Question

. $A=\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 1& 0\end{array}\right],B=\left[\begin{array}{ccc}1& 0& 0\\ 0& 3& 0\\ 0& 4& 5\end{array}\right]$,
$Tr\left(AB\right)=\lambda$ Tr ($A$).Tr ($B$), then $\lambda =$

• A. $1$
• B. $0$
• C. $\frac{6}{5}$
• D. $\frac{20}{27}$

Answer 1

Answer: (D)

$\frac{20}{27}$

Given ,$A=\left[\begin{array}{ccc}1& 1& 3\\ 4& 5& 6\\ 7& 1& 0\end{array}\right]$

$Tr\left(A\right)=1+5+0=6$
$B=\left[\begin{array}{ccc}1& 0& 0\\ 0& 3& 0\\ 0& 4& 5\end{array}\right]$

$Tr\left(B\right)=1+3+5=9$
Now,$AB=\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 1& 0\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& 3& 0\\ 0& 4& 5\end{array}\right]$
$\Rightarrow AB=\left[\begin{array}{ccc}1& 18& 15\\ 4& 39& 30\\ 7& 3& 0\end{array}\right]$
$Tr\left(AB\right)=1+39+0=40$
Also, given $Tr\left(AB\right)=\lambda Tr\left(A\right).Tr\left(B\right)$

$\Rightarrow \lambda =\frac{40}{54}=\frac{20}{27}$