Question

Find value of K such that system of equations are,(3k-8)x+3y+3z=0 ,3x+(3k-8)y+3z=0 , 3x+3y+(3k-8)z=0 has non- trivial solution

Answer 1

find the value of k so that the system of the equation

x + y + 3z = 0

4x + 3y + kz = 0

2x + y + 2z = 0

has non-trival solution

This is called a homogeneous system. It has only zeros

on the right side. The trivial solution for all

homogeneous systems is (x,y,z) = (0,0,0). But many

homogeneous systems have other, non-trivial solutions.

Make the augmented matrix:

[1 1 3 | 0]

[4 3 k | 0]

[2 1 2 | 0]

To get a 0 where the 4 is,

multiply row 1 temporarily

by -4 and add it to 1 times

row 2, then restore row 1:

-4[1 1 3 | 0]

1[4 3 k | 0]

[2 1 2 | 0]

[1 1 3 | 0]

[0 -1 -12+k | 0]

[2 1 2 | 0]

To get a 0 where the 2 on

the bottom left is, multiply

row 1 temporarily by -2 and

add it to 1 times row 3, then

restore row 1:

-2[1 1 3 | 0]

[0 -1 -12+k | 0]

1[2 1 2 | 0]

[1 1 3 | 0]

[0 -1 -12+k | 0]

[0 -1 -4 | 0]

To get a 0 where the -1 on

the 3rd row is, multiply row

2 temporarily by -1 and add

it to 1 times row 3, then

restore row 2:

[1 1 3 | 0]

-1[0 -1 -12+k | 0]

1[0 -1 -4 | 0]

[1 1 3 | 0]

[0 -1 -12+k | 0]

[0 0 8-k | 0]

Multiply row 2 by -1

[1 1 3 | 0]

[0 1 12-k | 0]

[0 0 8-k | 0]

The system will have

a non-trivial solution

if the bottom row has

only zeros, hence 8-k

must = 0. So

8 - k = 0

k = 8

That's the answer.

Answer 2

Given : system of equations of (3k-8)x+3y+3z=0 ,3x+(3k-8)y+3z=0 , 3x+3y+(3k-8)z=0  has non trivial solution

To find : Value of k

Solution:

(3k-8)x+3y+3z=0

3x+(3k-8)y+3z=0

3x+3y+(3k-8)z=0

No trivial solution if

$\left[\begin{array}{ccc}3k-8&3&3\\3&3k-8&3\\3&3&3k-8\end{array}\right] =0$

k = 2/3  , 11/3

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