Question

Let $A$ and $B$ are two matrices of same order $3\times 3$ given by $A=\left[\begin{array}{ccc}1& 3& \lambda +2\\ 2& 4& 6\\ 3& 5& 8\end{array}\right]$ $B=\left[\begin{array}{ccc}3& 2& 4\\ 3& 2& 5\\ 2& 1& 4\end{array}\right]$If $tr(AB{)}^t+tr(BA{)}^t=tr\left(AB\right)$ then the value of $2\lambda$ equals

• A. $103$
• B. $206$
• C. $-103$
• D. $-206$

Answer 1

The correct option is C $-103$

$A=\left[\begin{array}{ccc}1& 3& \lambda +2\\ 2& 4& 6\\ 3& 5& 8\end{array}\right],B=\left[\begin{array}{ccc}3& 2& 4\\ 3& 2& 5\\ 2& 1& 4\end{array}\right]$
$AB=\left[\begin{array}{ccc}1& 3& \lambda +2\\ 2& 4& 6\\ 3& 5& 8\end{array}\right]\left[\begin{array}{ccc}3& 2& 4\\ 3& 2& 5\\ 2& 1& 4\end{array}\right]$
$=\left[\begin{array}{ccc}16+2\lambda & \lambda +8& 4\lambda +27\\ 30& 18& 52\\ 40& 24& 69\end{array}\right]$
$\Rightarrow tr\left(AB\right)=2\lambda +16+18+69=2\lambda +103$
Now, given $tr(AB{)}^t+tr(BA{)}^t=tr\left(AB\right)$
$\Rightarrow tr\left(AB\right)+tr\left(BA\right)=tr\left(AB\right)(\because tr(A)=tr(A^T),tr(AB)=tr(BA\left)\right)$
$\Rightarrow tr\left(AB\right)=0$

$\Rightarrow 2\lambda =-103$