Question Number 141193 by Eric002 last updated on 16/May/21
$${A}=\begin{bmatrix}{\mathrm{2}\:\:}&{\mathrm{1}}&{−\mathrm{1}}\\{−\mathrm{3}}&{−\mathrm{1}}&{\mathrm{2}}\\{−\mathrm{2}}&{\mathrm{1}}&{\mathrm{2}}\end{bmatrix}{find}\:{the}\:{inverse}\:{of}\:\:{this}\:{matrix} \\ $$$$ \\ $$
Answered by bramlexs22 last updated on 16/May/21
$$\:{Cayley}−{Hamilton}\:{theorem} \\ $$$$\:\mid{A}−\lambda{I}\mid\:=\:\mathrm{0} \\ $$$$\Rightarrow\lambda^{\mathrm{3}} −\left({tr}\:{A}\right)\lambda^{\mathrm{2}} +\begin{pmatrix}{{minor}\:{of}\:{the}\:{terms}}\\{{on}\:{the}\:{leading}\:{diagonal}\:{of}\:{A}}\end{pmatrix}\:\lambda−{det}\left({A}\right)=\mathrm{0} \\ $$$$\Rightarrow\lambda^{\mathrm{3}} −\mathrm{3}\lambda^{\mathrm{2}} −\lambda−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{p}\left({A}\right)=\:\mathrm{0} \\ $$$$\Rightarrow\:\left[{A}^{\mathrm{3}} −\mathrm{3}{A}^{\mathrm{2}} −{A}−{I}=\:\mathrm{0}\:\right]×{A}^{−\mathrm{1}} \\ $$$$\Rightarrow{A}^{\mathrm{2}} −\mathrm{3}{A}−{I}−{A}^{−\mathrm{1}} \:=\:\mathrm{0} \\ $$$$\Rightarrow{A}^{−\mathrm{1}} \:=\:{A}^{\mathrm{2}} −\mathrm{3}{A}−{I}\: \\ $$$$ \\ $$
Commented by Eric002 last updated on 16/May/21
$${well}\:{done}\:{sir} \\ $$