cramer’s Rule. x = � Consider the linear system A � b , where A = [ a ij ] is an n × n invertible matrix,     x 1 b 1 x 2 b 2      and � x = � b =  . . .     . .     . .   x n b n The system has a unique solution given by x k = det( B k ) det( A ) for k = 1 , 2 , . . . , n , where B k is the matrix obtained from A by replacing the k th column vector of A by the column vector � b . x = � Proof. Since A is invertible, we know that the linear system A � b has a unique solution by Theorem 1.16. Let � x be this unique solution. () Linear Algebra November 8, 2018 6 / 19

Theorem 4.5. Cramer’s Rule Theorem 4.5 (continued 1) Proof (continued). Let X k be the matrix obtained from the n × n identity matrix by replacing its k th column vector by the column vector � x , so   1 0 0 · · · x 1 0 0 · · · 0 0 1 0 0 0 0 · · · x 2 · · ·     0 0 1 · · · x 3 0 0 · · · 0    .  .   X k = . .     0 0 0 0 0 0 · · · x k · · ·   .   .   .   0 0 0 · · · x n 0 0 · · · 1 We now compute the product AX k . If j � = k , then the j th column of AX k is the product of A and the j th column of the identity matrix, which is just x = � the j th column of A . If j = k , then the j th column of AX k is A � b . Thus AX k is the matrix obtained from A by replacing the k th column of A by the column vector � b . () Linear Algebra November 8, 2018 7 / 19

Theorem 4.5. Cramer’s Rule Theorem 4.5 (continued 1) Proof (continued). Let X k be the matrix obtained from the n × n identity matrix by replacing its k th column vector by the column vector � x , so   1 0 0 · · · x 1 0 0 · · · 0 0 1 0 0 0 0 · · · x 2 · · ·     0 0 1 · · · x 3 0 0 · · · 0    .  .   X k = . .     0 0 0 0 0 0 · · · x k · · ·   .   .   .   0 0 0 · · · x n 0 0 · · · 1 We now compute the product AX k . If j � = k , then the j th column of AX k is the product of A and the j th column of the identity matrix, which is just x = � the j th column of A . If j = k , then the j th column of AX k is A � b . Thus AX k is the matrix obtained from A by replacing the k th column of A by the column vector � b . () Linear Algebra November 8, 2018 7 / 19

Theorem 4.5. Cramer’s Rule Theorem 4.5 (continued 2) Proof (continued). That is, AX k is the matrix B k described in the statement of the theorem. From the equation AX k = B k and Theorem 4.4, “The Multiplicative Property,” we have det( A ) det( X k ) = det( B k ) . Computing det( X k ) by expanding by minors across the k th row (applying Theorem 4.2, “General Expansion by Minors”), we see that det( X k ) = x k and thus det( A ) x k = det( B k ) . () Linear Algebra November 8, 2018 8 / 19

Theorem 4.5. Cramer’s Rule Theorem 4.5 (continued 2) Proof (continued). That is, AX k is the matrix B k described in the statement of the theorem. From the equation AX k = B k and Theorem 4.4, “The Multiplicative Property,” we have det( A ) det( X k ) = det( B k ) . Computing det( X k ) by expanding by minors across the k th row (applying Theorem 4.2, “General Expansion by Minors”), we see that det( X k ) = x k and thus det( A ) x k = det( B k ) . Because A is invertible, we know that det( A ) � = 0 by Theorem 4.3, “Determinant Criterion for Invertibility,” and so x k = det( B k ) / det( A ) as claimed. () Linear Algebra November 8, 2018 8 / 19

Theorem 4.5. Cramer’s Rule Theorem 4.5 (continued 2) Proof (continued). That is, AX k is the matrix B k described in the statement of the theorem. From the equation AX k = B k and Theorem 4.4, “The Multiplicative Property,” we have det( A ) det( X k ) = det( B k ) . Computing det( X k ) by expanding by minors across the k th row (applying Theorem 4.2, “General Expansion by Minors”), we see that det( X k ) = x k and thus det( A ) x k = det( B k ) . Because A is invertible, we know that det( A ) � = 0 by Theorem 4.3, “Determinant Criterion for Invertibility,” and so x k = det( B k ) / det( A ) as claimed. () Linear Algebra November 8, 2018 8 / 19

Page 272 Number 26 Page 272 Number 26 Page 272 Number 26. Use Cramer’s Rule to solve 3 x 1 + = 5 x 2 0 . 2 x 1 + x 2 = � 3 � 5 � 5 1 � � 1 � and � Solution. We have A = b = . So B 1 = and 2 1 0 0 1 � 3 � 5 B 2 = . 2 0 () Linear Algebra November 8, 2018 9 / 19

Page 272 Number 26 Page 272 Number 26 Page 272 Number 26. Use Cramer’s Rule to solve 3 x 1 + = 5 x 2 0 . 2 x 1 + x 2 = � 3 � 5 � 5 1 � � 1 � and � Solution. We have A = b = . So B 1 = and 2 1 0 0 1 � 3 � 5 B 2 = . Next, det( A ) = (3)(1) − (1)(2) = 1, 2 0 det( B 1 ) = (5)(1) − (1)(0) = 5, and det( B 2 ) = (3)(0) − (5)(2) = − 10. So by Cramer’s Rule, x 1 = det( B 1 ) det( A ) = 5 1 = 5 and x 2 = det( B 2 ) det( A ) = − 10 = − 10 . 1 So x 1 = 5 and x 2 = − 10. � () Linear Algebra November 8, 2018 9 / 19