Question

Expand the values of x, if

(i)$\left|\begin{array}{cc}2& 4\\ 5& 1\end{array}\right|=\left|\begin{array}{cc}2x& 4\\ 6& x\end{array}\right|$

(ii) $\left|\begin{array}{cc}2& 3\\ 4& 5\end{array}\right|=\left|\begin{array}{cc}x& 3\\ 2x& 5\end{array}\right|$

Answer 1

Given, $\left|\begin{array}{cc}2& 4\\ 5& 1\end{array}\right|=\left|\begin{array}{cc}2x& 4\\ 6& x\end{array}\right|$
On expanding both determinants, we get
$2\times 1-5\times 4=2x\times x-6\times 4\Rightarrow 2-20=2x^2-24\Rightarrow 2x^2=-18+24\Rightarrow x^2=\frac{6}{2}=3\Rightarrow x=\pm \sqrt{3}$

Given, $\left|\begin{array}{cc}2& 3\\ 4& 5\end{array}\right|=\left|\begin{array}{cc}x& 3\\ 2x& 5\end{array}\right|$
On expanding both determinants, we get
$2\times 5-4\times 3=5\times x-3\times 2x\Rightarrow 10-12=5x-6x\Rightarrow -2=-x\Rightarrow x=2$