Let p, q and r be three distinct positive real numbers. If \(\rm D = \left| {\begin{array}{*{20}{c}} \rm p&\rm q&\rm r\\ \rm q&\rm r&\rm p\\ \rm r&\rm p&\rm q \end{array}} \right|,\) then which one of the following is correct?
The question asks us to evaluate the sign of a specific 3x3 determinant, denoted by \( \rm D \). The entries of the determinant are three distinct positive real numbers, p, q, and r. We need to determine if D is less than, greater than, less than or equal to, or greater than or equal to zero.
Now consider the terms inside the square brackets. Since p, q, and r are real numbers, the squares of the differences, \( (p-q)^2 \), \( (q-r)^2 \), and \( (r-p)^2 \), must be non-negative (greater than or equal to zero).
\( (p-q)^2 \ge 0 \)
\( (q-r)^2 \ge 0 \)
\( (r-p)^2 \ge 0 \)
Furthermore, the problem states that p, q, and r are distinct positive real numbers. This means that \( p \neq q \), \( q \neq r \), and \( r \neq p \). Therefore, the differences \( (p-q) \), \( (q-r) \), and \( (r-p) \) are all non-zero.
Since these differences are non-zero, their squares must be strictly positive:
\( (p-q)^2 > 0 \)
\( (q-r)^2 > 0 \)
\( (r-p)^2 > 0 \)
The sum of these three strictly positive terms, \( (p-q)^2 + (q-r)^2 + (r-p)^2 \), is also strictly positive. Multiplying by \( \frac{1}{2} \) (which is positive) maintains the strict positivity.
\( (p^2 + q^2 + r^2 - pq - qr - rp) > 0 \) (because p, q, r are distinct)
Therefore, their product is strictly positive:
\( p^3 + q^3 + r^3 - 3pqr > 0 \)
Finally, substitute this back into the expression for D:
\( \rm D = -(p^3 + q^3 + r^3 - 3pqr) \)
Since \( (p^3 + q^3 + r^3 - 3pqr) \) is strictly positive, \( \rm D \) must be strictly negative.
\( \rm D < 0 \)
Summary of Determinant Sign
For three distinct positive real numbers p, q, and r, the determinant \( \rm D \) is calculated as \( \rm D = 3pqr - p^3 - q^3 - r^3 = -(p^3 + q^3 + r^3 - 3pqr) \). Using the identity \( p^3 + q^3 + r^3 - 3pqr = (p+q+r)(p^2 + q^2 + r^2 - pq - qr - rp) \) and the fact that p, q, r are distinct, we showed that \( p^3 + q^3 + r^3 - 3pqr > 0 \). Consequently, \( \rm D \) must be less than 0.
Comparing this result with the given options:
Option 1: D \( \le \) 0 (Incorrect, D is strictly less than 0)
Option 2: D \( < \) 0 (Correct)
Option 3: D \( > \) 0 (Incorrect)
Option 4: D \( \ge \) 0 (Incorrect, D is strictly less than 0)
Property
Value/Sign
Reason
p, q, r
Distinct positive real numbers
Given in the problem
\( p+q+r \)
\( > 0 \)
Sum of positive numbers is positive
\( (p-q)^2 + (q-r)^2 + (r-p)^2 \)
\( > 0 \)
Sum of squares of non-zero differences (since p,q,r are distinct) is positive
\( p^2 + q^2 + r^2 - pq - qr - rp \)
\( > 0 \)
Equal to \( \frac{1}{2}[(p-q)^2 + (q-r)^2 + (r-p)^2] \), which is positive
\( p^3 + q^3 + r^3 - 3pqr \)
\( > 0 \)
Equal to \( (p+q+r)(p^2 + q^2 + r^2 - pq - qr - rp) \), which is a product of two positive terms
\( \rm D \)
\( < 0 \)
Equal to \( -(p^3 + q^3 + r^3 - 3pqr) \), which is the negative of a positive term
Revision Table: Determinants and Algebraic Identities
Concept
Description
Example/Formula
Determinant of a 3x3 Matrix
A scalar value calculated from the elements of a square matrix. It provides information about the matrix, such as invertibility.
The determinant in the problem is slightly different in the second and third rows' arrangement. However, the calculation leads to a form related to \( p^3+q^3+r^3 - 3pqr \).
The expression \( p^3+q^3+r^3 - 3pqr \) is zero if and only if \( p+q+r = 0 \) or \( p^2+q^2+r^2-pq-qr-rp = 0 \). For distinct real numbers, the second condition \( p^2+q^2+r^2-pq-qr-rp = 0 \) is equivalent to \( \frac{1}{2}((p-q)^2 + (q-r)^2 + (r-p)^2) = 0 \), which implies \( p-q=0, q-r=0, r-p=0 \), meaning \( p=q=r \). But the problem states p, q, and r are distinct. Also, since p, q, r are positive, \( p+q+r \) cannot be zero.
Therefore, for distinct positive real numbers p, q, r, the value \( p^3+q^3+r^3 - 3pqr \) is always strictly positive, leading to D being strictly negative.
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