Exams Joint Entrance Examination Mathematics Matrices and Determinants Question 347 Question 347 of 472 Medium MCQ
If the system of linear equations
x + y + z = 5
x + 2y + 2z = 6
x + 3y + \[\lambda \]z = \[\mu \], (\[\lambda \], \[\mu \] \[ \in \] R), has infinitely many solutions, then the value of \[\lambda \] + \[\mu \] is :
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x + y + z = 5
x + 2y + 2z = 6
x + 3y + \[\lambda \]z = \[\mu \] have infinite solution
\[\Delta \] = 0, \[\Delta \]x = \[\Delta \]y = \[\Delta \]z = 0
\[\Delta = \left| { ( 1 1 1 1 2 2 1 3 λ ) \begin{pmatrix} 1 & 1 & 1 \\
1 & 2 & 2 \\
1 & 3 & \lambda \\
\end{pmatrix} 1 1 1 1 2 3 1 2 λ } \right| = 0\]
\[ \Rightarrow 1(2\lambda - 6) - 1(\lambda - 2) + 1(3 - 2) = 0\]
\[ \Rightarrow 2\lambda - 6 - \lambda + 2 + 1 = 0\]
\[ \Rightarrow \lambda = 3\]
Now, \[\Delta x = \left| { ( 5 1 1 6 2 2 μ 3 3 ) \begin{pmatrix} 5 & 1 & 1 \\
6 & 2 & 2 \\
\mu & 3 & 3 \\
\end{pmatrix} 5 6 μ 1 2 3 1 2 3 } \right| = 0\],
\[\Delta \]y = \[\left| { ( 1 5 1 1 6 2 1 μ 3 ) \begin{pmatrix} 1 & 5 & 1 \\
1 & 6 & 2 \\
1 & \mu & 3 \\
\end{pmatrix} 1 1 1 5 6 μ 1 2 3 } \right| = 0\]
\[ \Rightarrow \left| { ( 1 5 1 0 1 1 0 μ − 5 2 ) \begin{pmatrix} 1 & 5 & 1 \\
0 & 1 & 1 \\
0 & {\mu - 5} & 2 \\
\end{pmatrix} 1 0 0 5 1 μ − 5 1 1 2 } \right| = 0\]
\[ \Rightarrow \mu = 7\]
\[\Delta z = \left| { ( 1 1 5 1 2 6 1 3 μ ) \begin{pmatrix} 1 & 1 & 5 \\
1 & 2 & 6 \\
1 & 3 & \mu \\
\end{pmatrix} 1 1 1 1 2 3 5 6 μ } \right| = \left| { ( 1 1 5 0 − 1 − 1 0 2 μ − 5 ) \begin{pmatrix} 1 & 1 & 5 \\
0 & { - 1} & { - 1} \\
0 & 2 & {\mu - 5} \\
\end{pmatrix} 1 0 0 1 − 1 2 5 − 1 μ − 5 } \right|\]
\[ \Rightarrow 1(5 - \mu + 2) = 0\]
\[ \Rightarrow \mu = 7\]
So, \[\lambda + \mu = 10\] Answer the question to unlock