Answer
(a) $$(AB)^{-1}=\left[ \begin {array}{cc} {35}& {17}\\ {4} & {10} \end {array} \right] $$
(b) $$(A^T)^{-1}=\left[ \begin {array}{cc} {2}& {-7}\\ {5} & {6} \end {array} \right]$$
(c) $$(2A)^{-1}=\left[ \begin {array}{cc} {1}& {\frac{5}{2}}\\ {-\frac{7}{2}} & {3} \end {array} \right]$$
Work Step by Step
(a) $$(AB)^{-1}=B^{-1}A^{-1}=\left[ \begin {array}{cc} {7}& {-3}\\ {2} & {0} \end {array} \right]\left[ \begin {array}{cc} {2}& {5}\\ {-7} & {6} \end {array} \right] =\left[ \begin {array}{cc} {35}& {17}\\ {4} & {10} \end {array} \right] $$
(b) $$(A^T)^{-1}=(A^{-1})^T=\left[ \begin {array}{cc} {2}& {-7}\\ {5} & {6} \end {array} \right]$$
(c) $$(2A)^{-1}=\frac{1}{2}A^{-1}=\frac{1}{2}\left[ \begin {array}{cc} {2}& {5}\\ {-7} & {6} \end {array} \right]=\left[ \begin {array}{cc} {1}& {\frac{5}{2}}\\ {-\frac{7}{2}} & {3} \end {array} \right]$$
