Question

If \( A = \begin{bmatrix} 2 & 3 \\ 1 & k \end{bmatrix} \) is a singular matrix, then the quadratic equation having the roots \( k \) and \( \frac{1}{k} \) is:

• A. \( 6x^2 + 13x + 6 = 0 \)
• B. \( 12x^2 - 25x + 12 = 0 \)
• C. \( 6x^2 - 13x + 6 = 0 \)
• D. \( 2x^2 - 5x + 2 = 0 \)

Hint:

To find the quadratic equation from given roots, use the formula \( x^2 - (r_1 + r_2)x + r_1r_2 = 0 \).

Answer 1

The Correct Option is C

We are given that the matrix \( A = \begin{bmatrix} 2 & 3 \\ 1 & k \end{bmatrix} \) is a singular matrix. For a matrix to be singular, its determinant must be zero. 
Step 1: The determinant of matrix \( A \) is: \[ \text{det}(A) = \left| \begin{matrix} 2 & 3 \\ 1 & k \end{matrix} \right| = (2)(k) - (1)(3) = 2k - 3. \] Since the matrix is singular, the determinant is zero: \[ 2k - 3 = 0. \] Solving for \( k \): \[ 2k = 3 \quad \Rightarrow \quad k = \frac{3}{2}. \] 
Step 2: Now, we need to find the quadratic equation whose roots are \( k = \frac{3}{2} \) and \( \frac{1}{k} = \frac{2}{3} \). For a quadratic equation, if the roots are \( r_1 \) and \( r_2 \), the equation can be written as: \[ x^2 - (r_1 + r_2)x + r_1r_2 = 0. \] Substitute \( r_1 = \frac{3}{2} \) and \( r_2 = \frac{2}{3} \): \[ r_1 + r_2 = \frac{3}{2} + \frac{2}{3} = \frac{9}{6} + \frac{4}{6} = \frac{13}{6}, \] \[ r_1r_2 = \frac{3}{2} \times \frac{2}{3} = 1. \] Thus, the quadratic equation is: \[ x^2 - \frac{13}{6}x + 1 = 0. \] Multiplying through by 6 to eliminate the fraction: \[ 6x^2 - 13x + 6 = 0. \] 
Conclusion: Therefore, the quadratic equation having the roots \( k = \frac{3}{2} \) and \( \frac{1}{k} = \frac{2}{3} \) is \( 6x^2 - 13x + 6 = 0 \).