If `|(x^2+x,x+1,x-2),(2x^2+3x-1,3x,3x-3),(x^2+2x+3,2x-1,2x-1)|=ax-12` then 'a' is equal to (1) 12 (2) 24 (3) -12 (4) -24

Options

A. 12
B. 24
C. `-12`
D. `-24`

Answer

To solve the problem, we need to find the value of \( a \) in the equation given by the determinant:

\[ \left| \begin{array}{ccc} x^2 + x & x + 1 & x - 2 \\ 2x^2 + 3x - 1 & 3x & 3x - 3 \\ x^2 + 2x + 3 & 2x - 1 & 2x - 1 \end{array} \right| = ax - 12 \]

### Step 1: Substitute \( x = 1 \)

To simplify the calculation, we substitute \( x = 1 \) into the determinant:

\[ \left| \begin{array}{ccc} 1^2 + 1 & 1 + 1 & 1 - 2 \\ 2(1^2) + 3(1) - 1 & 3(1) & 3(1) - 3 \\ 1^2 + 2(1) + 3 & 2(1) - 1 & 2(1) - 1 \end{array} \right| \]

Calculating each entry:

- First row: \( 1 + 1 = 2, 1 + 1 = 2, 1 - 2 = -1 \)
- Second row: \( 2 + 3 - 1 = 4, 3, 0 \)
- Third row: \( 1 + 2 + 3 = 6, 1, 1 \)

So, the determinant simplifies to:

\[ \left| \begin{array}{ccc} 2 & 2 & -1 \\ 4 & 3 & 0 \\ 6 & 1 & 1 \end{array} \right| \]

### Step 2: Calculate the Determinant

Now we will calculate the determinant using the formula for a \( 3 \times 3 \) 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:

\[ \text{Det} = 2 \left( 3 \cdot 1 - 0 \cdot 1 \right) - 2 \left( 4 \cdot 1 - 0 \cdot 6 \right) + (-1) \left( 4 \cdot 1 - 3 \cdot 6 \right) \]

Calculating each term:

1. \( 2(3 \cdot 1 - 0 \cdot 1) = 2 \cdot 3 = 6 \)
2. \( -2(4 \cdot 1 - 0 \cdot 6) = -2 \cdot 4 = -8 \)
3. \( -1(4 \cdot 1 - 3 \cdot 6) = -1(4 - 18) = -1(-14) = 14 \)

Combining these results:

\[ \text{Det} = 6 - 8 + 14 = 12 \]

### Step 3: Set the Determinant Equal to \( ax - 12 \)

Now we have:

\[ 12 = a(1) - 12 \]

### Step 4: Solve for \( a \)

Rearranging gives:

\[ 12 + 12 = a \implies a = 24 \]

### Final Answer

Thus, the value of \( a \) is \( \boxed{24} \).

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