Find value of determinant of `A =abs[[(a+1)(a+2),(a+2),1],[(a+3)(a+2),(a+3),1],[(a+3)(a+4),(a+4),1]]`

Options

A. `-2`
B. `(a+1)(a+2)(a+3)`
C. 1
D. `(a+4)(a+2)(a+3)`

Answer

To find the value of the determinant of the matrix

\[ A = \begin{bmatrix} (a+1)(a+2) & (a+2) & 1 \\ (a+3)(a+2) & (a+3) & 1 \\ (a+3)(a+4) & (a+4) & 1 \end{bmatrix} \]

we will perform row operations to simplify the determinant calculation.

### Step 1: Apply Row Operations
We will subtract the first row from the second and third rows.

\[ R_2 \rightarrow R_2 - R_1 \]
\[ R_3 \rightarrow R_3 - R_1 \]

After performing these operations, the matrix becomes:

\[ A = \begin{bmatrix} (a+1)(a+2) & (a+2) & 1 \\ (a+3)(a+2) - (a+1)(a+2) & (a+3) - (a+2) & 0 \\ (a+3)(a+4) - (a+1)(a+2) & (a+4) - (a+2) & 0 \end{bmatrix} \]

### Step 2: Simplify the Second Row
Calculating the elements of the second row:

- First element:
\[ (a+3)(a+2) - (a+1)(a+2) = (a+2)((a+3) - (a+1)) = (a+2)(2) = 2(a+2) \]
- Second element:
\[ (a+3) - (a+2) = 1 \]
- Third element:
\[ 0 \]

Thus, the second row becomes:
\[ [2(a+2), 1, 0] \]

### Step 3: Simplify the Third Row
Calculating the elements of the third row:

- First element:
\[ (a+3)(a+4) - (a+1)(a+2) = (a+4)(a+3) - (a^2 + 3a + 2) = a^2 + 7a + 12 - (a^2 + 3a + 2) = 4a + 10 \]
- Second element:
\[ (a+4) - (a+2) = 2 \]
- Third element:
\[ 0 \]

Thus, the third row becomes:
\[ [4a + 10, 2, 0] \]

### Step 4: Updated Matrix
Now, the matrix looks like this:

\[ A = \begin{bmatrix} (a+1)(a+2) & (a+2) & 1 \\ 2(a+2) & 1 & 0 \\ 4a + 10 & 2 & 0 \end{bmatrix} \]

### Step 5: Expand the Determinant
Now we can expand the determinant along the third column:

\[ \text{det}(A) = 1 \cdot \begin{vmatrix} 2(a+2) & 1 \\ 4a + 10 & 2 \end{vmatrix} \]

Calculating the 2x2 determinant:

\[ = 2(a+2) \cdot 2 - 1 \cdot (4a + 10) = 4(a + 2) - (4a + 10) = 4a + 8 - 4a - 10 = -2 \]

### Final Result
Thus, the value of the determinant of matrix \( A \) is

\[ \text{det}(A) = -2. \]