Find the values of x, if

Solution:

(i)

Given : $\left|\begin{array}{ll}2 & 4 \\ 5 & 1\end{array}\right|=\left|\begin{array}{cc}2 x & 4 \\ 6 & x\end{array}\right|$

$\Rightarrow 2-20=2 x^{2}-24$

$\Rightarrow-18=2 x^{2}-24$

$\Rightarrow 2 x^{2}=6$

$\Rightarrow x^{2}=3$

$\Rightarrow x=\pm \sqrt{3}$

(ii)

Given: $\left|\begin{array}{ll}2 & 3 \\ 4 & 5\end{array}\right|=\left|\begin{array}{cc}x & 3 \\ 2 x & 5\end{array}\right|$

$\Rightarrow 10-12=5 x-6 x$

$\Rightarrow-2=-x$

$\Rightarrow \quad x=2$

(iii)

Given: $\left|\begin{array}{ll}3 & x \\ x & 1\end{array}\right|=\left|\begin{array}{ll}3 & 2 \\ 4 & 1\end{array}\right|$

$\Rightarrow 3-x^{2}=3-8$

$\Rightarrow-x^{2}=-8$

$\Rightarrow x^{2}=8$

$\Rightarrow x=\pm 2 \sqrt{2}$

(iv)

Given : $\left|\begin{array}{cc}3 x & 7 \\ 2 & 4\end{array}\right|=10$

$\Rightarrow 12 x-14=10$

$\Rightarrow 12 x=24$

$\Rightarrow x=2$

(v)

Given : $\left|\begin{array}{ll}x+1 & x-1 \\ x-3 & x+2\end{array}\right|=\left|\begin{array}{cc}4 & -1 \\ 1 & 3\end{array}\right|$

$\Rightarrow(x+1)(x+2)-(x-3)(x-1)=12+1$

$\Rightarrow x^{2}+3 x+2-x^{2}+4 x-3=13$

$\Rightarrow 7 x-1=13$

$\Rightarrow 7 x=14$

$\Rightarrow x=2$

(vi)

Given : $\left|\begin{array}{cc}2 x & 5 \\ 8 & x\end{array}\right|=\left|\begin{array}{cc}6 & 5 \\ 8 & 3\end{array}\right|$

$\Rightarrow 2 x^{2}-40=18-40$

$\Rightarrow 2 x^{2}=18$

$\Rightarrow x^{2}=9$

$\Rightarrow x=\pm 3$