Question

The system of linear equations in real \( (x, y) \) given by \[ \begin{pmatrix} x \\ y \end{pmatrix} \begin{pmatrix} 2 & 5 - 2\alpha \\ \alpha & 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \] involves a real parameter \( \alpha \) and has infinitely many non-trivial solutions for special value(s) of \( \alpha \). Which one or more among the following options is/are non-trivial solution(s) of \( (x, y) \) for such special value(s) of \( \alpha \)?

• A. \( x = 2, \, y = -2 \)
• B. \( x = -1, \, y = 4 \)
• C. \( x = 1, \, y = 1 \)
• D. \( x = 4, \, y = -2 \)

Hint:

For systems of linear equations with a parameter, determine when the determinant of the coefficient matrix equals zero to find the special values of the parameter that give infinitely many solutions.

Answer 1

The Correct Option is A, B

The system of linear equations given is: \[ \begin{pmatrix} x \\ y \end{pmatrix} \begin{pmatrix} 2 & 5 - 2\alpha \\ \alpha & 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \] This system of equations will have infinitely many non-trivial solutions for special values of \( \alpha \) when the determinant of the coefficient matrix equals zero. The determinant of the coefficient matrix is: \[ \text{det} = \begin{vmatrix} 2 & 5 - 2\alpha \\ \alpha & 1 \end{vmatrix} = 2(1) - \alpha(5 - 2\alpha)
= 2 - \alpha(5 - 2\alpha)
= 2 - 5\alpha + 2\alpha^2.
\] For the system to have infinitely many solutions, the determinant must be zero: \[ 2 - 5\alpha + 2\alpha^2 = 0. \] This is a quadratic equation in \( \alpha \). Solving it: \[ 2\alpha^2 - 5\alpha + 2 = 0. \] Using the quadratic formula: \[ \alpha = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(2)}}{2(2)} = \frac{5 \pm \sqrt{25 - 16}}{4} = \frac{5 \pm 3}{4}. \] So, the two possible values of \( \alpha \) are: \[ \alpha = \frac{5 + 3}{4} = 2, \quad \alpha = \frac{5 - 3}{4} = \frac{1}{2}. \] Step 1: Check the solutions for \( \alpha = 2 \) For \( \alpha = 2 \), the system becomes: \[ \begin{pmatrix} 2 & 5 - 2(2)
2 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 1
2 & 1 \end{pmatrix}. \] This system has infinitely many solutions. Checking the options: - For \( x = 2, y = -2 \) (Option A): \[ 2(2) + 1(-2) = 4 - 2 = 0, \quad 2(2) + 1(-2) = 4 - 2 = 0. \] Thus, \( x = 2, y = -2 \) is a valid solution for \( \alpha = 2 \). Step 2: Check the solutions for \( \alpha = \frac{1}{2} \) For \( \alpha = \frac{1}{2} \), the system becomes: \[ \begin{pmatrix} 2 & 5 - 2(\frac{1}{2})
\frac{1}{2} & 1 \end{pmatrix} = \begin{pmatrix} 2 & 4
\frac{1}{2} & 1 \end{pmatrix}. \] This system also has infinitely many solutions. Checking the options: - For \( x = -1, y = 4 \) (Option B): \[ 2(-1) + 4(4) = -2 + 16 = 14, \quad \frac{1}{2}(-1) + 1(4) = -\frac{1}{2} + 4 = \frac{7}{2}. \] Thus, \( x = -1, y = 4 \) is a valid solution for \( \alpha = \frac{1}{2} \). Thus, the correct answers are (A) and (B).