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Let A and f(t) be as in Exercise 20. (a) Prove that f(t) = (A₁1-t) (A22-t) (Ann-t)+q(t), where q(t) is a polynomial of degree at most n-2. Hint: Apply mathematical induction to n. (b) Show that tr(A) = (-1)n-¹a an-1.

Let A and f(t) be as in Exercise 20. (a) Prove that f(t) = (A₁1-t) (A22-t) (Ann-t)+q(t), where q(t) is a polynomial of degree at most n-2. Hint: Apply mathematical induction to n. (b) Show that tr(A) = (-1)n-¹a an-1.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question21

20. Let A be an n × n matrix with characteristic polynomial ƒ(t) = (−1)″țn + An-1th-1 + + a₁t+ao. Prove that f(0) = ao = det(A). Deduce that A is invertible if and only if ao 0. 21. Let A and f(t) be as in Exercise 20. (a) Prove that f(t) = (A11 – t)(A22 − t) · · · (Ann −t)+q(t), where q(t) is a polynomial of degree at most n-2. Hint: Apply mathematical induction to n. (b) Show that tr(A) = (−1)n-¹an-1.
Transcribed Image Text:20. Let A be an n × n matrix with characteristic polynomial ƒ(t) = (−1)″țn + An-1th-1 + + a₁t+ao. Prove that f(0) = ao = det(A). Deduce that A is invertible if and only if ao 0. 21. Let A and f(t) be as in Exercise 20. (a) Prove that f(t) = (A11 – t)(A22 − t) · · · (Ann −t)+q(t), where q(t) is a polynomial of degree at most n-2. Hint: Apply mathematical induction to n. (b) Show that tr(A) = (−1)n-¹an-1.
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下午7:34 2月26日 週日 Homework help starts here! Thus, we have bartleby.com NOT FOR TEATS! Receive feat aime accome updates and special offers from paralepy. Mag & data rates may Now, we expand the product (B₁1 - t) (B22-t) (Bn-1,n-1- row, namely: f(t) = det tiji = det = ... det (B-tIn-1) = (B₁1 – t)(B22 – t) ... (Bn-1,n-1 – t) A₁1 - t A21 : An-1,1 where p(t) is a polynomial of degree at most n - 2. : An-1,1 An1 A₁1-t A21 +)(A Let A and f(t) be as in Exercise 20. (a) Prove that f(t) =... A12 A22 - t t) using the formula for the determinant of a matrix with a zero column or : An-1,2 An2 A12 A22 - t An-1,2 Solution (b) Using the result from part (a), we have: Expanding the determinant using cofactor expansion along the last row, we get : A1,n-1 A2,n-1 An-1,n-1-t Ann-1 A1,n-1 A2,n-1 n+j An-1,n-1-t n-1 f(t) = (-1)" (Ann - t) Cnn +(-1)+ AnjСn,j(Bjj – t) + p(t) j=1 - n-1 = (-1)" (Ann – t) det(B − tIn-1) + (−1)n+j AnjCn,j det (Bjj — tIn−1) + p(t) - j=1 Ain A2n n-1 (Ann – t) (−1)n-¹(n − 1)t + Σ(−1)n+j+¹ AnjCnj det (B − tIn−1) + p(t) j=1 = = (Ann- t) det(B − tIn−1) + q(t) where g(t) is a polynomial of degree at most n - 2, obtained by collecting the terms of degree at most n-2 in the above expression. An-1,n Ann -t (Ann- t) +p(t) Thus, we have shown that f(t) = (A11 – t) (A22 – t) (Ann- t) + q(t), as desired. Moreover, since q(t) has degree at most n - result holds for all n ≥ 1 by mathematical induction. 2, the f(t) = (A11 t) (A22 t) (Ann- t) + g(t) (+) Setting t = 0, we get: = ao f(0) = (−1)".0″+an-1 .On. +. + a₁.0 + ao On the other hand, we can expand f(t) as follows: f(t) (A +) + Ć ASK AN EXPERT 見 CHAT + 43% VX MATH SOLVER
Transcribed Image Text:下午7:34 2月26日 週日 Homework help starts here! Thus, we have bartleby.com NOT FOR TEATS! Receive feat aime accome updates and special offers from paralepy. Mag & data rates may Now, we expand the product (B₁1 - t) (B22-t) (Bn-1,n-1- row, namely: f(t) = det tiji = det = ... det (B-tIn-1) = (B₁1 – t)(B22 – t) ... (Bn-1,n-1 – t) A₁1 - t A21 : An-1,1 where p(t) is a polynomial of degree at most n - 2. : An-1,1 An1 A₁1-t A21 +)(A Let A and f(t) be as in Exercise 20. (a) Prove that f(t) =... A12 A22 - t t) using the formula for the determinant of a matrix with a zero column or : An-1,2 An2 A12 A22 - t An-1,2 Solution (b) Using the result from part (a), we have: Expanding the determinant using cofactor expansion along the last row, we get : A1,n-1 A2,n-1 An-1,n-1-t Ann-1 A1,n-1 A2,n-1 n+j An-1,n-1-t n-1 f(t) = (-1)" (Ann - t) Cnn +(-1)+ AnjСn,j(Bjj – t) + p(t) j=1 - n-1 = (-1)" (Ann – t) det(B − tIn-1) + (−1)n+j AnjCn,j det (Bjj — tIn−1) + p(t) - j=1 Ain A2n n-1 (Ann – t) (−1)n-¹(n − 1)t + Σ(−1)n+j+¹ AnjCnj det (B − tIn−1) + p(t) j=1 = = (Ann- t) det(B − tIn−1) + q(t) where g(t) is a polynomial of degree at most n - 2, obtained by collecting the terms of degree at most n-2 in the above expression. An-1,n Ann -t (Ann- t) +p(t) Thus, we have shown that f(t) = (A11 – t) (A22 – t) (Ann- t) + q(t), as desired. Moreover, since q(t) has degree at most n - result holds for all n ≥ 1 by mathematical induction. 2, the f(t) = (A11 t) (A22 t) (Ann- t) + g(t) (+) Setting t = 0, we get: = ao f(0) = (−1)".0″+an-1 .On. +. + a₁.0 + ao On the other hand, we can expand f(t) as follows: f(t) (A +) + Ć ASK AN EXPERT 見 CHAT + 43% VX MATH SOLVER
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