If \(Δ_1 = \begin{vmatrix} 1 & p & q \\ 1 & q & r \\ 1 & r & p\end{vmatrix} \ \text{and} \ Δ_2 = \begin{vmatrix} 1 & 1 & 1 \\ q & r & p \\ r & p & q \end{vmatrix}\) where p ≠ q ≠ r, then Δ 1+ Δ 2is
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.
| Determinant | Value |
|---|---|
| Δ1 | \(pq + qr + rp - (p^2 + q^2 + r^2)\) |
| Δ2 | \(pq + qr + rp - (p^2 + q^2 + r^2)\) |
| Δ1 + Δ2 | \(-2(p^2 + q^2 + r^2 - pq - qr - rp)\) |
| Simplified Sum | \(-((p-q)^2 + (q-r)^2 + (r-p)^2)\) |
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.
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