## Elementary Linear Algebra 7th Edition

First method: $$A^{-2}=(A^2)^{-1}=\left[ \begin {array}{ccc} {\frac {873}{4}}&57&-401 \\ {\frac {61}{4}}&4&-28\\ -{ \frac {1317}{4}}&-86&605\end {array} \right] .$$ Second method; $$A^{-2}=(A^{-1})^{2}=\left[ \begin {array}{ccc} {\frac {873}{4}}&57&-401 \\ {\frac {61}{4}}&4&-28\\ -{ \frac {1317}{4}}&-86&605\end {array} \right] .$$ One can see that the two methods are equivalent
First method; we find $A^2$ which is given by $$A^{2}=\left[ \begin {array}{ccc} 48&4&32\\ -29&48&-17 \\ 22&9&15\end {array} \right] .$$ By calculating the inverse of $A^2$, we have $$A^{-2}=(A^2)^{-1}=\left[ \begin {array}{ccc} {\frac {873}{4}}&57&-401 \\ {\frac {61}{4}}&4&-28\\ -{ \frac {1317}{4}}&-86&605\end {array} \right] .$$ Second method; we find $A^{-1}$ as follows $$A^{-1}= \left[ \begin {array}{ccc} -\frac{15}{2}&-2&14\\ -\frac{1}{2}&0&1 \\ \frac{23}{2}&3&-21\end {array} \right] .$$ Now, we have $$A^{-2}=(A^{-1})^{2}=\left[ \begin {array}{ccc} {\frac {873}{4}}&57&-401 \\ {\frac {61}{4}}&4&-28\\ -{ \frac {1317}{4}}&-86&605\end {array} \right] .$$ One can see that the two methods are equivalent.