The value of \(\begin{vmatrix}7! & 8! & 9!\\8! & 9! & 10!\\9! & 10! & 11!\end{vmatrix}\) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(2 \times 7! \times 8! \times 9!\)
Since \(n! = n \times (n - 1)!\), each entry is a multiple of the first entry of its row:
\(8! = 8 \times 7!,\quad 9! = 72 \times 7!\)
\(9! = 9 \times 8!,\quad 10! = 90 \times 8!\)
\(10! = 10 \times 9!,\quad 11! = 110 \times 9!\)
Taking out the common factors 7!, 8! and 9! from the three rows:
\( \displaystyle \Delta = 7!\,8!\,9!\begin{vmatrix}1 & 8 & 72\\1 & 9 & 90\\1 & 10 & 110\end{vmatrix} \)
Applying \(R_2 \to R_2 - R_1\) and \(R_3 \to R_3 - R_1\):
\( \displaystyle \Delta = 7!\,8!\,9!\begin{vmatrix}1 & 8 & 72\\0 & 1 & 18\\0 & 2 & 38\end{vmatrix} \)
Expanding along the first column:
\(\Delta = 7!\,8!\,9!\,(1 \times 38 - 18 \times 2)\)
\(\Delta = 2 \times 7! \times 8! \times 9!\)