Q. If matrix $\begin{bmatrix} 0 & 1 & - 2 \\ - 1 & 0 & 3 \\ \lambda & - 3 & 0 \end{bmatrix}$is singular, then $\lambda$ is equal to
A) $- 2$
B) $- 1$
C) 1
D) 2
Explanation
(d)
Given, matrix is singular.
Therefore, $\left| \begin{array}{r}
0\ \ \ \ \ 1\ - 2 \\
- 1\ \ \ \ 0\ \ \ \ \ 3 \\
\lambda\ - 3\ \ \ \ 0
\end{array} \right| = 0$
$\Rightarrow \ + 1(0 - 6) + \lambda(3) = 0$
$\Rightarrow - 6 + 3\lambda = 0$
$\Rightarrow \ \ \lambda = 2$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.