Question

Find K if the equation x+2y+2z = 0 , x-3y-3z = 0 , 2x+y+Kz = 0 have only trival solution​

Answer 1

Answer:

x + 2y +2z = 0

x = -(2y+2z)

x -3y-3z=0

-2y-2z-3y-3z=0

-5y-5z=0

5z= - 5y

z=-5y/5= - y

2x + y+Kz=0

Kz= -2x- y

K=-2x - y/ z

=-2(-2y-2z) - y /- y

=(4y+4z)-y/y

=4y+4z/y - 1

Step-by-step explanation:

Answer 2

Answer:

Therefore, k ≠ 1.

Step-by-step explanation:

Given equations are x+2y+z = 0, x-3y-3z = 0, 2x+y+kz = 0

We have to find the value of 'k'.

Trivial solution means |A| ≠ 0

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

This is in the form of AX = B

Here A = $\left[\begin{array}{ccc}1&2&2\\1&-3&-3\\2&1&k\end{array}\right]$

|A| ≠ 0

$\left|\begin{array}{ccc}1&2&2\\1&-3&-3\\2&1&k\end{array}\right|$  ≠ 0

$1(-3k+3)-2(k+6)+2(1+6)$ ≠ 0

$-3k+3-2k-12+14$ ≠ 0

$-5k+5$ ≠ 0

$-5k$ ≠ -5

k ≠ 1

Hence, k ≠ 1.