Question

Let K be positive real number and let $A=\left[\begin{array}{ccc}2K-1& 2\sqrt{K}& 2\sqrt{K}\\ 2\sqrt{K}& 1& -2K\\ -2\sqrt{K}& 2K& -1\end{array}\right]andB=\left[\begin{array}{ccc}0& 1-K& \sqrt{K-1}\\ K-1& 0& K-2\\ -\sqrt{K-1}& 2-K& 0\end{array}\right]$
If det(AdjA) + det(AdjB) = 729, then the value of K is

• A. 1
• B. -1
• C. 2

Answer 1

The correct option is A 1

$|A|=\left[\begin{array}{ccc}2K-1& 2\sqrt{K}& 2\sqrt{K}\\ 2\sqrt{K}& 1& -2K\\ -2\sqrt{K}& 2K& -1\end{array}\right]$ operate $R_2\to R_2+R_3$ and take (2K + 1) common from $R_2$

$|A|=(2K+1)\left[\begin{array}{ccc}2K-1& 2\sqrt{K}& 2\sqrt{K}\\ 0& 1& -1\\ -2\sqrt{K}& 2K& -1\end{array}\right]Operate{C}_2\to C_2+C_3|A|=\left|\begin{array}{ccc}2K-1& 4\sqrt{K}& 2\sqrt{K}\\ 0& 0& -1\\ -2\sqrt{K}& 2K-1& -1\end{array}\right|=(2K+1)(2K+1{)}^2=(2K+1{)}^3$
and |B| = 0 (skew-symmetric determinant of odd order)
$|A|=(2K+1{)}^3\Rightarrow |AdjA|=(|A|{)}^{n-1}\Rightarrow |A{|}^2=(2K+1{)}^3(2K+1{)}^{3}forn=3|AdjB|=|B{|}^2=0\Rightarrow (2K+1{)}^6=729=3^6\Rightarrow 2K+1=3\Rightarrow K=1$