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Question: \(\left| \begin{matrix} 1 & 1 & 1 \\ mC_{1} & m + 1C_{1} & m + 2C_{1} \\ mC_{2} & m + 1C_{2} & m + 2...

111mC1m+1C1m+2C1mC2m+1C2m+2C2\left| \begin{matrix} 1 & 1 & 1 \\ mC_{1} & m + 1C_{1} & m + 2C_{1} \\ mC_{2} & m + 1C_{2} & m + 2C_{2} \end{matrix} \right| =

A

m(m + 1)

B

m(m – 1)

C

1

D

0

Answer

1

Explanation

Solution

Put m = 2

= 1112C13C14C12C23C24C2\left| \begin{matrix} 1 & 1 & 1 \\ 2C_{1} & 3C_{1} & 4C_{1} \\ 2C_{2} & 3C_{2} & 4C_{2} \end{matrix} \right| = 111234136\left| \begin{matrix} 1 & 1 & 1 \\ 2 & 3 & 4 \\ 1 & 3 & 6 \end{matrix} \right|

= $\left| \begin{matrix} 0 & 0 & 1 \

  • 1 & - 1 & 4 \
  • 2 & - 3 & 6 \end{matrix} \right|$

[c1 → c1 – c2; c2 → c2 – c3] = 3 – 2 = 1