david c. lay algebra lineal y sus aplicaciones solucionario

David C. Lay Algebra Lineal Y Sus Aplicaciones Solucionario For Free Report This

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The general solution of ( 2 )A I x 0 is 3 1 31 3.1 x A basis for the eigenspace corresponding to 2 is 1 31 3;1 another is 11.3 15. For 1 2 3 0 1 2 3 0 3 ( 3 ) 1 2 3 0 0 0 0 0.2 4 6 0 0 0 0 0 A I: 0 Thus 1 2 32 3 0, x x x with 2x and 3x free.

David C. Lay Algebra Lineal Y Sus Aplicaciones Solucionario For Free Report This

Lay Cap 5 Date post: 25-Jul-2015 Category: Education View: 516 times Download: 15 times Download for free Report this document Share this document with a friend Transcript: 257 5.1 SOLUTIONS Notes: Exercises 16 reinforce the definitions of eigenvalues and eigenvectors.

The subsection on eigenvectors and difference equations, along with Exercises 33 and 34, refers to the chapter introductory example and anticipates discussions of dynamical systems in Sections 5.2 and 5.6.

The number 2 is an eigenvalue of A if and only if the equation 2A x x has a nontrivial solution.

Check 3 2 times the first entry of x: 2 (3 2)( 1 2) 3 2 2 2 1 2 2 This matches the first entry of,xA so 1 21 is an eigenvector of A, and the corresponding eigenvalue is 3 2.

Is Ax a multiple of x Compute 3 7 9 4 04 5 1 3 0.2 4 4 1 0 So 431 is an eigenvector of A for the eigenvalue 0.

Is Ax a multiple of x Compute 3 6 7 1 2 13 3 7 2 4 ( 2) 25 6 5 1 2 1 So 121 is an eigenvector of A for the eigenvalue 2.

To determine if 4 is an eigenvalue of A, decide if the matrix 4A I is invertible.

A I Invertibility can be checked in several ways, but since an eigenvector is needed in the event that one exists, the best strategy is to row reduce the augmented matrix for ( 4 )A I x 0: 1 0 1 0 1 0 1 0 1 0 1 02 1 1 0 0 1 1 0 0 1 1 03 4 1 0 0 4 4 0 0 0 0 0 The equation ( 4 )A I x 0 has a nontrivial solution, so 4 is an eigenvalue.

Any nonzero solution of ( 4 )A I x 0 is a corresponding eigenvector.

The entries in a solution satisfy 1 3 0x x and 2 3 0, x x with 3x free.

The general solution is not requested, so to save time, simply take any nonzero value for 3x to produce an eigenvector.

If 3 1,x then ( 1 1 1).,,x Note: The answer in the text is (1 1 1),,, written in this form to make the students wonder whether the more common answer given above is also correct.

This may initiate a class discussion of what answers are correct.

To determine if 3 is an eigenvalue of A, decide if the matrix 3A I is invertible.

A I Row reducing the augmented matrix (A 3 ) I 0 yields: 2 2 2 0 1 1 1 0 1 0 3 03 5 1 0 0 1 2 0 0 1 2 00 1 2 0 0 2 4 0 0 0 0 0 The equation ( 3 )A I x 0 has a nontrivial solution, so 3 is an eigenvalue.

Any nonzero solution of ( 3 )A I x 0 is a corresponding eigenvector.

The entries in a solution satisfy 1 33 0x x and 2 32 0, x x with 3x free.

If 3 1,x then (3 2 1).,,x 5.1 Solutions 259 9.

For 5 0 1 0 4 0 1 12 1 0 1 2 0 A I: The augmented matrix for ( )A I x 0 is 4 0 0.2 0 0 Thus 1 0x and 2x is free.

The general solution of ( )A I x 0 is 2 2,ex where 20,1 e and so 2e is a basis for the eigenspace corresponding to the eigenvalue 1.

For 5 0 5 0 0 0 5 52 1 0 5 2 4: A I The equation ( 5 )A I x 0 leads to 1 22 4 0, x x so that 1 22x x and 2x is free.

The general solution is 1 2 22 2 2 2.

So 21 is a basis for the eigenspace.

For 10 9 4 0 6 9 4 4.4 2 0 4 4 6: A I The augmented matrix for ( 4 )A I x 0 is 6 9 0 1 9 6 0.4 6 0 0 0 0 Thus 1 2(3 2)x x and 2x is free.

The general solution is 1 2 22 2 (3 2) 3 2.

A basis for the eigenspace corresponding to 4 is 3 2.1 Another choice is 3.

A I The augmented matrix for ( 10 )A I x 0 is 6 2 0 1 1 3 0.3 1 0 0 0 0 Thus 1 2( 1 3)x x and 2x is free.

The general solution is 1 2 22 2 (1 3) 1 3.

A basis for the eigenspace corresponding to 10 is 1 3.1 Another choice is 1.3 12.

For 7 4 1 0 6 4 13 1 0 1 3 2 A I: The augmented matrix for ( )A I x 0 is 6 4 0 1 2 3 0.3 2 0 0 0 0 Thus 1 2( 2 3)x x and 2x is free.

A basis for the eigenspace corresponding to 1 is 2 3.1 Another choice is 2.3 For 7 4 5 0 2 4 5 5.3 1 0 5 3 6: A I 260 CHAPTER 5 Eigenvalues and Eigenvectors The augmented matrix for ( 5 )A I x 0 is 2 4 0 1 2 0.3 6 0 0 0 0 Thus 1 22x x and 2x is free.

A basis for the eigenspace is 2.

For 1: 4 0 1 1 0 0 3 0 1 1 2 1 0 0 1 0 2 0 02 0 1 0 0 1 2 0 0 A I The equations for ( )A I x 0 are easy to solve: 1 31 3 02 0 x xx Row operations hardly seem necessary.

The general solution of ( )A I x 0 is 2 2,ex where 2 (0 1 0),,,e and so 2e provides a basis for the eigenspace.

For 2: 4 0 1 2 0 0 2 0 1 2 2 1 0 0 2 0 2 1 02 0 1 0 0 2 2 0 1 A I 2 0 1 0 2 0 1 0 0 1 2 0 ( 2 ) 2 1 0 0 0 1 1 0 0 1 02 0 1 0 0 0 0 0 0 0 0 0 A I1 1 0 So 1 3 2 3(1 2),x x x x, with 3x free.

The general solution of ( 2 )A I x 0 is 3 1 21.1 x A nice basis vector for the eigenspace is 12.2 For 3: 4 0 1 3 0 0 1 0 1 3 2 1 0 0 3 0 2 2 02 0 1 0 0 3 2 0 2 A I 1 0 1 0 1 0 1 0 0 1 0 ( 3 ) 2 2 0 0 0 2 2 0 0 1 02 0 2 0 0 0 0 0 0 0 0 0 1 1 0 A I So 1 3 2 3,, x x x x with 3x free.

A basis vector for the eigenspace is 11.1 5.1 Solutions 261 14.

For 1 0 1 2 0 0 3 0 1 2 ( 2 ) 2 1 3 0 0 2 0 1 1 0.4 13 1 0 0 2 4 13 3: A I A I The augmented matrix for ( 2), x 0A I or ( 2 ), x 0A I is 3 0 1 0 1 0 1 3 0 1 0 1 3 0 ( 2 ) 1 1 0 0 0 1 1 3 0 0 1 1 3 04 13 3 0 0 13 13 3 0 0 0 0 0 A I 0 Thus 1 3 2 3(1 3) (1 3),, x x x x with 3x free.

The general solution of ( 2 )A I x 0 is 3 1 31 3.1 x A basis for the eigenspace corresponding to 2 is 1 31 3;1 another is 11.3 15.

For 1 2 3 0 1 2 3 0 3 ( 3 ) 1 2 3 0 0 0 0 0.2 4 6 0 0 0 0 0 A I: 0 Thus 1 2 32 3 0, x x x with 2x and 3x free.