Search any question & find its solution
Question:
Answered & Verified by Expert
If the system of equations \(x+\lambda y+2=0\), \(\lambda x+y-2=0, \lambda x+\lambda y+3=0\) is consistent, then
Options:
Solution:
1523 Upvotes
Verified Answer
The correct answer is:
\(\lambda= \pm 1\)
The system of equations will be consistent if
\(\Delta=\left|\begin{array}{ccc}
1 & \lambda & 2 \\
\lambda & 1 & -2 \\
\lambda & \lambda & 3
\end{array}\right|=0\)
To evaluate \(\Delta\) we use \(R_1 \rightarrow R_1+R_2\) followed by \(\mathrm{C}_2 \rightarrow \mathrm{C}_2-\mathrm{C}_1\) to obtain
\(\begin{aligned}
\Delta & =\left|\begin{array}{ccc}
\lambda+1 & \lambda+1 & 0 \\
\lambda & 1 & -2 \\
\lambda & \lambda & 3
\end{array}\right|=\left|\begin{array}{ccc}
\lambda+1 & 0 & 0 \\
\lambda & 1-\lambda & -2 \\
\lambda & 0 & 3
\end{array}\right| \\
& =3(\lambda+1)(1-\lambda)=3\left(1-\lambda^2\right)
\end{aligned}\)
For the system to be consistent, we must have \(1-\lambda^2=0\) or \(\lambda= \pm 1\).
\(\Delta=\left|\begin{array}{ccc}
1 & \lambda & 2 \\
\lambda & 1 & -2 \\
\lambda & \lambda & 3
\end{array}\right|=0\)
To evaluate \(\Delta\) we use \(R_1 \rightarrow R_1+R_2\) followed by \(\mathrm{C}_2 \rightarrow \mathrm{C}_2-\mathrm{C}_1\) to obtain
\(\begin{aligned}
\Delta & =\left|\begin{array}{ccc}
\lambda+1 & \lambda+1 & 0 \\
\lambda & 1 & -2 \\
\lambda & \lambda & 3
\end{array}\right|=\left|\begin{array}{ccc}
\lambda+1 & 0 & 0 \\
\lambda & 1-\lambda & -2 \\
\lambda & 0 & 3
\end{array}\right| \\
& =3(\lambda+1)(1-\lambda)=3\left(1-\lambda^2\right)
\end{aligned}\)
For the system to be consistent, we must have \(1-\lambda^2=0\) or \(\lambda= \pm 1\).
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.