Question

System of equation

$x+3y+2z=6$

$x+\lambda y+2z=7$

$x+3y+2z=\mu$ has

• A. Unique solution if $\lambda =2,\mu \ne 6$
• B. infinitely many solution if $\lambda =4\mu =6$
• C. no solution if $\lambda =5,\mu =7$
• D. no solution if $\lambda =3,\mu =5$

Answer 1

Answer: (B), (C), (D)

The correct options are
B

infinitely many solution if $\lambda =4\mu =6$

C

no solution if $\lambda =5,\mu =7$

D

no solution if $\lambda =3,\mu =5$

$x+3y+2z=6$ ..............(i)

$x+\lambda y+2z=7$ ..............(ii)

$x+3y+2z=\mu$ ..............(iii)

(A) If $\lambda =2$, then D = 0, therefore unique solution is not possible

(B) If $\lambda =4,\mu =6$

$x+3y=6-2z$

$x+4y=7-2z$

$\therefore$ y = 1 and x = 3 - 2z

substituting in equation (iii)

$3-2z+3+2z=6$ is satisfied

$\therefore$ infinite solutions

(C) $\lambda =5,\mu =7$

consider equation (ii) and (iii)

$x+5y=7-2z$

$x+3y=7-2z$

$\therefore$ y = 0 $x=7-2z$ are solution

sub. in (i)

$7-2z+2z=6$ does not satisfy

$\therefore$ no solution

(D) If $\lambda =3,\mu =5$

then equation (i) and (ii) have no solution

$\therefore$ no solution