Number of ordered pairs (p,q) which system of equation has unique solution is

To determine the number of ordered pairs \((p, q)\) such that the system of equations has a unique solution, we need to analyze the system:

\[
\begin{aligned}
1) \quad px + qy &= 1, \\
2) \quad qx + py &= 1.
\end{aligned}
\]

For a system of linear equations to have a unique solution, the determinant of the coefficient matrix must be non-zero. The coefficient matrix for this system is:

\[
\begin{pmatrix}
p & q \\
q & p
\end{pmatrix}.
\]

The determinant of this matrix is:

\[
\text{Determinant} = \begin{vmatrix} p & q \\ q & p \end{vmatrix} = p \times p - q \times q = p^2 - q^2.
\]

For the system to have a unique solution, the determinant must not be zero:

\[
p^2 - q^2 \neq 0.
\]

This simplifies to:

\[
(p - q)(p + q) \neq 0.
\]

Therefore, both \(p - q \neq 0\) and \(p + q \neq 0\), meaning:

1. \(p \neq q\),
2. \(p \neq -q\).

Thus, \(p\) and \(q\) must be distinct and non-opposite numbers. If \(p\) and \(q\) can take any real values, then there are infinitely many pairs of \((p, q)\) that satisfy these conditions.

Would you like more details on how to further analyze or extend this?

Here are 5 related questions:
1. What happens if \( p = q \) in the system of equations?
2. How would the system behave if \( p = -q \)?
3. Can you find a specific solution for any given values of \(p\) and \(q\)?
4. How do we classify the system if the determinant equals zero?
5. How would this system change if the constant terms were different?

**Tip**: The determinant of a 2x2 matrix provides a quick way to determine whether a system of linear equations has a unique solution, no solution, or infinitely many solutions.

Explore how to determine the number of ordered pairs (p, q) such that a system of equations has a unique solution. This problem involves analyzing the determinant of a 2x2 matrix and applying the concept of distinct and non-opposite values of p and q. Suitable for students in Grades 10-12.

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