Exams Joint Entrance Examination Mathematics Matrices and Determinants Question 454 Question 454 of 472 Medium MCQ
If the system of equations
\[\[\begin{gathered} 11 x+y+\lambda z=-5 \\ 2 x+3 y+5 z=3 \\ 8 x-19 y-39 z=\mu \end{gathered}\]\]
has infinitely many solutions, then \[\lambda^4-\mu\] is equal to :
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\[ \begin{aligned}
& 11 x+y+\lambda z=-5 \\
& 2 x+3 y+5 z=3 \\
& 8 x-19 y-39 z=\mu \\
& \Delta=0 \Rightarrow\left|\[\begin{array}{|ccccccc|}11 & 1 & \lambda \\ 2 & 3 & 5 \\ 8 & -19 & -39\end{array}\end{aligned} \]\right|=0 \\
\& 11(-39.3+19.5)-1(-39.2-40)+\lambda(-38-24)=0 \\
\& =11(-117+95)-1(-118)-62 \lambda=0 \\
\& =-242+118=62 \lambda \\
\& \Rightarrow \lambda=-2 \\
\& \Delta 2=0 \\
\& \Rightarrow\left|\[ 11 1 − 5 2 3 3 8 − 19 μ \begin{array}{|ccccccc|}11 & 1 & -5 \\ 2 & 3 & 3 \\ 8 & -19 & \mu\end{array} 11 2 8 1 3 − 19 − 5 3 μ \]\right|=0
\]
\[ 11 ( 3 μ + 57 ) − 1 ( 2 μ − 24 ) − 5 ( − 38 − 24 ) = 0 33 μ + 627 − 2 μ + 24 + 310 = 0 μ = − 31 ⇒ λ 4 − 31 = 16 + 31 = 47 \begin{aligned}
& 11(3 \mu+57)-1(2 \mu-24)-5(-38-24)=0 \\
& 33 \mu+627-2 \mu+24+310=0 \\
& \mu=-31 \\
& \Rightarrow \lambda^4-31 \\
& =16+31 \\
& =47
\end{aligned} 11 ( 3 μ + 57 ) − 1 ( 2 μ − 24 ) − 5 ( − 38 − 24 ) = 0 33 μ + 627 − 2 μ + 24 + 310 = 0 μ = − 31 ⇒ λ 4 − 31 = 16 + 31 = 47 \] Answer the question to unlock