Question

If \( \left| \begin{array}{cc} 2x & 5 \\ 4 & x \end{array} \right| = \left| \begin{array}{cc} 3 & 5 \\ 4 & 6 \end{array} \right| \), then the value of \( x \) is:

• A. $\frac{3}{2}$
• B. $6$
• C. $3$
• D. $\pm 3$

Hint:

Don’t forget to solve quadratic equations completely — always check for $\pm$ when square roots are involved.

Answer 1

The Correct Option is D

The determinant of a \( 2 \times 2 \) matrix \( \left| \begin{array}{cc} a & b \\ c & d \end{array} \right| \) is calculated as \( ad - bc \). Applying this to the left side: \( \left| \begin{array}{cc} 2x & 5 \\ 4 & x \end{array} \right| = (2x)(x) - (4)(5) = 2x^2 - 20 \). For the right side: \( \left| \begin{array}{cc} 3 & 5 \\ 4 & 6 \end{array} \right| = (3)(6) - (4)(5) = 18 - 20 = -2 \). Equating the two sides: \( 2x^2 - 20 = -2 \). This simplifies to \( 2x^2 = 18 \), then \( x^2 = 9 \), and finally \( x = \pm 3 \).