a
Let $A=\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5\end{array}\right]$
We have,
$|A|=1(10-0)-2(0-0)+3(0-0)=10$
Now,
$A_{11}=10-0, A_{12}=-(0-0)=0, A_{13}=0-0=0$
$A_{21}=-(10-0)=-10, A_{22}=5-0=5, A_{23}=-(0-0)=0$
$A_{31}=8-6=2, A_{32}=-(4-0)=-4, A_{33}=2-0=2$
$\therefore a d j A=\left[\begin{array}{ccc}10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2\end{array}\right]$
$\therefore A^{-1}=\frac{1}{|A|} a d j A=\frac{1}{10}\left[\begin{array}{ccc}10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2\end{array}\right]$