Question

If the system of equations of 3x – 2y + z = 0, kx – 14y + 15z = 0, x +2y + 3z = 0 has non trivial solution
then k =
(A) 29
(B) 26
(C) 23
(D) 19​

Answer 1

Given  : system of equations of 3x – 2y + z = 0, kx – 14y + 15z = 0, x +2y + 3z = 0  has non trivial solution

To find : Value of k

Solution:

3x – 2y + z = 0,

kx – 14y + 15z = 0,

x +2y + 3z = 0

No trivial solution if

$\left[\begin{array}{ccc}3&-2&1\\k&-14&15\\1&2&3\end{array}\right] = 0$

=> 3( -42 - 30) -(-2)( 3k - 15)  +  1(2k + 14)  = 0

=> -216  + 6k - 30  + 2k + 14 = 0

=> 8k  = 232

=> k = 29

Value of k = 29

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Answer 2

Given:

3x – 2y + z = 0

kx – 14y + 15z = 0

x +2y + 3z = 0

To find:

k=?

Solution:

3x – 2y + z = 0

kx – 14y + 15z = 0

x +2y + 3z = 0

solve by non trivial method:

$\left[\begin{array}{ccc}3&-2&1\\k&-14&15\\1&2&3\end{array}\right]=0$

$\to [3(-42-30)-(-2)(3k-15)+1(2k+14)]=0\\\\\to [3(-42-30)+2(3k-15)+1(2k+14)]=0\\\\\to [3(-72)+6k-30+2k+14)]=0\\\\\to [3(-72)+6k-30+2k+14)]=0\\\\\to [-216+6k+2k-16]=0\\\\\to [-232+8k]=0\\\\\to 8k-232=0\\\\\to 8k=232\\\\\to k= \frac{232}{8}\\\\\to \boxed{k= 29}$

The final answer is "29".