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MA122 Mock Midterm Answers (Full Solutions will NOT be posted; use the MAC’s drop-in help centre if you have any questions.) **** Please remember that the mock test was meant as a means of providing an extra set of practice questions and basis for a review class. Do not study for the midterm based solely on the topics covered by the mock test! Go back through notes/labs/homework to ensure you have reviewed all concepts discussed in the course. 1. A is a subspace of R4 2 3 2 3 2 3 2. (a) ! x = 415 + t 4 4 5, t 2 R 0 1 3. (a) p9 902 4. (a) p4 21 p (b) (b) 1 1 ; 1; 2 2 (b) 821 3x 3y + 6z = 27 (or any multiple, such as x + y (c) the line is parallel to the plane 2z = 9) 0 5. — — – ! use properties of cross product ! 6. — — – ! use ! v1 (k1 ! v1 + k2 ! v2 + k3 ! v3 ) = ! v1 0 to show that k1 = 0 ! similarily, show that k2 = 0 and k3 = 0 2 7. — — – ! use property that k! xk =! x ! x to show LS = 2 3 1=3 6 7 6 2 7 6 7 8. (a) ! x =6 7 6 1 7 4 5 0 9. (a) a = 2 2 2 6 16 6 ! (b) v = 6 36 4 (b) a = 1 3s 6 1+s 6 6 6 ! s 10. (a) x = 6 6 6 t 4 t (b) for ex: 2t 3 2 2 3 2 0 7 6 7 6 627 6 07 7 6 7 6 7 + 26 7 + 6 37 637 6 5 4 5 4 6 1 3 3 (c) a 2 R, a 6= 2 6 t 7 7 6 7 6 7 6 7=6 7 6 7 6 5 4 3 2 1 6 17 7 6 7 6 7 0 7 + s6 6 7 6 7 6 05 4 0 3 3 2 2 0 3 7 27 7 7 27 5 0 (justify each answer!) 3 2 17 7 7 07 7, s; t 2 R 7 17 5 6 17 7 6 7 6 7 1 7 + t6 6 7 6 7 6 05 4 0 1 = RS if s = 0 = t then a sol’n would be ! x = ( 1; 1; 0; 0; 0) if s = 1 and t = 2 then a sol’n would be ! x = ( 8; 0; 1; 2; 2)