1332W10T2S

# 1332W10T2S - Version 1 Question 1. [4 points] a ) An...

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Unformatted text preview: Version 1 Question 1. [4 points] a ) An equilibrium point x * of the differential equation dx dt = F ( x ) satisfies F ( x * ) = 0. We have F ( x * ) = 25 x *- ( x * ) 3 = x * (25- ( x * ) 2 ) , so the equilibria are , 5 ,- 5 . b ) We have F ( x ) = 25- 3 x 2 . The derivative test gives F (0) = 25 > = 0 the equilibrium is unstable F (5) =- 50 > 0 = 5 the equilibrium point is stable and F (- 5) = F (5) =- 50 > 0 = - 5 the equilibrium point is stable. We could also draw the graph of the function and determine the stability using the regions of increase or decrease. c ) See Figure 1. Question 2. [6 points] a ) It is easy to verify that f (2) = 8- 24 + 26- 10 = 0. b ) Dividing f ( x ) by x- x 1 = x- 2 gives g ( x ) = f ( x ) x- 2 = x 2- 4 x + 5. g ( x ) = 0 implies x 2 / 3 = 1 2 (4 - 4) = 2 i . c ) x 2 x 3 = (2- i )(2 + i ) = 4- 2 i + 2 i- i 2 = 5 . d ) x 2 x 3 = 2- i 2 + i = (2- i )(2- i ) (2- i )(2 + i ) = 4- 2 i- 2 i + i 2 5 = 3 5- 4 5 i. e ) (3 + i )(3 + i ) = 9 + 3 i + 3 i- i 2 = 8 + 6 i. f ) Using the relationship e i = cos( ) + i sin( ) we find e i = cos(1) + i sin(1) . Question 3. [5 points] (a) The determinant of B is 1 det( B ) = 1 4 0 + 2 1 c + 3 5- 1 1 5- 2 - 3 4 c so det( B ) =- 10 c- 5 . The matrix B is invertible is and only if det( B ) 6 = 0, ie c 6 =- 1 / 2. (b) The associated augmented matrix is M = 5 3 1 1 2 1 3- 1 a . (i) The system has a unique solution if the determinant is not zero. det( M ) = 5 2 a + 1 (- 1) 1 + 3 3 1- 5 (- 1) 1- 1 3 a- 3 2 1 so det( M ) = 7 a + 7 . There will be a unique solution if a 6 =- 1. (ii) To have an infinite number of solutions, we first need a =- 1, so that the matrix is noninvertible. Then the system becomes 5 x + 3 y + z = 1 x + 2 y + z = 2 3 x- y- z = b Subtracting the first line from the second and adding the first line to the third, we have 5 x + 3 y + z = 1- 4 x- y = 8 x + 2 y = b + 1 Adding twice the second row to the third, we have 5 x + 3 y + z = 1- 4 x- y = 0 = b + 1 Thus, the system will have an infinite number of solutions if a =- 1 and b =- 1. (iii) From the previous calculations, if a =- 1 and b 6 =- 1, the system has no solutions. 2 Question 4. [6 points] (a) The eigenvalues of A are the numbers where the polynomial det( A- I ) equals 0. So lets evaluate that determinant. A- I = 5- - 2 1 3- , so det( A- I ) = (5- )(3- )- 1(- 2) = 2- 8 + 17 . The roots of this polynomial are =- (- 8) p (- 8) 2- 4 * 17 2 = 8 - 4 2 = 4 i, as claimed. (b) The eigenvector corresponding to an eigenvalue is a nonzero solution v to ( A- I ) v = . Here is 4 + i , so A- I = 1- i- 2 1- 1- i ....
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## This note was uploaded on 03/19/2011 for the course MAT 1332 taught by Professor Munteanu during the Spring '07 term at University of Ottawa.

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1332W10T2S - Version 1 Question 1. [4 points] a ) An...

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