Search any question & find its solution
Question:
Answered & Verified by Expert
If the matrix $\left[\begin{array}{ccc}1 & 3 & \lambda+2 \\ 2 & 4 & 8 \\ 3 & 5 & 10\end{array}\right]$ is singular, then $\lambda=$
Options:
Solution:
2690 Upvotes
Verified Answer
The correct answer is:
4
$|\mathrm{A}|=0$ as the matrix $\mathrm{A}$ is singular
$$
\therefore|\mathrm{A}|=\left|\begin{array}{ccc}
1 & 3 & \lambda+2 \\
2 & 4 & 8 \\
3 & 5 & 10
\end{array}\right|=0
$$
Apply $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-2 \mathrm{R}_{1}$ and $\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-3 \mathrm{R}_{1}$
and expand. $-2(4-3 \lambda)+4(4-2 \lambda)=0$
$$
\Rightarrow 8-2 \lambda=0 \Rightarrow \lambda=4
$$
For $\lambda=4,$ the second and the third column are proportional.
$$
\therefore|\mathrm{A}|=\left|\begin{array}{ccc}
1 & 3 & \lambda+2 \\
2 & 4 & 8 \\
3 & 5 & 10
\end{array}\right|=0
$$
Apply $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-2 \mathrm{R}_{1}$ and $\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-3 \mathrm{R}_{1}$
and expand. $-2(4-3 \lambda)+4(4-2 \lambda)=0$
$$
\Rightarrow 8-2 \lambda=0 \Rightarrow \lambda=4
$$
For $\lambda=4,$ the second and the third column are proportional.
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.