Calculating Determinants Δ₁ and Δ₂

We are given two determinants, Δ₁ and Δ₂, involving variables p, q, and r, where p ≠ q ≠ r.

The first determinant is:

\(Δ_1 = \begin{vmatrix} 1 & p & q \\ 1 & q & r \\ 1 & r & p\end{vmatrix}\)

We can calculate Δ₁ by expanding along the first column:

\(Δ_1 = 1 \cdot \begin{vmatrix} q & r \\ r & p \end{vmatrix} - 1 \cdot \begin{vmatrix} p & q \\ r & p \end{vmatrix} + 1 \cdot \begin{vmatrix} p & q \\ q & r \end{vmatrix}\)

\(Δ_1 = 1(qp - r^2) - 1(p^2 - qr) + 1(pr - q^2)\)

\(Δ_1 = pq - r^2 - p^2 + qr + pr - q^2\)

Rearranging the terms, we get:

\(Δ_1 = pq + qr + rp - (p^2 + q^2 + r^2)\)

The second determinant is:

\(Δ_2 = \begin{vmatrix} 1 & 1 & 1 \\ q & r & p \\ r & p & q \end{vmatrix}\)

We can calculate Δ₂ by expanding along the first row:

\(Δ_2 = 1 \cdot \begin{vmatrix} r & p \\ p & q \end{vmatrix} - 1 \cdot \begin{vmatrix} q & p \\ r & q \end{vmatrix} + 1 \cdot \begin{vmatrix} q & r \\ r & p \end{vmatrix}\)

\(Δ_2 = 1(rq - p^2) - 1(q^2 - rp) + 1(qp - r^2)\)

\(Δ_2 = qr - p^2 - q^2 + rp + qp - r^2\)

Rearranging the terms, we get:

\(Δ_2 = pq + qr + rp - (p^2 + q^2 + r^2)\)

Finding the Sum Δ₁ + Δ₂

Now we need to find the sum Δ₁ + Δ₂:

\(Δ_1 + Δ_2 = (pq + qr + rp - (p^2 + q^2 + r^2)) + (pq + qr + rp - (p^2 + q^2 + r^2))\)

\(Δ_1 + Δ_2 = 2(pq + qr + rp - (p^2 + q^2 + r^2))\)

\(Δ_1 + Δ_2 = -2(p^2 + q^2 + r^2 - pq - qr - rp)\)

Analyzing the Result

Let's look at the expression \(p^2 + q^2 + r^2 - pq - qr - rp\). We can rewrite this expression using algebraic identities. Consider the expression \((p-q)^2 + (q-r)^2 + (r-p)^2\):

\((p-q)^2 + (q-r)^2 + (r-p)^2 = (p^2 - 2pq + q^2) + (q^2 - 2qr + r^2) + (r^2 - 2rp + p^2)\)

\(= p^2 - 2pq + q^2 + q^2 - 2qr + r^2 + r^2 - 2rp + p^2\)

\(= 2p^2 + 2q^2 + 2r^2 - 2pq - 2qr - 2rp\)

\(= 2(p^2 + q^2 + r^2 - pq - qr - rp)\)

So, \(p^2 + q^2 + r^2 - pq - qr - rp = \frac{1}{2}((p-q)^2 + (q-r)^2 + (r-p)^2)\).

Now substitute this back into the expression for Δ₁ + Δ₂:

\(Δ_1 + Δ_2 = -2 \left( \frac{1}{2}((p-q)^2 + (q-r)^2 + (r-p)^2) \right)\)

\(Δ_1 + Δ_2 = -((p-q)^2 + (q-r)^2 + (r-p)^2)\)

We are given that p ≠ q, q ≠ r, and r ≠ p. This means that \(p-q \ne 0\), \(q-r \ne 0\), and \(r-p \ne 0\). Therefore, the squares \((p-q)^2\), \((q-r)^2\), and \((r-p)^2\) are all positive numbers.

• \((p-q)^2 > 0\)
• \((q-r)^2 > 0\)
• \((r-p)^2 > 0\)

The sum of these positive numbers, \((p-q)^2 + (q-r)^2 + (r-p)^2\), must also be positive.

\((p-q)^2 + (q-r)^2 + (r-p)^2 > 0\)

Finally, Δ₁ + Δ₂ is the negative of this positive sum:

\(Δ_1 + Δ_2 = -((p-q)^2 + (q-r)^2 + (r-p)^2)\)

This means Δ₁ + Δ₂ is always a negative number.

Conclusion

Based on the calculation and analysis, the sum Δ₁ + Δ₂ is always negative when p ≠ q ≠ r.

Since p, q, and r are distinct, \((p-q)^2 + (q-r)^2 + (r-p)^2\) is strictly positive, making the sum Δ₁ + Δ₂ always negative.