The value of the determinant `Delta=|(1!,2!,3!),(2!,3!,4!),(3!,4!,5!)|` is

Options

A. `2!`
B. `3!`
C. `4!`
D. `5!`

Answer

To find the value of the determinant

\[ \Delta = \begin{vmatrix} 1! & 2! & 3! \\ 2! & 3! & 4! \\ 3! & 4! & 5! \end{vmatrix} \]

we will first calculate the factorial values and then evaluate the determinant.

### Step 1: Calculate the factorial values

We know:
- \(1! = 1\)
- \(2! = 2\)
- \(3! = 6\)
- \(4! = 24\)
- \(5! = 120\)

Substituting these values into the determinant gives:

\[ \Delta = \begin{vmatrix} 1 & 2 & 6 \\ 2 & 6 & 24 \\ 6 & 24 & 120 \end{vmatrix} \]

### Step 2: Factor out common terms from each row

We can factor out common terms from each row:
- From the first row, we can factor out \(1\) (no change).
- From the second row, we can factor out \(2\).
- From the third row, we can factor out \(6\).

This gives us:

\[ \Delta = 1 \cdot 2 \cdot 6 \cdot \begin{vmatrix} 1 & 2 & 6 \\ 1 & 3 & 12 \\ 1 & 4 & 20 \end{vmatrix} \]

### Step 3: Simplify the determinant

Now we have:

\[ \Delta = 12 \cdot \begin{vmatrix} 1 & 2 & 6 \\ 1 & 3 & 12 \\ 1 & 4 & 20 \end{vmatrix} \]

### Step 4: Expand the determinant

We can expand the determinant along the first column:

\[ \Delta = 12 \cdot \left( 1 \cdot \begin{vmatrix} 3 & 12 \\ 4 & 20 \end{vmatrix} - 1 \cdot \begin{vmatrix} 2 & 6 \\ 4 & 20 \end{vmatrix} + 1 \cdot \begin{vmatrix} 2 & 6 \\ 3 & 12 \end{vmatrix} \right) \]

Calculating each of these 2x2 determinants:

1. \(\begin{vmatrix} 3 & 12 \\ 4 & 20 \end{vmatrix} = (3 \cdot 20) - (12 \cdot 4) = 60 - 48 = 12\)

2. \(\begin{vmatrix} 2 & 6 \\ 4 & 20 \end{vmatrix} = (2 \cdot 20) - (6 \cdot 4) = 40 - 24 = 16\)

3. \(\begin{vmatrix} 2 & 6 \\ 3 & 12 \end{vmatrix} = (2 \cdot 12) - (6 \cdot 3) = 24 - 18 = 6\)

### Step 5: Substitute back into the determinant

Now substituting back:

\[ \Delta = 12 \cdot (1 \cdot 12 - 1 \cdot 16 + 1 \cdot 6) = 12 \cdot (12 - 16 + 6) = 12 \cdot 2 = 24 \]

### Final Result

Thus, the value of the determinant \(\Delta\) is:

\[ \Delta = 24 \]