Let \lambda be a real number for which the system of linear equations

x+y+z=6

4x+\lambda y-\lambda z=\lambda -2

3x+2 y-4 z= -5

has infinitely many solutions. Then \lambda is a root of the quadratic equation :

  • Option 1)

    \lambda^{2}+3\lambda-4=0

  • Option 2)

    \lambda^{2}-3\lambda-4=0

  • Option 3)

    \lambda^{2}+\lambda-6=0

  • Option 4)

    \lambda^{2}-\lambda-6=0

 

Answers (1)

linear equations

x+y+z=6

4x+\lambda y-\lambda z=\lambda -2

3x+2 y-4 z= -5

\Delta =\begin{vmatrix} 1 &1 &1 \\ 4& \lambda & -\lambda\\ 3& 2 &-4 \end{vmatrix}

    =(-4\lambda+2\lambda)-1(-16+3\lambda)+1(8-3\lambda)

   =-2\lambda+16-3\lambda+8-3\lambda

  =-8\lambda+24

Now, using cramers law for infinite solution 

\Delta ,\Delta _1,\Delta _1 all will be zero     

-8\lambda+24=0

\lambda=3

Now put \lambda=3 in options and check for the correct one 

(1)\lambda^{2}+3\lambda-4=(3)^{2}+3.3-4\neq 0

(2)\lambda^{2}-3\lambda-4=(3)^{2}-3.3-4\neq 0

(3)\lambda^{2}+\lambda-6=(3)^{2}+3-6\neq 0

(4)\lambda^{2}-\lambda-6=(3)^{2}-3-6=0

So, option (4) is correct.

 

 


Option 1)

\lambda^{2}+3\lambda-4=0

Option 2)

\lambda^{2}-3\lambda-4=0

Option 3)

\lambda^{2}+\lambda-6=0

Option 4)

\lambda^{2}-\lambda-6=0

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