3f-fall2010-homework12solutions

3f-fall2010-homework12solutions - Math 3F — Homework #12...

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Unformatted text preview: Math 3F — Homework #12 Solutions 1 −2 0 4 −2 3 2 −1 and B = −1 5 0 , compute: 1. If A = 3 2 1 3 6 12 (a) 2A + B 1 −2 0 4 −2 3 2 −1 + −1 5 0 2A + B = 2 3 1 3 6 12 2 2 −4 0 4 −2 3 4 −2 + −1 5 0 = 6 4 2 6 6 12 6 −6 3 9 −2 . = 5 10 3 8 4 −2 3 − 4 −1 5 0 6 12 16 −8 12 − −4 20 0 24 48 −12 −1 . −5 (b) A − 4B 1 −2 0 2 −1 A − 4B = 3 2 1 3 1 −2 0 2 −1 = 3 2 1 3 −15 6 7 −18 = −22 −3 (c) AB 1 −2 0 4 −2 3 6 −12 3 2 −1 −1 5 0 = 4 3 7 . AB = 3 2 1 3 6 12 25 4 12 4 −9 11 1 −2 0 4 −2 3 2 −1 = 14 12 −5 . 5 0 3 BA = −1 13 −8 5 2 1 3 6 12 (d) BA ( 2. If A = 1 + i − 1 + 3i 3 + 2i 2−i ( A − 2B = ) and B = ( i 3 2 −2i ) ) , find: (a) A − 2B −2 ( )( ) 1 + i −1 + 3i 2i 6 = − 3 + 2i 2−i 4 −4i ( ) 1−i − 7 + 3i = . − 1 + 2i 2 + 3 i 1 + i −1 + 3i 3 + 2i 2−i ( i 3 2 −2i ) (b) 3A + B )( ) 1 + i −1 + 3 i i 3 3A + B = 3 + 3 + 2i 2−i 2 −2i ( )( ) 3 + 3i −3 + 9i i 3 = + 9 + 6i 6 − 3i 2 −2i ( ) 3 + 4i 9i = . 11 + 6i 6 − 5i (c) AB ( AB = (d) BA ( BA = i 3 2 −2i ( )( 1 + i − 1 + 3i 3 + 2i 2−i ) = ( 8 + 7i 3 − 4i 6 − 4i −4 + 2i ) . 1 + i −1 + 3i 3 + 2i 2−i )( i 3 2 −2i ) = ( −3 + 7i 9 + 5i 2+i 7 + 2i ) . ( ) ( ) 4 3 −2 2t ′ 3. Verify that ⃗ = x e satisfies the equation ⃗ = x ⃗. x 2 2 −2 Let’s start from the right side. ( ) ( )( ) () () 3 −2 3 −2 4 8 4 2t 2t ⃗= x e= e =2 e2t = ⃗ ′ . x 2 −2 2 −2 2 4 2 ) ( ) () ( ) 1 1 1 2 −1 t ′ t ⃗+ x et . te satisfies ⃗ = x 4. Check that ⃗ = x e +2 −1 3 −2 0 1 Again, start from the right side. ( ) ( ) ( ) (( ) ())( ) 2 −1 1 2 −1 1 1 1 ⃗+ x et = et + 2 tet + et 3 −2 −1 3 −2 0 1 −1 () () ( ) 2 2 1 t t = e+ te + et 3 2 −1 () () 3 2 t = e+ tet = ⃗ ′ . x 2 2 ( ( 5. Find the eigenvalues and corresponding eigenvectors for 3 −2 4 −1 ) . Solve the characteristic equation: ( ) 3−r −2 det = (3 − r)(−1 − r) + 8 = 0 4 −1 − r That is, r2 −2r +5 = 0. Using the quadratic equation, we get r = 1±2i. For r1 = 1 + 2i: ( ) ( ) 2 − 2i −2 0 1 − i −1 0 −→ , 4 −2 − 2i 0 0 00 1 so a corresponding eigenvector is ⃗1 = v 1−i ( ) 1 responding to r2 = 1 − 2i is ⃗2 = v . 1+i ( ) . An eigenvector cor- ( 6. Find the eigenvalues and corresponding eigenvectors for −2 1 1 −2 ) . Again, start with the characteristic equation: ( ) −2 − r 1 det = (−2 − r)2 − 1 = r2 + 4r + 3 = 0 1 −2 − r Factoring, we see that r1 = −1 and r2 = −3. For r1 = −1: ( ) ( ) −1 10 1 −1 0 −→ , 1 −1 0 0 00 ( so a corresponding eigenvector is ⃗1 = v ( 110 110 ) −→ ( so a corresponding eigenvector is ⃗2 = v ( 1 1 ) . For r2 = −3: ) , 110 000 1 −1 ) . 7. Find the general solution of each system of equations. ( ) 3 −2 ′ (a) ⃗ = x ⃗ x 2 −2 We first find the eigenvalues: ( ) 3−r −2 det = (3 − r)(−2 − r) + 4 = r2 − r − 2 = 0. 2 −2 − r The eigenavlues are r1 = −1 and r2 = 2. Now find some eigenvectors. For r1 = −1: ( ) ( ) 4 −2 0 2 −1 0 −→ , 2 −1 0 0 00 ( so ⃗1 = v 1 2 ) . Now, for r2 = 2: ( ( so ⃗2 = v ) . Hence, the general solution is ( ⃗ = c1 x ′ 1 −2 0 2 −4 0 ) −→ ( 1 −2 0 0 00 ) , 2 1 1 2 ) e −t ( + c2 2 1 ) e−2t . ) 1 1 (b) ⃗ = x ⃗ x 4 −2 Again, eigenvalues first: ( ) 1−r 1 det = (1 − r)(−2 − r) − 4 = 0, 4 −2 − r ( so the eigenvalues are r1 = −3 and r2 = 2. For r1 = −3: ) ( ) ( 410 410 −→ , 410 000 ( so ⃗1 = v 1 −4 ( ( so ⃗2 = v ) . For r2 = 2: −1 10 4 −4 0 ) −→ ( 1 −1 0 0 00 ) , 1 1 ) . Hence, the general solution is ( ⃗ = c1 x 1 −4 ) −3t ( + c2 e 1 1 ) e2t . ...
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This note was uploaded on 01/14/2011 for the course MATH 3F taught by Professor Williamson during the Fall '09 term at Laney College.

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