Question

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:

• A. 37
• B. -6
• C. -4
• D. -9

Answer 1

Answer: (C)

-4

Answer 2

The equation of a normal to the parabola $y^2 = 4x$ (where $a=1$) is given by $tx + y - t^3 - 2t = 0$. For the normal to pass through (9, -6), we substitute these coordinates into the equation: $9t - 6 - t^3 - 2t = 0$, which simplifies to $t^3 - 7t + 6 = 0$. Factoring this cubic equation gives $(t-1)(t-2)(t+3) = 0$, so the roots are $t=1, 2, -3$. The slopes of the normals are $m = -t$, which are $m_1 = -1, m_2 = -2, m_3 = 3$. Ordering these slopes by their absolute values: $|-1|=1, |-2|=2, |3|=3$. Thus, $m_1 = -1, m_2 = -2, m_3 = 3$. The diagonal elements of matrix A are $a_{ii} = m_i + i$. $a_{11} = m_1 + 1 = -1 + 1 = 0$ $a_{22} = m_2 + 2 = -2 + 2 = 0$ $a_{33} = m_3 + 3 = 3 + 3 = 6$ The matrix A is:

$$A = \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 6 \end{pmatrix}$$

The determinant of A is: $\det(A) = 0(0-1) - 1(6-1) + 1(1-0) = 0 - 5 + 1 = -4$.