In a square matrix A of order 3,a_{ii} = i + | vedclass

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(C) The equation of a normal to the parabola $y^2 = 4ax$ with $a = 1$ is $y = mx - 2am - am^3$,which simplifies to $y = mx - 2m - m^3$.Since the norma

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$-4$

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(C) The equation of a normal to the parabola $y^2 = 4ax$ with $a = 1$ is $y = mx - 2am - am^3$,which simplifies to $y = mx - 2m - m^3$.Since the normal passes through $(9, -6)$,we have $-6 = 9m - 2m - m^3$,which gives $m^3 - 7m - 6 = 0$.Factoring the cubic equation,we get $(m + 1)(m + 2)(m - 3) = 0$,so the slopes are $m = -1, -2, 3$.The absolute values are $|-1| = 1$,$|-2| = 2$,and $|3| = 3$. Arranging them in increasing order,we have $m_1 = -1, m_2 = -2, m_3 = 3$.The diagonal elements are $a_{11} = 1 + m_1 = 1 - 1 = 0$,$a_{22} = 2 + m_2 = 2 - 2 = 0$,and $a_{33} = 3 + m_3 = 3 + 3 = 6$.The matrix $A$ is $\begin{bmatrix} 0 & 1 & 1 \ 1 & 0 & 1 \ 1 & 1 & 6 \end{bmatrix}$.Calculating the determinant: $\det(A) = 0(0 - 1) - 1(6 - 1) + 1(1 - 0) = 0 - 5 + 1 = -4$.

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