If the system of equations 11 x+y+lambda z=-5 | vedclass

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(C) For a system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix $ D $ must be $ 0 $,and the augmented

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$ 47 $

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(C) For a system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix $ D $ must be $ 0 $,and the augmented determinants $ D_1, D_2, D_3 $ must also be $ 0 $.$ D = \begin{vmatrix} 11 & 1 & \lambda \ 2 & 3 & 5 \ 8 & -19 & -39 \end{vmatrix} = 0 $$ 11(-117 + 95) - 1(-78 - 40) + \lambda(-38 - 24) = 0 $$ 11(-22) + 118 - 62\lambda = 0 $$ -242 + 118 = 62\lambda $$ 62\lambda = -124 \Rightarrow \lambda = -2 $Now,for $ D_1 = 0 $:$ D_1 = \begin{vmatrix} -5 & 1 & -2 \ 3 & 3 & 5 \ \mu & -19 & -39 \end{vmatrix} = 0 $$ -5(-117 + 95) - 1(-117 - 5\mu) - 2(-57 - 3\mu) = 0 $$ -5(-22) + 117 + 5\mu + 114 + 6\mu = 0 $$ 110 + 231 + 11\mu = 0 $$ 11\mu = -341 \Rightarrow \mu = -31 $Finally,calculate $ \lambda^4 - \mu $:$ \lambda^4 - \mu = (-2)^4 - (-31) = 16 + 31 = 47 $

If the system of equations
$ 11 x+y+\lambda z=-5 $
$ 2 x+3 y+5 z=3 $
$ 8 x-19 y-39 z=\mu $
has infinitely many solutions,then $ \lambda^4-\mu $ is equal to :

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If the system of equations$ 11 x+y+\lambda z=-5 $$ 2 x+3 y+5 z=3 $$ 8 x-19 y-39 z=\mu $has infinitely many solutions,then $ \lambda^4-\mu $ is equal to :