The value of $\begin{vmatrix} 7! & 8! & 9! \\ 8! & 9! & 10! \\ 9! & 10! & 11! \end{vmatrix}$ is:

The determinant is:

$\begin{vmatrix} 7! & 8! & 9! \\ 8! & 9! & 10! \\ 9! & 10! & 11! \end{vmatrix}$

Using the factorial relationships where $8! = 8 \times 7!$, $9! = 9 \times 8!$, and $10! = 10 \times 9!$:

From Row 1, factor out $7!$:

• $7! = 7! \times 1$
• $8! = 7! \times 8$
• $9! = 7! \times 72$

From Row 2, factor out $8!$:

• $8! = 8! \times 1$
• $9! = 8! \times 9$
• $10! = 8! \times 90$

From Row 3, factor out $9!$:

• $9! = 9! \times 1$
• $10! = 9! \times 10$
• $11! = 9! \times 110$

The determinant becomes:

$= 7! \cdot 8! \cdot 9! \begin{vmatrix} 1 & 8 & 72 \\ 1 & 9 & 90 \\ 1 & 10 & 110 \end{vmatrix}$

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Applying row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$:

$= 7! \cdot 8! \cdot 9! \begin{vmatrix} 1 & 8 & 72 \\ 0 & 1 & 18 \\ 0 & 2 & 38 \end{vmatrix}$

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Expanding along the first column:

$= 7! \cdot 8! \cdot 9! \times 1 \times \begin{vmatrix} 1 & 18 \\ 2 & 38 \end{vmatrix}$

$= 7! \cdot 8! \cdot 9! \times [(1)(38) - (18)(2)]$

$= 7! \cdot 8! \cdot 9! \times [38 - 36]$

$= 7! \cdot 8! \cdot 9! \times 2$

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Therefore, the value of the determinant is $2 \times 7! \times 8! \times 9!$