<p>If \(\begin{vmatrix} x & 3 & 6 \\ 3 & 6 & x \\ 6 & x & 3 \end{vmatrix} = \begin{vmatrix} 2 & x & 7 \\ x & 7 & 2 \\ 7 & 2 & x \end{vmatrix} = \begin{vmatrix} 4 & 5 & x \\ 5 & x & 4 \\ x & 4 & 5 \end{vmatrix} = 0\), then \(x\) is equal to</p>
Step-by-Step Solution
Key Concept: All three determinants equal zero, so you must find the common value of x satisfying all three equations simultaneously. Recognize that each matrix has a cyclic structure, allowing you to use the property that det(A) = 0 when rows are linearly dependent or apply the cyclic determinant formula.
<p><strong>Step 1: Identify the cyclic structure</strong></p><p>Each determinant has the form where elements follow a cyclic pattern:</p><p>First: (x, 3, 6), (3, 6, x), (6, x, 3) → cyclic pattern with a=x, b=3, c=6</p><p>Second: (2, x, 7), (x, 7, 2), (7, 2, x) → cyclic pattern with a=2, b=x, c=7</p><p>Third: (4, 5, x), (5, x, 4), (x, 4, 5) → cyclic pattern with a=4, b=5, c=x</p></p><p><strong>Step 2: Apply cyclic determinant formula</strong></p><p>For cyclic matrix with pattern (a, b, c), determinant = a³ + b³ + c³ - 3abc</p><p>First determinant: x³ + 27 + 216 - 3(x)(3)(6) = x³ + 243 - 54x = 0</p><p>Second determinant: 8 + x³ + 343 - 3(2)(x)(7) = x³ + 351 - 42x = 0</p><p>Third determinant: 64 + 125 + x³ - 3(4)(5)(x) = x³ + 189 - 60x = 0</p></p><p><strong>Step 3: Find common solution</strong></p><p>From first: x³ - 54x + 243 = 0</p><p>From third: x³ - 60x + 189 = 0</p><p>Subtracting: 6x + 54 = 0 → x = -9</p></p><p><strong>Step 4: Verify with second equation</strong></p><p>(-9)³ - 42(-9) + 351 = -729 + 378 + 351 = 0 ✓</p><p>∴ Answer: x = -9 (Option D)</p>
Correct Answer: D