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If the system of linear equations
\mathrm{2 x+3 y-z=-2}
\mathrm{x+y+z=4 }
\mathrm{x-y+|\lambda| z=4 \lambda-4 }
where \mathrm{\lambda \in \mathbb{R} }  has no solution, then

Option: 1

\lambda=7


Option: 2

\lambda=-7


Option: 3

\lambda=8


Option: 4

\lambda^{2}=1


Answers (1)

best_answer

For No solution,
\mathrm{\Delta = 0\:\quad But\: \quad \Delta _{x}\,\: or\, \: \Delta _{y}\: \: or\, \: \Delta _{z}} is non zero

\mathrm{\Rightarrow \Delta = \begin{vmatrix} 2 &3 &-1 \\ 1&1 &1 \\ 1 &-1 & \left | \lambda \right | \end{vmatrix}= 0}

\mathrm{\Rightarrow 2\left | \lambda \right |+3+1+1+2-3\left | \lambda \right |= 0\: \: \Rightarrow \: \: \left | \lambda \right |= 7\: \: \Rightarrow\: \lambda= \pm 7}

\mathrm{\Delta _{z}= \begin{vmatrix} 2 &3 &-2 \\ 1& 1& 4\\ 1& -1 &4\lambda-4 \end{vmatrix}= 8\lambda-8+12+2+2+8-12\lambda+12}
                                               \mathrm{= -4\lambda+28}


 \mathrm{For\: \: \lambda= 7,\Delta _{z}= 0,\: \: \: \: For\: \: \lambda= -7,\Delta _{z}\neq 0}
\mathrm{So\: \: \lambda= -7,} No solution

Option (B)

Posted by

shivangi.shekhar

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