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Problem 3. Assume that A is a n x n real matrix. We define the exponential of A, as the series e^ = 1+A+ where I is the n x n identity matrix. a) Show that for 1 +²+ A = +. - (88) b where a and b are 2 real numbers, we have 1 n! A" +... e - (89). 0 eb b) Show that if B € M₂ (R) is similar to a diagonal matrix D with a transition matrix denoted by P (that is B = PDP 1) then eB Pepp 1. c) Show that if A and B are 2 matrices such that AB = BA, then eA¹B = eAB

Problem 3. Assume that A is a n x n real matrix. We define the exponential of A, as the series e^ = 1+A+ where I is the n x n identity matrix. a) Show that for 1 +²+ A = +. - (88) b where a and b are 2 real numbers, we have 1 n! A" +... e - (89). 0 eb b) Show that if B € M₂ (R) is similar to a diagonal matrix D with a transition matrix denoted by P (that is B = PDP 1) then eB Pepp 1. c) Show that if A and B are 2 matrices such that AB = BA, then eA¹B = eAB

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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Problem 3. Assume that A is a n x n real matrix. We define the exponential of A, as the series 1 e^= 1+A+A² where I is the n x n identity matrix. a) Show that for A +. where a and b are 2 real numbers, we have eª (8 = a = (8 %) 1 n! 0 eb 0 2). A" +... b) Show that if B € M₂ (R) is similar to a diagonal matrix D with a transition matrix denoted by P (that is B = PDP ¹) then eB Pepp 1 c) Show that if A and B are 2 matrices such that AB = BA, then eA¹B = eA eB
Transcribed Image Text:Problem 3. Assume that A is a n x n real matrix. We define the exponential of A, as the series 1 e^= 1+A+A² where I is the n x n identity matrix. a) Show that for A +. where a and b are 2 real numbers, we have eª (8 = a = (8 %) 1 n! 0 eb 0 2). A" +... b) Show that if B € M₂ (R) is similar to a diagonal matrix D with a transition matrix denoted by P (that is B = PDP ¹) then eB Pepp 1 c) Show that if A and B are 2 matrices such that AB = BA, then eA¹B = eA eB
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