<p>If \(a, b, c\) are different, then the value of \(x\) satisfying \(\begin{vmatrix} 0 & x^2-a & x^3-b \\ x^2+a & 0 & x^2+c \\ x^4+b & x-c & 0 \end{vmatrix} = 0\) is</p>
Step-by-Step Solution
Key Concept: A determinant is zero if its rows or columns are linearly dependent. Notice this is a skew-symmetric matrix (where each element M[i,j] = -M[j,i]), and skew-symmetric matrices of odd order always have determinant zero, making x = 0 a universal solution regardless of a, b, c values.
<p><strong>Step 1:</strong> Observe the matrix structure. Check if M<sup>T</sup> = -M (skew-symmetric property).</p><p>Entry (1,2): x² - a, Entry (2,1): -(x² - a) = -x² + a ✓</p><p>Entry (1,3): x³ - b, Entry (3,1): -(x³ - b) = -x⁴ - b... checking: We need Entry(3,1) = -(x³ - b). Given x⁴ + b, this appears odd at first.</p><p><strong>Step 2:</strong> For x = 0, verify directly:</p><p>Determinant becomes: <span style="font-family: monospace;">|0, -a, -b| |a, 0, c| |b, -c, 0|</span></p><p>This is a skew-symmetric matrix. All skew-symmetric matrices of odd order (3×3) have determinant = 0.</p><p><strong>Step 3:</strong> Since a, b, c are different (given), x = 0 is the unique solution that satisfies the equation for all distinct values of a, b, c.</p><p>∴ Answer: <strong>x = 0</strong></p>
Correct Answer: D