Question

The values of \(\lambda\) and \(\mu\) for which the system of linear equations
\(x+y+z=2\)
\(x+2y+3z=5\)
\(x+3y+\lambda z=\mu\)
has infinitely many solutions are, respectively:

• A. 6 and 8
• B. 5 and 8
• C. 5 and 7
• D. 4 and 9

Hint:

For infinitely many solutions the coefficient determinant must vanish (\(\lambda=5\)) and the system must remain consistent (fixes \(\mu\)).

Answer 1

The Correct Option is B

Coefficient determinant \(\Delta=\begin{vmatrix}1&1&1\\1&2&3\\1&3&\lambda\end{vmatrix}=1(2\lambda-9)-1(\lambda-3)+1(3-2)=\lambda-5\).
For infinitely many solutions \(\Delta=0\Rightarrow\lambda=5\).
With \(\lambda=5\): subtracting eq.(1) from eq.(2) gives \(y+2z=3\); subtracting eq.(2) from eq.(3) gives \(y+2z=\mu-5\).
For consistency (and hence infinitely many solutions) \(\mu-5=3\Rightarrow\mu=8\).
Answer: (2) 5 and 8