This preview has intentionally blurred parts. Sign up to view the full document

View Full Document

Unformatted Document Excerpt

Chapter 6 ISM: Linear Algebra Chapter 6 6.1 1. Fails to be invertible; since det 1 2 3 6 = 6- 6 = 0. 2. Invertible; since det 2 3 4 5 = 10- 12 =- 2. 3. Invertible; since det 3 5 7 11 = 33- 35 =- 2. 4. Fails to be invertible; since det 1 4 2 8 = 8- 8 = 0. 5. Invertible; since det 2 5 7 11 7 5 = 2 11 5 + 0 + 0--- 0 = 110. 6. Invertible; since det 6 5 4 3 2 1 = 6 4 1 + 0 + 0--- 0 = 24. 7. This matrix is clearly not invertible, so the determinant must be zero. 8. This matrix fails to be invertible, since the det( A ) = 0 . 9. Invertible; since det 1 2 7 8 3 6 5 4 = 0 + 3 6 + 2 7 5- 7 4- 2 8 6 =- 36. 10. Invertible; since det 1 1 1 1 2 3 1 3 6 = 1 2 6 + 1 3 1 + 1 1 3- 3 3 1- 2 1 1- 6 1 1 = 1. 11. det k 2 3 4 6 = 0 when 4 k 6 = 6 , or k 6 = 3 2 . 12. det 1 k k 4 6 = 0 when k 2 6 = 4 , or k 6 = 2 ,- 2. 13. det k 3 5 2 6 4 = 8 k, so k 6 = 0 will ensure that this matrix is invertible. 290 ISM: Linear Algebra Section 6.1 14. det 4 3 k 2 1 = 0 , so the matrix will never be invertible, no matter which k is chosen. 15. det k 1 2 3 4 5 6 7 = 6 k- 3 . This matrix is invertible when k 6 = 1 2 . 16. det 1 2 3 4 k 5 6 7 8 = 60 + 84 + 8 k- 18 k- 35- 64 = 45- 10 k. So this matrix is invertible when k 6 = 4 . 5. 17. det 1 1 1 1 k- 1 1 k 2 1 = 2 k 2- 2 = 2( k 2- 1) = 2( k- 1)( k + 1) . So k cannot be 1 or -1. 18. det 1 k 3 2 k 5 9 7 5 = 30 + 21 k- 18 k 2 =- 3( k- 2)(6 k + 5) . So k cannot be 2 or- 5 6 . 19. det 1 1 k 1 k k k k k =- k 3 + 2 k 2- k =- k ( k- 1) 2 . So k cannot be 0 or 1. 20. det 1 k 1 1 k + 1 k + 2 1 k + 2 2 k + 4 = ( k + 1)(2 k + 4) + k ( k + 2) + ( k + 2)- ( k + 1)- k (2 k + 4)- ( k + 2)( k + 2) = ( k + 1)(3 k + 6)- (3 k 2 + 9 k + 5) = 1 . Thus, A will always be invertible, no matter the value of k , meaning that k can have any value. 21. det k 1 1 1 k 1 1 1 k = k 3- 3 k + 2 = ( k- 1) 2 ( k + 2) . So k cannot be -2 or 1. 22. det cos k 1- sin k 2 sin k cos k = 2 cos 2 k + 2 sin 2 k = 2 . So k can have any value. 23. det( A- I 2 ) = det 1- 2 4- = (1- )(4- ) = 0 if is 1 or 4. 24. det( A- I 2 ) = det 2- 1- = (2- )(- ) = 0 if is 2 or 0. 291 Chapter 6 ISM: Linear Algebra 25. det( A- I 2 ) = det 4- 2 4 6- = (4- )(6- )- 8 = ( - 8)( - 2) = 0 if is 2 or 8. 26. det( A- I 2 ) = det 4- 2 2 7- = (4- )(7- )- 4 = ( - 8)( - 3) = 0 if is 3 or 8. 27. A- I 3 is a lower triangular matrix with the diagonal entries (2- ) , (3- ) and (4- ).... View Full Document

End of Preview

Sign up now to access the rest of the document