For the system of linear equations: [x - 2y = 1,x - y + kz = - 2,ky + 4z = 6,k R], conside…

Question

For the system of linear equations:

$$x - 2y = 1,x - y + kz =  - 2,ky + 4z = 6,k \in R$$,

consider the following statements :

(A) The system has unique solution if $$k \ne 2,k \ne  - 2$$.

(B) The system has unique solution if k = -2

(C) The system has unique solution if k = 2

(D) The system has no solution if k = 2

(E) The system has infinite number of solutions if k $$ \ne $$ -2.

Which of the following statements are correct?

Accepted Answer

$x - 2y + 0.z = 1$

$x - y + kz =  - 2$

$0.x + ky + 4z = 6$

$$\Delta  = \left| {\begin{pmatrix} 1 & { - 2} & 0  \
   1 & { - 1} & k  \
   0 & k & 4  \

\end{pmatrix} } \right| = 4 - {k^2}$$

For unique solution $$4 - {k^2} \ne 0$$

$$ \Rightarrow $$ k $$ \ne $$ $$ \pm $$ 2

For k = 2 :

$x - 2y + 0.z = 1$

$x - y + 2z =  - 2$

$0.x + 2y + 4z = 6$

$$\Delta x = \left| {\begin{pmatrix} 1 & { - 2} & 0  \
   2 & { - 1} & 2  \
   6 & 2 & 4  \

\end{pmatrix} } \right| = ( - 8) + 2[ - 20]$$

$$\Delta x =  - 48 \ne 0$$

For k = 2, $$\Delta x \ne 0$$

$$ \therefore $$ For K = 2; The system has no solution.

Solutions · 1