a
(a) Given, $\left( {\begin{array}{*{20}{c}}4&2&2\\{ - 5}&0&\alpha \\1&{ - 2}&3\end{array}} \right)\, = \,10{A^{ - 1}}$
$ \Rightarrow $ $\left( {\begin{array}{*{20}{c}}4&2&2\\{ - 5}&0&\alpha \\1&{ - 2}&3\end{array}} \right)\,\,\left( {\begin{array}{*{20}{c}}1&{ - 1}&1\\2&1&{ - 3}\\1&1&1\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{10}&0&0\\0&{10}&0\\0&0&{10}\end{array}} \right)$
$ \Rightarrow $ $ - 5 + \alpha = 0 \Rightarrow \alpha = 5$
(Equating the element of $ 2nd$ row and first column).