1. Key Concept: Properties of Determinants

The most efficient way to evaluate a determinant, especially for a $3 \times 3$ matrix or larger, is by using its properties to simplify it. The key property we will utilize here is that the value of a determinant remains unchanged if we apply elementary row operations of the type $R_i \to R_i + k R_j$ (or column operations $C_i \to C_i + k C_j$). Our goal is to create as many zeros as possible in a particular row or column, which significantly simplifies the expansion of the determinant.

2. Step-by-Step Solution

Step 1: Write down the given determinant. Let the given determinant be $\Delta$. $\Delta = \left| {\begin{matrix} {(a + 1)(a + 2)} & {a + 2} & 1 \\ {(a + 2)(a + 3)} & {a + 3} & 1 \\ {(a + 3)(a + 4)} & {a + 4} & 1 \\ \end{matrix} } \right|$

Step 2: Apply row operations to introduce zeros.

• Why this step? Observe that the third column ($C_3$) consists entirely of 1s. This is a clear indicator that we can perform row operations to make two of these elements zero, without changing the determinant's value. This will make the determinant expansion much simpler.
• We will apply the following row operations:
  1. $R_2 \to R_2 - R_1$
  2. $R_3 \to R_3 - R_1$

Let's calculate the new elements for $R_2'$ and $R_3'$:

• For the new Row 2 ($R_2'$):

  • First element ($a_{21}'$): $(a + 2)(a + 3) - (a + 1)(a + 2) = (a + 2)[(a + 3) - (a + 1)]$ $= (a + 2)[a + 3 - a - 1] = (a + 2)(2) = 2(a + 2)$
  • Second element ($a_{22}'$): $(a + 3) - (a + 2) = 1$
  • Third element ($a_{23}'$): $1 - 1 = 0$

• For the new Row 3 ($R_3'$):

  • First element ($a_{31}'$): $(a + 3)(a + 4) - (a + 1)(a + 2)$ $= (a^2 + 7a + 12) - (a^2 + 3a + 2)$ $= a^2 + 7a + 12 - a^2 - 3a - 2 = 4a + 10$
  • Second element ($a_{32}'$): $(a + 4) - (a + 2) = 2$
  • Third element ($a_{33}'$): $1 - 1 = 0$

Now, substitute these new elements back into the determinant: $\Delta = \left| {\begin{matrix} {(a + 1)(a + 2)} & {a + 2} & 1 \\ {2(a + 2)} & 1 & 0 \\ {4a + 10} & 2 & 0 \\ \end{matrix} } \right|$

Step 3: Expand the determinant along the column with zeros.

• Why this step? We have successfully created two zeros in the third column ($C_3$). Expanding the determinant along this column will significantly reduce the number of terms we need to calculate, as any term multiplied by zero will vanish.
• The expansion formula for a $3 \times 3$ determinant along the third column is: $\Delta = a_{13} C_{13} + a_{23} C_{23} + a_{33} C_{33}$ where $a_{ij}$ are the elements and $C_{ij}$ are their respective cofactors.
• In our determinant, $a_{13} = 1$, $a_{23} = 0$, and $a_{33} = 0$.
• Therefore, the expansion simplifies to: $\Delta = 1 \cdot C_{13} + 0 \cdot C_{23} + 0 \cdot C_{33} = 1 \cdot C_{13}$
• The cofactor $C_{13}$ is given by $(-1)^{1+3}$ times the minor obtained by deleting the 1st row and 3rd column: $C_{13} = (-1)^4 \left| {\begin{matrix} {2(a + 2)} & 1 \\ {4a + 10} & 2 \\ \end{matrix} } \right| = 1 \cdot \left( (2(a + 2))(2) - (1)(4a + 10) \right)$
• Now, calculate the value of this $2 \times 2$ determinant: $\Delta = (2(a + 2))(2) - (1)(4a + 10)$ $\Delta = 4(a + 2) - (4a + 10)$

Step 4: Simplify the expression to find the final value. $\Delta = 4a + 8 - 4a - 10$ $\Delta = (4a - 4a) + (8 - 10)$ $\Delta = 0 - 2$ $\Delta = -2$

The value of the determinant is $-2$.

3. Tips for Success & Common Mistakes

• Strategic Simplification: Always look for patterns (like a column or row of 1s) that allow you to create zeros using row/column operations. This is almost always faster and less prone to errors than direct expansion.
• Algebraic Accuracy: Be meticulous with your algebraic manipulations, especially when simplifying expressions like $(a+2)(a+3) - (a+1)(a+2)$. Factoring common terms, like $(a+2)$ in this case, can prevent mistakes and simplify calculations.
• Sign Errors: Remember the alternating signs when expanding determinants (e.g., for cofactor $C_{ij}$, it's $(-1)^{i+j} M_{ij}$). While expanding along a column with zeros, only the non-zero term's sign matters.
• Check Your Work: If time permits, quickly re-evaluate one or two tricky steps to catch any arithmetic or algebraic errors.

4. Summary and Key Takeaway

This problem beautifully illustrates the power of determinant properties. By strategically applying elementary row operations ($R_i \to R_i - R_j$), we transformed a seemingly complex determinant involving quadratic expressions into a much simpler form with zeros. This allowed us to expand the determinant along a column containing zeros, drastically simplifying the final calculation to a constant value of $-2$. The key takeaway is to always seek opportunities to simplify the determinant using row/column operations before resorting to direct expansion.