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Question Number 141193 by Eric002 last updated on 16/May/21
![A= [((2 ),1,(−1)),((−3),(−1),2),((−2),1,2) ]find the inverse of this matrix](Q141193.png)
$${A}=\begin{bmatrix}{\mathrm{2}\:\:}&{\mathrm{1}}&{−\mathrm{1}}\\{−\mathrm{3}}&{−\mathrm{1}}&{\mathrm{2}}\\{−\mathrm{2}}&{\mathrm{1}}&{\mathrm{2}}\end{bmatrix}{find}\:{the}\:{inverse}\:{of}\:\:{this}\:{matrix} \\ $$$$ \\ $$
Answered by bramlexs22 last updated on 16/May/21
![Cayley−Hamilton theorem ∣A−λI∣ = 0 ⇒λ^3 −(tr A)λ^2 + (((minor of the terms)),((on the leading diagonal of A)) ) λ−det(A)=0 ⇒λ^3 −3λ^2 −λ−1=0 ⇒p(A)= 0 ⇒ [A^3 −3A^2 −A−I= 0 ]×A^(−1) ⇒A^2 −3A−I−A^(−1) = 0 ⇒A^(−1) = A^2 −3A−I](Q141194.png)
$$\:{Cayley}−{Hamilton}\:{theorem} \\ $$$$\:\mid{A}−\lambda{I}\mid\:=\:\mathrm{0} \\ $$$$\Rightarrow\lambda^{\mathrm{3}} −\left({tr}\:{A}\right)\lambda^{\mathrm{2}} +\begin{pmatrix}{{minor}\:{of}\:{the}\:{terms}}\\{{on}\:{the}\:{leading}\:{diagonal}\:{of}\:{A}}\end{pmatrix}\:\lambda−{det}\left({A}\right)=\mathrm{0} \\ $$$$\Rightarrow\lambda^{\mathrm{3}} −\mathrm{3}\lambda^{\mathrm{2}} −\lambda−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{p}\left({A}\right)=\:\mathrm{0} \\ $$$$\Rightarrow\:\left[{A}^{\mathrm{3}} −\mathrm{3}{A}^{\mathrm{2}} −{A}−{I}=\:\mathrm{0}\:\right]×{A}^{−\mathrm{1}} \\ $$$$\Rightarrow{A}^{\mathrm{2}} −\mathrm{3}{A}−{I}−{A}^{−\mathrm{1}} \:=\:\mathrm{0} \\ $$$$\Rightarrow{A}^{−\mathrm{1}} \:=\:{A}^{\mathrm{2}} −\mathrm{3}{A}−{I}\: \\ $$$$ \\ $$
Commented by Eric002 last updated on 16/May/21
$${well}\:{done}\:{sir} \\ $$