Question

if the system of equations 2x+3ky+(3k+4)z=0,x+(k+4)y+(4k+2)z=0 ,x+2(k+4y)+(3k+4)z=0 has non trival solution then k =​

Answer 1

Given : system of equations of 2x+3ky+(3k+4)z=0,x+(k+4)y+(4k+2)z=0 ,x+2(k+4y)+(3k+4)z=0  has non trivial solution

To find : Value of k

Solution:

2x+3ky+(3k+4)z=0

x+(k+4)y+(4k+2)z=0

x+(2 k+4)y +(3k+4)z=0

No trivial solution if

$\left[\begin{array}{ccc}2&3&3k+4\\1&k+4&4k+2\\1&2k+4&3k+4\end{array}\right] =0$

Solve to find k

2(3k² + 16k + 16 - (8k² + 20k + 8)) - 3(3k + 4 - 4k - 2) + (3k + 4) (2k + 4 - k - 4) =0

=> 2(-5k²-4k + 8) - 3(-k + 2) + (3k + 4)(k) = 0

=> -10k² - 8k + 16 +3k - 6 + 3k² + 4k = 0

=> -7k²  - k + 10 = 0

=> 7k² + k - 10 = 0

=> k =( -7 ± √1 + 280)/2(7)

=> k = ( - 7  ± √ 281)/14

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