If $A\cdot \left[\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\\ 3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}4\end{array}\right]=I$, then the trace of $A^T$ is 

Let $B=\left[\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\\ 3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}4\end{array}\right],A=\left[\begin{array}{l}a\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}b\\ c\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}d\end{array}\right]$

Now, $\left[\begin{array}{l}a\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}b\\ c\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}d\end{array}\right]\left[\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\\ 3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}4\end{array}\right]=\left[\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right]$

$\Rightarrow a+3b=1$                                                              $\text{ ...(i) }$

$2a+4b=0$                                                                $\text{ ...(ii) }$

$c+3d=0$                                                                  $\text{ ...(iii) }$

$2c+4d=1$                                                                $\dots \text{ (iv) }$

Solving $\text{ (i) and (ii), }$ we get $a=-2;b=1$

Solving $\text{ (iii) and (iv), }$we get $c=\frac{3}{2};d=\frac{-1}{2}$

Now, $A^T=\left[\begin{array}{cc}-2& 3/2\\ 1& -1/2\end{array}\right]$

So, $\mathrm{tr}\left(A^T\right)=\frac{-5}{2}$