a
(a) ${A^{ - 1}} = \frac{{adj\,(A)}}{{|A|}} = \frac{1}{{|A|}}.\,adj\,(A)$
$A = \left[ {\begin{array}{*{20}{c}}
0&1&2 \\
1&2&3 \\
3&1&1\end{array}} \right]$; $|A| = 0-1(1-9)+2(1-6)=8-10$
$|A|= -2 \ne 0$
$Adj\,\,A = \left[ {\begin{array}{*{20}{c}}{{A_{11}}}&{{A_{21}}}&{{A_{31}}}\\{{A_{12}}}&{{A_{22}}}&{{A_{32}}}\\{{A_{13}}}&{{A_{23}}}&{{A_{33}}}\end{array}} \right]$
${A_{11}} = {( - 1)^{1 + 1}}[(2)\,(1) - (3)(1)] = - 1$
${A_{12}} = 8,$ $\frac{1}{{1 - a}} - 1 + \frac{1}{{1 - b}} + \frac{1}{{1 - c}} = 0$,
${A_{21}} = 1$, ${A_{22}} = - 6$
${A_{23}} = 3$,${A_{31}} = - 1$, ${A_{32}} = 2$, ${A_{33}} = - 1$
$\therefore $ ${A^{ - 1}} = \frac{1}{{ - 2}}\,\left[ {\begin{array}{*{20}{c}}{ - 1}&1&{ - 1}\\8&{ - 6}&2\\{ - 5}&3&{ - 1}\end{array}} \right]$$ = \left[ {\begin{array}{*{20}{c}}{1/2}&{ - 1/2}&{1/2}\\{ - 4}&3&{ - 1}\\{5/2}&{ - 3/2}&{1/2}\end{array}} \right]$.