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Question Number 98661 by bemath last updated on 15/Jun/20
![using cayley − hamilton theorem what is the inverse of matrix A= [((0 1 −1)),((1 2 2)),((0 1 −1)) ]](Q98661.png)
$$\mathrm{using}\:\mathrm{cayley}\:−\:\mathrm{hamilton} \\ $$$$\mathrm{theorem}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{inverse}\:\mathrm{of} \\ $$$$\mathrm{matrix}\:\mathrm{A}=\:\begin{bmatrix}{\mathrm{0}\:\:\:\:\mathrm{1}\:\:\:−\mathrm{1}}\\{\mathrm{1}\:\:\:\:\mathrm{2}\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:\:\:\mathrm{1}\:\:\:−\mathrm{1}}\end{bmatrix}\: \\ $$
Commented by john santu last updated on 15/Jun/20
![we first compute the characteristic equation ∣A−λI∣=0 [((−λ 1 −1 )),(( 1 2−λ 2)),(( 0 1 −1−λ)) ]= 0 = −λ{(2−λ)(−1−λ)−2}−(−1−λ)−1 = λ{λ^2 −λ−4}+1+λ−1 = λ^3 −λ^2 −3λ the cayley−hamilton theorem states that a matrix satisfies its own characteristic equation ⇒A^3 −A^2 −3A = 0 A(A^2 −A−3I) = 0 ⇔I = (1/3)A(A−I) ⇔A^(−1) = (1/3)(A−I) ⇔A^(−1) = (1/3) [((−1 1 −1)),(( 1 1 2)),(( 0 1 −2)) ]](Q98666.png)
$$\mathrm{we}\:\mathrm{first}\:\mathrm{compute}\:\mathrm{the}\:\mathrm{characteristic} \\ $$$$\mathrm{equation}\:\mid\mathrm{A}−\lambda\mathrm{I}\mid=\mathrm{0} \\ $$$$\begin{bmatrix}{−\lambda\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:−\mathrm{1}\:}\\{\:\:\:\mathrm{1}\:\:\:\:\:\:\mathrm{2}−\lambda\:\:\:\:\:\:\:\:\:\:\mathrm{2}}\\{\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:−\mathrm{1}−\lambda}\end{bmatrix}=\:\mathrm{0} \\ $$$$=\:−\lambda\left\{\left(\mathrm{2}−\lambda\right)\left(−\mathrm{1}−\lambda\right)−\mathrm{2}\right\}−\left(−\mathrm{1}−\lambda\right)−\mathrm{1} \\ $$$$=\:\lambda\left\{\lambda^{\mathrm{2}} −\lambda−\mathrm{4}\right\}+\mathrm{1}+\lambda−\mathrm{1} \\ $$$$=\:\lambda^{\mathrm{3}} −\lambda^{\mathrm{2}} −\mathrm{3}\lambda\: \\ $$$$\mathrm{the}\:\mathrm{cayley}−\mathrm{hamilton}\:\mathrm{theorem} \\ $$$$\mathrm{states}\:\mathrm{that}\:\mathrm{a}\:\mathrm{matrix}\:\mathrm{satisfies}\:\mathrm{its} \\ $$$$\mathrm{own}\:\mathrm{characteristic}\:\mathrm{equation}\: \\ $$$$\Rightarrow\mathrm{A}^{\mathrm{3}} −\mathrm{A}^{\mathrm{2}} −\mathrm{3A}\:=\:\mathrm{0} \\ $$$$\mathrm{A}\left(\mathrm{A}^{\mathrm{2}} −\mathrm{A}−\mathrm{3I}\right)\:=\:\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{I}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\mathrm{A}\left(\mathrm{A}−\mathrm{I}\right) \\ $$$$\Leftrightarrow\mathrm{A}^{−\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{A}−\mathrm{I}\right)\: \\ $$$$\Leftrightarrow\mathrm{A}^{−\mathrm{1}} =\:\frac{\mathrm{1}}{\mathrm{3}}\:\begin{bmatrix}{−\mathrm{1}\:\:\:\:\mathrm{1}\:\:\:\:\:−\mathrm{1}}\\{\:\:\:\:\mathrm{1}\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\mathrm{2}}\\{\:\:\:\:\mathrm{0}\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:−\mathrm{2}}\end{bmatrix} \\ $$