Q. The value of determinant $\left| \begin{matrix} a + b & a + 2b & a + 3b \\ a + 2b & a + 3b & a + 4b \\ a + 4b & a + 5b & a + 6b \end{matrix} \right|$ is
A) $a^{2} + b^{2} + c^{2} - 3abc$
B) $3ab$
C) $3a + 5b$
D) 0
Explanation
(d)
$\left| \begin{matrix}
a + b & a + 2b & a + 3b \\
a + 2b & a + 3b & a + 4b \\
a + 4b & a + 5b & a + 6b
\end{matrix} \right|$
$= \left| \begin{matrix}
a + b & a + 2b & a + 3b \\
b & b & b \\
2b & 2b & 2b
\end{matrix} \right|\ \ \ \binom{R_{2} \rightarrow R_{2} - R_{1}}{R_{3} \rightarrow R_{3} - R_{2}}$
$=$ 0 $(\because\ R_{2}$ and $R_{3}$ are proportional)
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.