SECTION A

Q.1) If $$\left|\begin{array}{ll}2 & 4 \\ 5 & 1\end{array}\right|=\left|\begin{array}{cc}2 x & 4 \\ 6 & x\end{array}\right|$$, then the possible value(s) of ‘ x ‘ is/are
(a) 3
(b) $$\sqrt{3}$$
(c ) $$-\sqrt{3}$$
(d) $$\sqrt{3},-\sqrt{3}$$

Solution:

1. Solve the determinant equation

Step 1: Calculate the determinants of both matrices
The determinant of a $$2 \times 2$$ matrix $$\left|\begin{array}{ll}a & b \\ c & d\end{array}\right|$$ is calculated as $$a d-b c$$.
For the left matrix:
$$
\left|\begin{array}{ll}
2 & 4 \\
5 & 1
\end{array}\right|=(2)(1)-(4)(5)=2-20=-18
$$

For the right matrix:
$$
\left|\begin{array}{cc}
2 x & 4 \\
6 & x
\end{array}\right|=(2 x)(x)-(4)(6)=2 x^2-24
$$

Step 2: Set the determinants equal and solve for x
$$
\begin{aligned}
-18 & =2 x^2-24 \\
-18+24 & =2 x^2 \\
6 & =2 x^2 \\
3 & =x^2 \\
x & = \pm \sqrt{3}
\end{aligned}
$$

Step 3: Match the result with the given options
The possible values for $$x$$ are $$\sqrt{3}$$ and $$-\sqrt{3}$$, which corresponds to option (d).

Answer:
(d) $$\sqrt{3},-\sqrt{3}$$