With frame
| \[\tilde A = \left(\genfrac{}{}{0pt}{}{a}{c} \genfrac{}{}{0pt}{}{b}{d} \right) \qquad \tilde{A}^{-1} = \frac{1}{ad-bc}\left({\genfrac{}{}{0pt}{}{d}{-b}}{\genfrac{}{}{0pt}{}{-c}{a}}\right)^\top=\frac{1}{ad-bc}\left({\genfrac{}{}{0pt}{}{d}{-c}}{\genfrac{}{}{0pt}{}{-b}{a}}\right)\] |
\begin{eqnarray*}\tilde{A}&=&\left(\begin{array}{ccc}1&0&1\\ 2&1&0\\ 3&0&1\end{array}\right)\qquad\tilde{A}^{-1}=\mbox{ ?}\\ \det(\tilde{A})&=&\left|\begin{array}{ccc}1&0&1\\ 2&1&0\\ 3&0&1 \end{array}\right|=+\left|{\genfrac{}{}{0pt}{}{1}{3}}{\genfrac{}{}{0pt}{}{1}{1}}\right|=-2\end{eqnarray*}
| \[\begin{array}{lllll} A_{11}=+ \left| {\genfrac{}{}{0pt}{}{1}{0}} {\genfrac{}{}{0pt}{}{0}{1}}\right| =1 &\qquad& A_{12}=- \left| { \genfrac{}{}{0pt}{}{2}{3}} {\genfrac{}{}{0pt}{}{0}{1} }\right| =-2 &\qquad& A_{13}=+ \left| {\genfrac{}{}{0pt}{}{2}{3}} {\genfrac{}{}{0pt}{}{1}{0}}\right| =-3\\ \\ A_{21}=- \left| {\genfrac{}{}{0pt}{}{0}{0}} {\genfrac{}{}{0pt}{}{1}{1}}\right| =0 &\qquad& A_{22}=+ \left| { \genfrac{}{}{0pt}{}{1}{3}} {\genfrac{}{}{0pt}{}{1}{1}}\right| =-2 &\qquad& A_{23}=- \left| {\genfrac{}{}{0pt}{}{1}{3}} {\genfrac{}{}{0pt}{}{0}{0} }\right| =0\\ \\ A_{31}=+ \left| {\genfrac{}{}{0pt}{}{0}{1}} {\genfrac{}{}{0pt}{}{1}{0} }\right| =-1 &\qquad&A_{32}=- \left| {\genfrac{}{}{0pt}{}{1}{2}} {\genfrac{}{}{0pt}{}{1}{0}}\right| =+2 &\qquad& A_{33}=+ \left| {\genfrac{}{}{0pt}{}{1}{2}} {\genfrac{}{}{0pt}{}{0}{1}}\right| =1 \end{array}\] |
| \[\Rightarrow\;\;\tilde{A}^{-1}=\frac{1}{-2}\left(\begin{array}{ccc}+1&-2&-3\\ 0&-2&0\\ -1&+2&1\end{array}\right)^\top=\left(\begin{array}{ccc}-\frac{1}{2}&0&\frac{1}{2}\\ +1&1&-1\\ \frac{3}{2}&0&-\frac{1}{2}\end{array}\right)\] |
Test:
| \[\tilde{A}\tilde{A}^{-1}=\left(\begin{array}{ccc}1&0&0\\ 0&1&0\\ 0&0&1\end{array}\right)\;\;\mbox{(o.k.)}\] |
If \(\tilde A\) is an \(N \times N\) matrix with \(\det(\tilde A)\neq0\) then there exists an inverse with \(\tilde{A}\tilde{A}^{-1}=\tilde{A}^{-1}\tilde{A}=\tilde{I}\) and vice versa. If \(\tilde{A}^{-1}\) exists, then \(\det(\tilde A)\neq0\). Thus \(\det(\tilde{A}^{-1})=\frac{1}{\det(\tilde{A})}\) because \(\det(\tilde{A} \tilde{A}^{-1})=1=\det(\tilde I)\)
Regular matrix, inverse matrix
© J. Carstensen (Math for MS)