Question

If p, q, r are distinct non zero real numbers and $f(x) = \begin{vmatrix} p^2+x^2 & pq & pr \\ pq & q^2+x^2 & qr \\ pr & qr & r^2+x^2 \end{vmatrix}$ , then the number of distinct (both real and imaginary included) roots of f(x) = 0 are

• A. 4
• B. 5
• C. 2
• D. 3

Answer 1

Answer: (D)

3

Answer 2

The given determinant is: $f(x) = \begin{vmatrix} p^2+x^2 & pq & pr \\ pq & q^2+x^2 & qr \\ pr & qr & r^2+x^2 \end{vmatrix}$ This matrix can be written as $M = vv^T + x^2 I$, where $v = \begin{pmatrix} p \\ q \\ r \end{pmatrix}$ and $I$ is the 3x3 identity matrix. The determinant of such a matrix is given by $\det(x^2 I + vv^T) = (x^2)^n + v^T v \cdot (x^2)^{n-1}$, where $n$ is the order of the matrix. Here, $n=3$. So, $f(x) = (x^2)^3 + (p^2+q^2+r^2) \cdot (x^2)^{3-1}$ $f(x) = x^6 + (p^2+q^2+r^2) x^4$ $f(x) = x^4 (x^2 + p^2+q^2+r^2)$

To find the roots of $f(x)=0$, we set: $x^4 (x^2 + p^2+q^2+r^2) = 0$

This equation yields two sets of roots:

1. $x^4 = 0 \implies x = 0$ (This is a root with multiplicity 4).
2. $x^2 + p^2+q^2+r^2 = 0 \implies x^2 = -(p^2+q^2+r^2)$. Since $p, q, r$ are distinct non-zero real numbers, $p^2 > 0, q^2 > 0, r^2 > 0$. Thus, $p^2+q^2+r^2 > 0$. Let $K = p^2+q^2+r^2$. Then $x^2 = -K$. The roots are $x = \pm \sqrt{-K} = \pm i\sqrt{K} = \pm i\sqrt{p^2+q^2+r^2}$. These are two distinct imaginary roots.

The roots of $f(x)=0$ are $0, 0, 0, 0, i\sqrt{p^2+q^2+r^2}, -i\sqrt{p^2+q^2+r^2}$. The distinct roots are $0$, $i\sqrt{p^2+q^2+r^2}$, and $-i\sqrt{p^2+q^2+r^2}$. Therefore, there are 3 distinct roots.