Answer
$|A^{T}|=14$ , $|A^{2}|=196$ , $|A A^{T}|=196$ , $|2 A|=56$ and $|A^{-1}|=1/14$
Work Step by Step
Given the matrix
$A=\left[\begin{array}{ll}6 & -11 \\ 4 & -5\end{array}\right]$
We have
$A^{T}=\left[\begin{array}{cc}6 & 4 \\ -11 & -5\end{array}\right], A^{2}=\left[\begin{array}{cc}-8 & -11 \\ 4 & -19\end{array}\right]$ and $2 A=\left[\begin{array}{cc}12 & -22 \\ 8 & -10\end{array}\right]$
Also we have
$|A|=-30+44=14,\left|A^{T}\right|=-30+44=14$ and $\left|A^{2}\right|=8 *$ $19+11 * 4=196$
We obtain
$\quad\left|A A^{T}\right|=|A|\left|A^{T}\right|=14 * 14=196 \quad
and
\quad|2 A|=-12 * 10+8 *$
$22=56$
Since $|A|=14$, then the matrix A is non-singular and invertible. So, we have $A A^{-1} =I$ and then $|A A^{-1}| =1$
Since $|A A^{-1}|= |A| |A^{-1}|$, then $|A| |A^{-1}| =1$
Thus $\left|A^{-1}\right|=\frac{1}{|A|}=\frac{1}{14}$
