# ch6 - Chapter 6 ISM Linear Algebra Chapter 6 6.1 1 Fails to...

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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 0 11 7 0 0 5 = 2 · 11 · 5 + 0 + 0 - 0 - 0 - 0 = 110. 6. Invertible; since det 6 0 0 5 4 0 3 2 1 = 6 · 4 · 1 + 0 + 0 - 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 0 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 0 2 6 0 0 4 = 8 k, so k 6 = 0 will ensure that this matrix is invertible. 290

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ISM: Linear Algebra Section 6.1 14. det 4 0 0 3 k 0 2 1 0 = 0 , so the matrix will never be invertible, no matter which k is chosen. 15. det 0 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 0 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 0 2 0 sin k 0 cos k = 2 cos 2 k + 2 sin 2 k = 2 . So k can have any value. 23. det( A - λI 2 ) = det 1 - λ 2 0 4 - λ = (1 - λ )(4 - λ ) = 0 if λ is 1 or 4. 24. det( A - λI 2 ) = det 2 - λ 0 1 0 - λ = (2 - λ )( - λ ) = 0 if λ is 2 or 0. 291
Chapter 6 ISM: Linear Algebra 25. det( A - λI 2 ) = det 4 - λ 2 4 6 - λ = (4 - λ )(6 - λ ) - 8 = ( λ - 8)( λ - 2) = 0 if λ is 2 or 8. 26. det( A - λI 2 ) = det 4 - λ 2 2 7 - λ = (4 - λ )(7 - λ ) - 4 = ( λ - 8)( λ - 3) = 0 if λ is 3 or 8. 27. A - λI 3 is a lower triangular matrix with the diagonal entries (2 - λ ) , (3 - λ ) and (4 - λ ). Now, det( A - λI 3 ) = (2 - λ )(3 - λ )(4 - λ ) = 0 if λ is 2, 3 or 4. 28. A - λI 3 is an upper triangular matrix with the diagonal entries (2 - λ ) , (3 - λ ) and (5 - λ ). Now, det( A - λI 3 ) = (2 - λ )(3 - λ )(5 - λ ) = 0 if λ is 2, 3 or 5. 29. det( A - λI 3 ) = det 3 - λ 5 6 0 4 - λ 2 0 2 7 - λ = (3 - λ )( λ - 8)( λ - 3) = 0 if λ is 3 or 8. 30. det( A - λI 3 ) = det 4 - λ 2 0 4 6 - λ 0 5 2 3 - λ = (4 - λ )(6 - λ )(3 - λ ) - 8(3 - λ ) = (3 - λ )(8 - λ )(2 - λ ) = 0 if λ is 3, 8 or 2. 31. This matrix is upper triangular, so the determinant is the product of the diagonal entries, which is 24.

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