Question

For what value of k
X+y+z=2
2x+y-z=3
3x+2y+kz=4
Has a unique solution

Answer 1

from 1st equation

x=2-y-z

substituting this value in equation 2x+y-z=3

2(2-y-z)+y-z=3

4-2y-2z=3

1=2y+2z

1=2(y+z)

y+z=1/2

z = 0.5-y

substituting this value in equation 3x+2y+kz=4

from above we know that
x = 2-y-z
z =0.5-y

3x+2y+kz =4
3(2-y-z)+2y+k(0.5-y) = 4
6-3y-3z+2y+0.5k-ky = 4
6-y-3z+0.5k-ky = 4
6-y-3(0.5-y)+0.5k-ky = 4
6-y-1.5+3y+0.5k-ky = 4
4.5+2y+0.5k-ky = 4

from all the equations given

3x+2y = 5 = 4-kz
so
kz =(-1)

Answer 2

Answer:

2

Step-by-step explanation:

x + y + z = 2,2x + y -z = 3

so x+y+z+1 = 2x+ y-z

z+1 = x-z

2z +1 = x   .........(1)

2x + y -z = 3

y = 3+z-2x

y = 3+z - 2(2z+1)

y = 3+z-4z+1

y=4+3z ...........(2)

so,

x + y + z = 2 ,2x + y -z = 3 (adding them)

x+y+z+2x+y-z = 3+2

3x + 2y = 5   ...........(3)

3(2z+1) + 2(4+3z) = 5 .............using (1) and (2)

6z +3 + 8 + 6z = 5

12z = 5-11 = -6

z = -6/12 = -1/2   .........(4)

3x + 2y + kz = 4

5 +kz = 4   .............using(3)

kz  = 4-5 = -1

k (-1/2) = -1  ..............using(4)

k = -1 *2/-1 = 2

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