Q&A - Ask Doubts and Get Answers
Q

I have a doubt, kindly clarify. - Algebra - BITSAT-3

The value of \lambda\, \, and \, \, \mu  for which the system of equations x+y+z=6, x+2y+3z=10  and  x+2y+\lambda z=\mu  have no solution are

  • Option 1)

    \lambda = 3,\mu =10

  • Option 2)

    \lambda = 3,\mu \ne 10

  • Option 3)

    \lambda \neq 3,\mu =10

  • Option 4)

    \lambda \neq 3,\mu \ne 10

 
Answers (1)
112 Views

As learnt in

Inconsistent system of linear equation -

If the system of equations has no solutions

-

 

For no solution \Delta =0

\begin{vmatrix} 1 & 1 & 1\\ 1 & 2 & 3\\ 1 & 2 & \lambda \end{vmatrix}=0

1(2 \lambda-6)-1(\lambda-3)+1 \times 0=0

2 \lambda-6-\lambda+3=0

\lambda-3=0\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\lambda=3

 

Also,

\Delta_{3}\neq0\:\:\:\:\:\:\:\:\:\:\:\:\:\begin{vmatrix} 1 & 1 & 6\\ 1 & 2 & 10\\ 1 & 2 & \mu \end{vmatrix}\neq 0

1(2 \lambda -20)-1(2 \mu -10)+6 \times 0 \neq 0

\Delta_{1} \neq 0 \begin{vmatrix}6 & 1 & 1\\ 10 & 2 & 3\\ \mu & 2 & 3\end{vmatrix}=6 \times 0-1 (3_{0}-3 \mu)+1(2_{0}-2\mu)\neq 0

\mu -10 \neq 0\:\:\:\; \Rightarrow\mu \neq10

 


Option 1)

\lambda = 3,\mu =10

This option is incorrect.

Option 2)

\lambda = 3,\mu \ne 10

This option is correct.

Option 3)

\lambda \neq 3,\mu =10

This option is incorrect.

Option 4)

\lambda \neq 3,\mu \ne 10

This option is incorrect.

Exams
Articles
Questions