Question

the no. of values of K for which the system of equations (K-1)x+(3K+1)y+2Kz = 0,
(K-1)x+(4K-2)y+(K+3)z=0 and 2x+(3k+1)y+(3K-1)z=0 has a common non zero solution is..

(a) 0
(b) 1
(c) 2
(d) 3

Answer 1

Okay, in here you gotta use Cramer's rule. (studied that yet? (matrices, determinants)

Answer 2

(k-1)x+(3k+1)y+2kz=0
(k-1)x+(4k-2)y+(k+3)z=0
2x+(3k+1)y+(3k-1)z=0
\[\left[\begin{matrix}(k-1) & (3k+1)&2k \\ (k-1) & (4k-2)& (k+3)\\2x &(3k+1)&(3k-1)\end{matrix}\right]\left(\begin{matrix}x \\ y\\z\end{matrix}\right)=\left(\begin{matrix}0 \\ 0\\0\end{matrix}\right)\]
let Au=b, where
A=\[\left[\begin{matrix}(k-1) & (3k+1)&2k \\ (k-1) & (4k-2)& (k+3)\\2x &(3k+1)&(3k-1)\end{matrix}\right]\]
u=\[\left(\begin{matrix}x \\ y\\z\end{matrix}\right)\]
b=\[\left(\begin{matrix}0 \\ 0\\0\end{matrix}\right)\]
To find u. Use,\[u=A^-1b\]
recall, \[A^-1 = (1/detA)(adjA)\]
then solve

Answer 3

Gaussian elimination can be used here

Answer 4

\[ \det A = 6 k^3-34 k^2+46 k+6 = 0\]
\[k = 3, (4 - \sqrt19)/3, (4 + \sqrt19)/3\]

Answer 5

Hence, k =3 we can find x,y and z
\[\left[\begin{matrix}2 & 10&6 \\ 2 & 10&6\\2&10&8\end{matrix}\right]\left(\begin{matrix}x \\ y\\z\end{matrix}\right)=0\]
x=0, y=0, z=0

Answer 6

thx. :)

Answer 7

is this correct ?

Answer 8

then again i used Gaussian elimination and reduced row echelon augumented matrix.
k = 1, x=0, y=0 and z=0