HW9solutions

# HW9solutions - Section 6.1 2 Invertible because det 2 4 3...

This preview shows page 1. Sign up to view the full content.

Section 6.1 2. Invertible because det ± 2 3 4 5 ² = - 2 . 8. Not invertible because the determinant is 0. 16. det 1 2 3 4 k 5 6 7 8 = 60 + 84 + 8 k - 18 k - 35 - 64 = 45 - 10 k. The matrix is invertible when k 6 = 4 . 5 . 30. det( A - λI ) = 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 ) , which is 0 is λ is 3, 8, or 2. 40. Repeatedly expand down the ﬁrst column, to get that the determinant is 5(4)(3) det ± 0 2 1 0 ² = - 120. 46. If A = ± a b c d ² , then A - 1 = 1 det a ± d - b - c a ² , so det A - 1 = ³ 1 det A ´ 2 det ± d - b - c a ² = 1 det A . 56. a. We expand M n down the ﬁrst column and ﬁnd that d n = d n - 1 if n is odd, or d n = - d n - 1 if n is even. We can write this in one formula as d n = ( - 1) n - 1 d n - 1 . b. The ﬁrst eight terms are 1 , - 1 , - 1 , 1 , 1 , - 1 , - 1 , and 1. c. d 100 = d 4 = 1 . Section 6.2 8. Using a sequence of four row swaps, we turn this into an upper triangular matrix with diagonal entries 1 , 1 , 1 , 1
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Ask a homework question - tutors are online