Question

If $\begin{vmatrix}x^{2}+x&x+1&x-2\\ 2x^{2}+3x-1&3x&3x-3\\ x^{2}+2x+3&2x-1&2x-1\end{vmatrix}=ax-12,$ then 'a' is equal to :

• A. 12
• B. 24
• C. -12
• D. -24

Answer 1

The Correct Option is B

 To solve the problem, we need to evaluate the determinant of the given 3x3 matrix and equate it to the expression \(ax - 12\).

The matrix given is:

\(x^2 + x\)	\(x + 1\)	\(x - 2\)
\(2x^2 + 3x - 1\)	\(3x\)	\(3x - 3\)
\(x^2 + 2x + 3\)	\(2x - 1\)	\(2x - 1\)

We will calculate the determinant \(\begin{vmatrix} x^2+x & x+1 & x-2 \\ 2x^2+3x-1 & 3x & 3x-3 \\ x^2+2x+3 & 2x-1 & 2x-1 \end{vmatrix}\).

1. Use the formula for the determinant of a 3x3 matrix: 
   \[ \text{Det}(A) = a(ei − fh) − b(di − fg) + c(dh − eg) \] where:
   • \(a = x^2 + x\), \(b = x + 1\), \(c = x - 2\)
   • \(d = 2x^2 + 3x - 1\), \(e = 3x\), \(f = 3x - 3\)
   • \(g = x^2 + 2x + 3\), \(h = 2x - 1\), \(i = 2x - 1\)
2. Substitute these values into the determinant formula: 
    

\[\begin{aligned} \text{Det}(A) &= (x^2 + x)((3x)(2x - 1) - (3x - 3)(2x - 1)) \\ &\quad - (x + 1)((2x^2 + 3x - 1)(2x - 1) - (3x - 3)(x^2 + 2x + 3)) \\ &\quad + (x - 2)((2x^2 + 3x - 1)(2x - 1) - (3x)(x^2 + 2x + 3)) \\ \end{aligned}\]

1. First, calculate each of the sub-expressions: 
   
   • \((3x)(2x - 1) = 6x^2 - 3x\)
   • \((3x - 3)(2x - 1) = 6x^2 - 3x - 6x + 3\)
   • \((2x^2 + 3x - 1)(2x - 1) = 4x^3 + 6x^2 - 2x - 2x^2 - 3x + 1\)
   • \((3x)(x^2 + 2x + 3) = 3x^3 + 6x^2 + 9x\)
2. For each column, compute the simplified terms.
3. Expand and simplify to obtain a polynomial in terms of \( x \).
4. Equate the resultant polynomial with \( ax - 12 \) to identify 'a'.

Upon solving, it turns out: \( a = 24 \). Thus the correct answer is 24.