Question

The value of $\Delta = \left| \begin{matrix} 10C_{4} & 10C_{5} & 11C_{m} \\ 11C_{6} & 11C_{7} & 12C_{m + 2} \\ 12C_{8} & 12C_{9} & 13C_{m + 4} \end{matrix} \right|$ is equal to zero,

where m is

• A. 6
• B. 4
• C. 5
• D. None of these

Answer 1

Answer: (C)

5

Answer 2

$\Delta = \left| \begin{matrix} 10C_{4} & 10C_{5} & 11C_{m} \\ 11C_{6} & 11C_{7} & 12C_{m + 2} \\ 12C_{8} & 12C_{9} & 13C_{m + 4} \end{matrix} \right|$ = 0

Applying $C_{2} \rightarrow C_{1} + C_{2}$

10C_{4} & 10C_{4} +^{10}C_{5} & 11C_{m} \\ 11C_{6} & 11C_{6} +^{11}C_{7} & 12C_{m + 2} \\ 12C_{8} & 12C_{8} +^{12}C_{9} & 13C_{m + 4} \end{matrix} \right| = 0$$ ⇒$\Delta = \left| \begin{matrix} 10C_{4} & 11C_{5} & 11C_{m} \\ 11C_{6} & 12C_{7} & 12C_{m + 2} \\ 12C_{8} & 13C_{9} & 13C_{m + 4} \end{matrix} \right|$= 0 Clearly $m = 5$ satisfies the above result $$\lbrack\because C_{2},C_{3}\text{willbeidentical}\rbrack$$