Question

please give me this question answer​

Answer 1

Answer:

x = 2

Step-by-step explanation:

$Given:\left[\begin{array}{ccc}3\sqrt{6} &-4\sqrt{2}\\5\sqrt{3} &x\end{array}\right]$

$=>3\sqrt{6}x-(-4\sqrt{2})*5\sqrt{3}$

$=>3\sqrt{6}x + 4\sqrt{2} * 5\sqrt{3}$

$=>3\sqrt{6}x + 20\sqrt{6}$

Given, value of determinant is 26√6.

$=>3\sqrt{6}x + 20\sqrt{6} = 26\sqrt{6}$

$=>3\sqrt{6}x = 6\sqrt{6}$

$=>x = 2$

Hope it helps!