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Question Number 34305 by abdo mathsup 649 cc last updated on 03/May/18

let] A = (((1 1 0)),((1 1 1)) ) (0 1 1 1)find the caractetistic polynom of A 2) calculate A^n

$$\left.{let}\right]\:{A}\:=\begin{pmatrix}{\mathrm{1}\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{1}\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{0}\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\mathrm{1}\:\right. \\ $$$$\left.\mathrm{1}\right){find}\:{the}\:{caractetistic}\:{polynom}\:{of}\:{A} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{A}^{{n}} \\ $$

Commented by prof Abdo imad last updated on 06/May/18

p_c (A)=det(A −XI)= determinant (((1−X 1 0 )),((1 1−X 1))) ∣ 0 1−X 1−X∣∣ =(1−X)((1−X)^2 −1+X) −1(1−X) +0 =(1−X)(X^2 −2X +1−1+X) −1 +X =(1−X)(X^2 −X) +X−1 =−(X−1)^2 X +X −1 =(X−1)( 1−X(X−1)) =(X−1)( −X^2 +X +1)=(1−X)(X^2 −X−1) p_c (A)=0 ⇔ X =1 or X^2 −X −1 =0 Δ= 1−4(−1)=5 ⇒ X_1 = ((1+(√5))/2) X_2 =((1−(√5))/2) .the proper value are λ_1 =1 λ_2 =((1+(√5))/2) and λ_3 = ((1−(√5))/2) v(λ_1 )=ker( A −I) ={ u/(A−I)u =0 } be continued.....

$${p}_{{c}} \left({A}\right)={det}\left({A}\:−{XI}\right)=\begin{vmatrix}{\mathrm{1}−{X}\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\mathrm{0}\:}\\{\mathrm{1}\:\:\:\:\:\:\:\mathrm{1}−{X}\:\:\:\:\:\mathrm{1}}\end{vmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\mathrm{0}\:\:\:\:\mathrm{1}−{X}\:\:\:\mathrm{1}−{X}\mid\mid \\ $$$$=\left(\mathrm{1}−{X}\right)\left(\left(\mathrm{1}−{X}\right)^{\mathrm{2}} \:−\mathrm{1}+{X}\right)\:−\mathrm{1}\left(\mathrm{1}−{X}\right)\:+\mathrm{0} \\ $$$$=\left(\mathrm{1}−{X}\right)\left({X}^{\mathrm{2}} −\mathrm{2}{X}\:+\mathrm{1}−\mathrm{1}+{X}\right)\:−\mathrm{1}\:+{X} \\ $$$$=\left(\mathrm{1}−{X}\right)\left({X}^{\mathrm{2}} \:−{X}\right)\:\:+{X}−\mathrm{1} \\ $$$$=−\left({X}−\mathrm{1}\right)^{\mathrm{2}} {X}\:\:+{X}\:−\mathrm{1}\: \\ $$$$=\left({X}−\mathrm{1}\right)\left(\:\mathrm{1}−{X}\left({X}−\mathrm{1}\right)\right) \\ $$$$=\left({X}−\mathrm{1}\right)\left(\:−{X}^{\mathrm{2}} \:\:+{X}\:+\mathrm{1}\right)=\left(\mathrm{1}−{X}\right)\left({X}^{\mathrm{2}} \:−{X}−\mathrm{1}\right) \\ $$$${p}_{{c}} \left({A}\right)=\mathrm{0}\:\Leftrightarrow\:{X}\:=\mathrm{1}\:{or}\:{X}^{\mathrm{2}} \:−{X}\:−\mathrm{1}\:=\mathrm{0} \\ $$$$\Delta=\:\mathrm{1}−\mathrm{4}\left(−\mathrm{1}\right)=\mathrm{5}\:\Rightarrow\:{X}_{\mathrm{1}} =\:\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${X}_{\mathrm{2}} \:=\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:.{the}\:{proper}\:{value}\:{are}\:\lambda_{\mathrm{1}} =\mathrm{1} \\ $$$$\lambda_{\mathrm{2}} =\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:{and}\:\lambda_{\mathrm{3}} =\:\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${v}\left(\lambda_{\mathrm{1}} \right)={ker}\left(\:{A}\:−{I}\right)\:=\left\{\:{u}/\left({A}−{I}\right){u}\:=\mathrm{0}\:\right\} \\ $$$${be}\:{continued}.....\:\:\:\:\:\:\:\:\:\:\:\: \\ $$

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