Question

The correct option for \( x \) which satisfies the following equation is \[ \begin{vmatrix} x & 2 & 3 \\ 4 & x & 6 \\ x & 8 & 9 \end{vmatrix} = \begin{vmatrix} 102 & 18 & 36 \\ 1 & 3 & 4 \\ 17 & 3 & 6 \end{vmatrix} \]

• A. \( 3 \pm \sqrt{5} \)
• B. \( \frac{3 \pm \sqrt{5}}{2} \)
• C. \( 2(3 \pm \sqrt{5}) \)
• D. \( 3 \pm 2\sqrt{5} \)

Hint:

When calculating determinants, break them down into smaller 2x2 matrices and simplify systematically.

Answer 1

The Correct Option is A

Step 1: Understanding the determinant.
We are given the determinant of two 3x3 matrices. The equation can be solved by equating the determinants of both matrices. \[ \begin{vmatrix} x & 2 & 3 \\ 4 & x & 6 \\ x & 8 & 9 \end{vmatrix} = x \begin{vmatrix} x & 6 \\ 8 & 9 \end{vmatrix} - 2 \begin{vmatrix} 4 & 6 \\ x & 9 \end{vmatrix} + 3 \begin{vmatrix} 4 & x \\ x & 8 \end{vmatrix} \] First, we calculate the 2x2 determinants: \[ \begin{vmatrix} x & 6 \\ 8 & 9 \end{vmatrix} = x \cdot 9 - 6 \cdot 8 = 9x - 48 \] \[ \begin{vmatrix} 4 & 6 \\ x & 9 \end{vmatrix} = 4 \cdot 9 - 6 \cdot x = 36 - 6x \] \[ \begin{vmatrix} 4 & x \\ x & 8 \end{vmatrix} = 4 \cdot 8 - x \cdot x = 32 - x^2 \] Substitute these into the determinant equation: \[ \text{determinant} = x(9x - 48) - 2(36 - 6x) + 3(32 - x^2) \] Simplifying the expression: \[ = x(9x - 48) - 2(36 - 6x) + 3(32 - x^2) \] \[ = 9x^2 - 48x - 72 + 12x + 96 - 3x^2 \] \[ = 6x^2 - 36x + 24 \] Step 2: Solve for the value of \( x \).
Now solve for \( x \) by equating the determinant of the second matrix: \[ \begin{vmatrix} 102 & 18 & 36 \\ 1 & 3 & 4 \\ 17 & 3 & 6 \end{vmatrix} \] The determinant of the right-hand matrix: \[ = 102 \begin{vmatrix} 3 & 4 \\ 3 & 6 \end{vmatrix} - 18 \begin{vmatrix} 1 & 4 \\ 17 & 6 \end{vmatrix} + 36 \begin{vmatrix} 1 & 3 \\ 17 & 3 \end{vmatrix} \] Now calculating the 2x2 determinants: \[ \begin{vmatrix} 3 & 4 \\ 3 & 6 \end{vmatrix} = 3 \cdot 6 - 4 \cdot 3 = 18 - 12 = 6 \] \[ \begin{vmatrix} 1 & 4 \\ 17 & 6 \end{vmatrix} = 1 \cdot 6 - 4 \cdot 17 = 6 - 68 = -62 \] \[ \begin{vmatrix} 1 & 3 \\ 17 & 3 \end{vmatrix} = 1 \cdot 3 - 3 \cdot 17 = 3 - 51 = -48 \] Substitute these back into the determinant expression: \[ = 102 \cdot 6 - 18 \cdot (-62) + 36 \cdot (-48) \] \[ = 612 + 1116 - 1728 = 0 \] Thus, the determinant equals 0. Hence, the correct option for \( x \) is \( 3 \pm \sqrt{5} \). 
Final Answer: \[ \boxed{3 \pm \sqrt{5}} \]