Question

7. Use the Gauss-Jordan Algorithm to determine the points of intersection for each of the following systems of planes: • If the planes have no point of intersection, just say so: • If they have a single point of intersection, tell me what it is: . If they have infinitely many points of intersection, depict a general form for the points of intersection and determine any three points of intersection, + y + 2 -By+22 -2y+324 (b) 2x H5y 4 6 points 2+5y 5 3a +9y You must use the Gauss-Jordan Algorithm as described in my Module 2 notes and depict clearly, in the manner done there, your elementary operations, and don't forget to give me three points of intersection the system is dependent Х. W 29

Answer 1

7) Gauss- Jordan Algooithm m. a 1. x+y+z=2 x -2y +33 34 = 5. ant Sy in matrix & A X B 1 2 X 1 4 1 -2 3 os 5 cs 2 o 3 to take we For finding finding a, j, a have augu matăx . C = CA i B] mented ر A8 1 2 ſ : 2 1 1 1 1 R₂ R₃ O ; 5 2 is 4 1 -2 3 . 1 -2 3 un 0 2 un - - 1 2 1 1 1 1 R2 R, -2R, 0 -2/3 123 O 3 -2 : 1 2 3 4 1 -2 3 4 - R₃-R. -IR3 3 Il 2 -2/3 : V3 a 1 0 1 1 i 2 -2/3: 13 -2/3 0 - -3 o

R3-R₂ [ -2/3 2 1/3 0. non consistent Since, of 1, the system is is no set of values which satisfy all three equation at once. So, there (6) - 3y +23=2 anit sy 4 30+9y-z=s. writing in matrix form co X a -3 4 O g Q 5 3 3 g -1 Augumented matrix a [A:B 0 3 al Rhea O -3 a tan *** S 3 9 -1,5 sla - sla O / R2 IRI 0 Oi2 -2/3 :-?13 - 5 2 l 2 O -3 a 5 3 g

L 1/2 R3 sla Rz-R, - 512 0:2 -2/3.-23 V2 -1/3 = -1/3 a -1/3 -2/3 -1/3 : 53 - 3 R3 - 1 R2 s l2 0! 2 -2/3 : -2/3 0 1 0 o As we can see that 2 = 0 on & Oy + O2 is satisfied So, the planes have infinitely many for finding them, intersection points, Bott us 54 2=k. EQ: ý cestou 23. f x + 2 =k old - at sy 4 =) K =) ** *[3+3*]2 - 1 - 22 and [a= - adtak 3 ly 2 X 3 X 2. 12 5(-2+2k) 5(-2+24) the solution is in Whaaa we can express PR 51-2+k) -atak 3 k ]

for three lang) points of intersection :- ; ) let K=2 12 = h = olin 5(2) 5(-2+4) le ? sa) Š JO y -d +262) -2+9 3 3 LA) 2. let K=.3. 24 s 3 12 SX4 4. 3 5. ។ -2 + 6 =; 4 4. (B) 3 f 3 = 3. 17:) Let: K=4 ra ią 5x6 (3344) 1 (C) 2 ỹ = lu 11 4 three pointy of A, B and C represents any intersection.