<p>Let <br>\[\Delta = \begin{vmatrix} x^2+x & x+1 & x-2 \\ 2x^2+3x-1 & 3x & 3x-3 \\ x^2+2x+3 & 2x-1 & 2x-1 \end{vmatrix} = ax - 12\]<br>Then the value of \(a\) is:</p>
Step-by-Step Solution
Key Concept: Apply column operations to simplify the determinant into a form where we can factor out (ax - 12), then expand to find the coefficient a. Row/column operations preserve the determinant value, allowing us to reduce the matrix systematically.
Step 1: Apply the column operation $C_3 \to C_3 - C_2$.
The elements of the third column become:
$(x-2) - (x+1) = -3$
$(3x-3) - (3x) = -3$
$(2x-1) - (2x-1) = 0$
The determinant $\Delta$ is transformed into:
$$ \Delta = \begin{vmatrix} x^2+x & x+1 & -3 \\ 2x^2+3x-1 & 3x & -3 \\ x^2+2x+3 & 2x-1 & 0 \end{vmatrix} $$
Step 2: Factor out $-3$ from the third column.
$$ \Delta = -3 \begin{vmatrix} x^2+x & x+1 & 1 \\ 2x^2+3x-1 & 3x & 1 \\ x^2+2x+3 & 2x-1 & 0 \end{vmatrix} $$
Step 3: Evaluate the determinant.
Expanding the determinant yields:
$$ \Delta = 24x - 12 $$
Step 4: Compare the result with the given form $\Delta = ax - 12$.
Comparing $24x - 12$ with $ax - 12$, we conclude that $a = 24$.
Correct Answer: 24