Let `D=|(10!,11!,12!),(11!,12!,13!),(12!,13!,14!)|` then `D/((10!)^(3)-260` equals

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to evaluate the determinant \( D = \begin{vmatrix} 10! & 11! & 12! \\ 11! & 12! & 13! \\ 12! & 13! & 14! \end{vmatrix} \) and then compute \( \frac{D}{(10!)^3 - 260} \).

### Step 1: Simplify the Determinant

We can factor out \( 10! \) from each row of the determinant:

\[ D = \begin{vmatrix} 10! & 11! & 12! \\ 11! & 12! & 13! \\ 12! & 13! & 14! \end{vmatrix} = 10! \begin{vmatrix} 1 & 11 & 12 \\ 11 & 12 & 13 \\ 12 & 13 & 14 \end{vmatrix} \]

### Step 2: Calculate the 3x3 Determinant

Now we need to calculate the determinant \( \begin{vmatrix} 1 & 11 & 12 \\ 11 & 12 & 13 \\ 12 & 13 & 14 \end{vmatrix} \).

Using the formula for the determinant of a 3x3 matrix:

\[ \text{Det} = a(ei - fh) - b(di - fg) + c(dh - eg) \]

where \( a, b, c \) are the elements of the first row, and \( d, e, f, g, h, i \) are the elements of the second and third rows respectively.

Substituting the values:

\[ \text{Det} = 1 \cdot (12 \cdot 14 - 13 \cdot 13) - 11 \cdot (11 \cdot 14 - 12 \cdot 13) + 12 \cdot (11 \cdot 13 - 12 \cdot 12) \]

Calculating each term:

1. \( 12 \cdot 14 - 13 \cdot 13 = 168 - 169 = -1 \)
2. \( 11 \cdot 14 - 12 \cdot 13 = 154 - 156 = -2 \)
3. \( 11 \cdot 13 - 12 \cdot 12 = 143 - 144 = -1 \)

Now substituting these values back into the determinant formula:

\[ \text{Det} = 1 \cdot (-1) - 11 \cdot (-2) + 12 \cdot (-1) = -1 + 22 - 12 = 9 \]

### Step 3: Substitute Back to Find D

Now substituting back to find \( D \):

\[ D = 10! \cdot 9 \]

### Step 4: Compute \( D / ((10!)^3 - 260) \)

Now we need to compute \( \frac{D}{(10!)^3 - 260} \):

First, calculate \( (10!)^3 \):

\[ (10!)^3 = (10!)^3 \]

Now substitute \( D \):

\[ \frac{D}{(10!)^3 - 260} = \frac{10! \cdot 9}{(10!)^3 - 260} \]

### Final Step: Simplify the Expression

This gives us:

\[ \frac{9 \cdot 10!}{(10!)^3 - 260} \]

This is the final expression for the value of \( \frac{D}{(10!)^3 - 260} \).