Q. If [ ] denotes the greatest integer less than or equal to the real number under consideration and $- 1 \leq x < 0;0 \leq y < 1;1 \leq \mathcal{z} < 2,$ then the value of the determinant $\left| \begin{matrix} \lbrack x\rbrack + 1 & \lbrack y\rbrack & \mathcal{\lbrack z\rbrack} \\ \lbrack x\rbrack & \lbrack y\rbrack + 1 & \mathcal{\lbrack z\rbrack} \\ \lbrack x\rbrack & \lbrack y\rbrack & \left\lbrack \mathcal{z} \right\rbrack + 1 \end{matrix} \right|$ is
A) $\lbrack x\rbrack$
B) $\lbrack y\rbrack$
C) $\mathcal{\lbrack z\rbrack}$
D) None of these
Explanation
(c)
Since, $- 1 \leq x < 0$
$\therefore\lbrack x\rbrack = - 1$
Also, $0 \leq y < 1\ \Rightarrow \lbrack y\rbrack = 0$
and $1 \leq \mathcal{z} < 2\ \Rightarrow \left\lbrack \mathcal{z} \right\rbrack = 1$
$\therefore$ Given determinant becomes
$\left| \begin{matrix}
0 & 0 & 1 \\
- 1 & 1 & 1 \\
- 1 & 0 & 2
\end{matrix} \right| = 1 = \lbrack\mathcal{z\rbrack}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.