Question

For what values of k , the system of linear equations x + y + z = 2 2x + y -z = 3 3x + 2y + kz = 4 has a unique solution ?

Answer 1

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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