Week 1 try me exercise solution PDF

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Week 1 Try me Exercises Solution All solutions are here but I still haven’t carefully proof read it. So there may be some silly typos. Contact email: [email protected] Question 1: Consider the system of linear equations

x  y  2 x y  2 Determine the value(s) of

such the system has:

1) infinitely many solutions; 2) exactly one solution; 3) no solution. Solution: The augmented matrix is

   1 2  1  2   Which reduces to (performing row operation)

2    1 2  R1  R2 1   2    R1 R2  R2 1    2  1  2    1 2        0   1  2  2 Solve 2 1  0 and 2   2  0  2  2  0 2 1  0 (  1)(  1)  0 2  2   1   1,   1 1) For the system to have infinitely many solutions if a) and b) satisfy: a)  2 1  0 b) 2  2  0 1 Week 1 Try me Exercises Solution

Semester 1 2019

Thus, from the above calculation   1 . 2) For the system to have exactly one solution if a) satisfy: a)  2 1  0 Thus, from the above calculation   1 and   1 . 3) For the system to have no solution if a) and b) satisfy: a)  2 1  0 b) 2  2  0 Thus, from the above calculation   1 . Question 2: Give the solution set of 4 x  5 y  7 z  3 in parametric form Solution:

3 4 5 Solving foe 𝑧, we get z    x  y. Thus one parametric form could be 7 7 7

3 4 5 x  s , y  t and z    s  t , where s and t are real numbers. 7 7 7 (Note: the solution is not unique.)

Question 3: Determine a polynomial whose graph passes through the given points. 1,0  , 2,2 , 3,5 Solution: p( x)  a0  a1  a2x 2 Substituting x  1,2,3 into 𝑝(𝑥) and equating the results to the respective 𝑦 values.

p (1)  a 0  a 1(1)  a 2(1) 2  a 0  a 1  a 2  0 p (2)  a 0  a 1(2)  a 2(2) 2  a 0  2a 1  4a 2  2 p (3)  a 0  a 1(3)  a 2(3) 2  a 0  3a 1  9a 2  5 2 Week 1 Try me Exercises Solution

Semester 1 2019

The associated augmented matrix:

1 1 1 0  1 2 4 2   1 3 9 5  1 1 1 0  0 1 3 2   R  R  R ,  R  R  R 2 2 1 3 3   1  0 2 8 5 1 1 1 0  0 1 3 2  2R  R  R 2 3 3    0 0 2 1   1 1 1 0   1  0 1 3 2  2 R3  R3  1 0 0 1   2 The corresponding system is

a1  a 2  a 3  0 a2  3a3  2 a3 

1 2

Use back substitution:

a1=1,

1 1 a2  , a3  2 2

Therefore

p (x )  1

1 1 x  x 2. 2 2

3 Week 1 Try me Exercises Solution

Semester 1 2019...