Using properties of determinants, show that `palpha^(2) + 2qalpha + r=0`, given that p, q and r are not in G. P.

Answer

To show that \( p\alpha^2 + 2q\alpha + r = 0 \) given that \( p, q, \) and \( r \) are not in geometric progression (G.P.), we will use properties of determinants.

### Step-by-Step Solution:

1. **Set up the Determinant**:
We start with the determinant:
\[ D = \begin{vmatrix} p & 1 & 0 \\ q & 1 & 0 \\ r & 1 & 0 \\ \end{vmatrix} \]
This determinant is given to be equal to zero.

2. **Factor Out Constants**:
We can factor out \( \frac{1}{p} \) from the first row and \( \frac{1}{q} \) from the second row:
\[ D = \frac{1}{pq} \begin{vmatrix} p & 1 & 0 \\ q & 1 & 0 \\ r & 1 & 0 \\ \end{vmatrix} \]
This gives us:
\[ D = \frac{1}{pq} \begin{vmatrix} p & q & 0 \\ p\alpha + q & q & 0 \\ p\alpha + r & r & 0 \\ \end{vmatrix} \]

3. **Row Transformation**:
We perform the row operation \( R_2 \rightarrow R_2 + \alpha R_1 \):
\[ D = \frac{1}{pq} \begin{vmatrix} p & q & 0 \\ p\alpha + q & q & 0 \\ p\alpha + r & r & 0 \\ \end{vmatrix} \]

4. **Simplifying the Determinant**:
After performing the row operation, we get:
\[ D = \frac{1}{pq} \begin{vmatrix} p & q & 0 \\ p\alpha + q & q & 0 \\ p\alpha + r & r & 0 \\ \end{vmatrix} \]

5. **Further Row Transformation**:
We will now transform the third row \( R_3 \rightarrow R_3 - R_2 \):
\[ D = \frac{1}{pq} \begin{vmatrix} p & q & 0 \\ p\alpha + q & q & 0 \\ 0 & r-q & 0 \\ \end{vmatrix} \]

6. **Expanding the Determinant**:
The determinant simplifies to:
\[ D = \frac{1}{pq} \cdot (p\alpha + q)(r - q) \]

7. **Setting the Determinant to Zero**:
Since we know \( D = 0 \), we have:
\[ (p\alpha + q)(r - q) = 0 \]

8. **Analyzing the Factors**:
For the product to be zero, either \( p\alpha + q = 0 \) or \( r - q = 0 \). Since \( p, q, r \) are not in G.P., we cannot have \( r = q \).

9. **Conclusion**:
Therefore, we must have:
\[ p\alpha + q = 0 \implies p\alpha^2 + 2q\alpha + r = 0 \]
This shows that \( p\alpha^2 + 2q\alpha + r = 0 \).