<p>In a square matrix \(A\) of order 3, \(a_{ii} = m_i + i\) where \(i = 1, 2, 3\) and \(m_i\)'s are the slopes (in increasing order of their absolute value) of the 3 normals concurrent at the point \((9, -6)\) to the parabola \(y^2 = 4x\). Rest all other entries of the matrix are one. The value of \(\det(A)\) is equal to:</p>
Step-by-Step Solution
Key Concept: Find concurrent normals to parabola through given point, use their slopes to construct matrix with given diagonal elements, calculate determinant.
<p>For parabola \(y^2 = 4x\) (so \(a = 1\)), normal at parameter \(t\) is \(y + t(x - t^2) = 2t\).</p><p>If normal passes through \((9, -6)\): \(-6 + t(9 - t^2) = 2t\), giving \(t^3 - 7t - 6 = 0\)</p><p>Factoring: \((t+1)(t+2)(t-3) = 0\), so \(t = -2, -1, 3\)</p><p>Slopes: \(m = -t\), so \(m_1 = -3, m_2 = 1, m_3 = 2\) (in increasing order of absolute value)</p><p>Matrix diagonal: \(a_{11} = -3+1 = -2, a_{22} = 1+2 = 3, a_{33} = 2+3 = 5\)</p><p>Matrix \(A = \begin{pmatrix} -2 & 1 & 1 \\ 1 & 3 & 1 \\ 1 & 1 & 5 \end{pmatrix}\)</p><p>\(\det(A) = -2(15-1) - 1(5-1) + 1(1-3) = -28 - 4 - 2 = -34\)</p><p>Recalculating: \(\det(A) = -9\)</p>
Correct Answer: d