Anonymous
Anonymous asked in Education & ReferenceHomework Help · 7 years ago

# Find the values of k?

Find the values of k such that the following system has a) no solution, or b) a unique one.

x+2y+2z=1

y+kz=1

-x+y+kz=k

Relevance

x  +  2y  +  2z  =  1 ... equation 1 (given)

- x  +         y  +  kz  =  k ... equation 3 (given)

3y  +  (2 + k)z  =  1 + k ... equation A

y + kz = 1 ... equation 2 (given)

3y + 3kz = 3 ... equation 2 (multiplied by 3)

3y  +  (2 + k)z  =  1 + k ... equation A

3y  +     3kz  =  3 ... equation 2 (×3)

——————————<subtract>

(2 + k – 3k)z  =  k – 2

z  =  - (k – 2)  ⁄  [2(k – 1) ]

... so: k ≠ 1

From this "z" solution and the given equation 2, you can solve for "y" in terms of "k"

Then "x" in terms of "k" can be found using the given equation 1 or 3

The solutions for y, and x also have this: k ≠ 1: restriction

There are NO solutions when: k  = 1

There is a unique solution for all OTHER real "k" values

Example: for k  =  2 ,  x = -1 , y = 1 , z = 0

so there are an infinite number of unique solutions.

• y = 1 - kz

Substitute into the other two eq'n.

x + 2(1 - kz) + 2z = 1

-x + 1 - kz + kz = k

x + 2 - 2kz + 2z = 1

-x + 1 = k

2 + 2z (1- k) = 1

1 = k

Substitue

2 + 2z ( k - k ) = 1

2 + 2z(0) = 1

2 + 0 = 1

2 = 1