# Matrices Assignment 4 1

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Subject
Mathematics
School
Syracuse University
Type
Homework
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Matrices Assignment.
Student’s Name:
Institutional Affiliation:

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Matrices 2
1) The augmented matrix of the given system of equations will be;
A =

  
 
2) -6R
1
+-6R
2
R
2
on
 

will give out:
  


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Running head: Matrices 1 Matrices Assignment. Student’s Name: Institutional Affiliation: 2 Matrices 1) The augmented matrix of the given system of equations will be; 0 43 5 3 A = (58 0 25 23 ) 2 1 −36 −17 2) -6R1+-6R2→ R2 1 4 2 on (−3 −4 4 ) will give out: 6 5 −7 1 4 2 (−9 −28 −8) 6 5 −7 3 Matrices 3) On the given augmented matrix 1 −3 −4 −5 = ( 4 −11 4 3) −4 8 3 3 First: R2-4R1→ R2 1 −3 −4 −5 =( 0 1 20 17 ) −4 8 3 3 Second: R3+4R1→ R3 1 = (0 0 −3 −4 −5 1 20 17 ) −4 −13 −17 Third: R3+4R2→ R3 1 = (0 0 −3 −4 −5 1 20 17 ) → The final answer. 0 67 51 4 Matrices 4) a) Augmented Matrix is: −3 6 0 33 = ( 0 11 −5 58) 0 0 0 0 b) -1⁄3R1→R2 1 = (0 0 −2 0 −11 11 −5 58 ) 0 0 0 c) +5R1+R2→ 1 = (5 0 −2 0 −11 1 −5 3 ) 0 0 0 d) The reduced matrix is 1 = (0 0 0 0 1 0 0 0 −5/11 58/11 ) 0 x = -5/11 x = 58/11 =0 e) The system of equation has no solution. f) x = No Solution y=No Solution. 5 Matrices −1 1 1 0 5) = ( 4 −3 −1 −4) 1 1 1 0 R1/-1→R1 −1 1 1 0 = ( 4 −3 −1 −4) 1 1 1 0 R2-4 R1→R2 followed by R3-1R1→R3 1 = (0 0 −1 −1 0 1 3 −4) 2 2 0 R3/-4→R3 1 = (0 0 0 2 1 3 2 1 −4 −4) −2 R1-2R3→R1 followed by R2-3R3-R2 1 = (0 0 Hence; X=0 Y=2 Z=-2 0 0 1 0 0 1 0 2) −2 6 Matrices 8 −3 ℎ 6) ( ) −32 12 8 8x-3y=h -32x+12y=8 Divide the second equation by 4 8x-3y=h -8x+3y=2 Add the 2 equations 0=h+2 h = -2 7) There is no solution when there is ...
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