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If-A-4-3-1-0-use-the-fact-that-A-2-4A-3I-2-and-mathematical-induction-to-prove-A-n-3-n-1-2-A-3-3-n-2-I-if-n-1-

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Question Number 40352 by scientist last updated on 20/Jul/18
If A= [((4 −3)),((1 0)) ]use the fact that A^2 =4A−3I_2 and mathematical induction to prove A^n =(((3^n −1))/2)A +((3−3^n )/2)I if n≥1
$${If}\:{A}=\begin{bmatrix}{\mathrm{4}\:\:\:\:\:−\mathrm{3}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\end{bmatrix}{use}\:{the}\:{fact}\:{that}\: \\ $$$${A}^{\mathrm{2}} =\mathrm{4}{A}−\mathrm{3}{I}_{\mathrm{2}} \:\:{and}\:{mathematical}\:{induction}\:{to}\:{prove} \\ $$$$\:\:{A}^{{n}} =\frac{\left(\mathrm{3}^{{n}} −\mathrm{1}\right)}{\mathrm{2}}{A}\:\:+\frac{\mathrm{3}−\mathrm{3}^{{n}} }{\mathrm{2}}{I}\:\:{if}\:{n}\geqslant\mathrm{1} \\ $$
Commented by prof Abdo imad last updated on 20/Jul/18
the caracteristic polynom of A is p_c (A)=det(A−xI)= determinant (((4−x −3)),((1 −x))) =−x(4−x)+3 =x^2 −4x +3 kayley hamilton theorem give A^2 −4A +3I =0 ⇒A^2 =4A−3I ⇒ let prove by recurrence that A^(2n) =((3^(2n) −1)/2) A +((3−3^(2n) )/2) I the equality is true for n=0 A^(2n+2) =A^2 {((3^(2n) −1)/2) A +((3−3^(2n) )/2) I} =(4A−3I){((3^(2n) −1)/2) A +((3−3^(2n) )/2) I} =2(3^(2n) −1)A^2 +2(3−3^(2n) )A −3((3^(2n) −1)/2) A −3((3−3^(2n) )/2)I =2(3^(2n) −1)( 4A−3I) + ((4(3−3^(2n) )−33^(2n) +3)/2) A −3((3−3^(2n) )/2) I =8(3^(2n) −1)A −6(3^(2n) −1)I + ((15−7 3^(2n) )/2) A −3 ((3−3^(2n) )/2) I = ((16(3^(2n) −1)+15−73^(2n) )/2) A −((12(3^(2n) −1)+3(3−3^(2n) ))/2) A =((9.3^(2n) −1)/2) A −((9 3^(2n) −3)/2) I =((3^(2n+2) −1)/2) A+((3−3^(2n+2) )/2) I the relation is true at term (n+1) also we must prove by recurrence that A^(2n+1) =((3^(2n+1) −1)/2) A +((3−3^(2n+1) )/2) I .
$${the}\:{caracteristic}\:{polynom}\:{of}\:{A}\:{is} \\ $$$${p}_{{c}} \left({A}\right)={det}\left({A}−{xI}\right)=\begin{vmatrix}{\mathrm{4}−{x}\:\:\:\:\:\:−\mathrm{3}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:−{x}}\end{vmatrix} \\ $$$$=−{x}\left(\mathrm{4}−{x}\right)+\mathrm{3}\:={x}^{\mathrm{2}} \:−\mathrm{4}{x}\:+\mathrm{3} \\ $$$${kayley}\:{hamilton}\:{theorem}\:{give}\: \\ $$$${A}^{\mathrm{2}} −\mathrm{4}{A}\:+\mathrm{3}{I}\:=\mathrm{0}\:\Rightarrow{A}^{\mathrm{2}} \:=\mathrm{4}{A}−\mathrm{3}{I}\:\:\Rightarrow \\ $$$${let}\:{prove}\:{by}\:{recurrence}\:{that} \\ $$$${A}^{\mathrm{2}{n}} \:=\frac{\mathrm{3}^{\mathrm{2}{n}} −\mathrm{1}}{\mathrm{2}}\:{A}\:+\frac{\mathrm{3}−\mathrm{3}^{\mathrm{2}{n}} }{\mathrm{2}}\:{I}\:\: \\ $$$${the}\:{equality}\:{is}\:{true}\:{for}\:{n}=\mathrm{0} \\ $$$${A}^{\mathrm{2}{n}+\mathrm{2}} ={A}^{\mathrm{2}} \:\left\{\frac{\mathrm{3}^{\mathrm{2}{n}} −\mathrm{1}}{\mathrm{2}}\:{A}\:+\frac{\mathrm{3}−\mathrm{3}^{\mathrm{2}{n}} }{\mathrm{2}}\:{I}\right\} \\ $$$$=\left(\mathrm{4}{A}−\mathrm{3}{I}\right)\left\{\frac{\mathrm{3}^{\mathrm{2}{n}} −\mathrm{1}}{\mathrm{2}}\:{A}\:+\frac{\mathrm{3}−\mathrm{3}^{\mathrm{2}{n}} }{\mathrm{2}}\:{I}\right\} \\ $$$$=\mathrm{2}\left(\mathrm{3}^{\mathrm{2}{n}} −\mathrm{1}\right){A}^{\mathrm{2}} \:+\mathrm{2}\left(\mathrm{3}−\mathrm{3}^{\mathrm{2}{n}} \right){A}\:−\mathrm{3}\frac{\mathrm{3}^{\mathrm{2}{n}} −\mathrm{1}}{\mathrm{2}}\:{A} \\ $$$$−\mathrm{3}\frac{\mathrm{3}−\mathrm{3}^{\mathrm{2}{n}} }{\mathrm{2}}{I}\: \\ $$$$=\mathrm{2}\left(\mathrm{3}^{\mathrm{2}{n}} −\mathrm{1}\right)\left(\:\mathrm{4}{A}−\mathrm{3}{I}\right)\:+\:\frac{\mathrm{4}\left(\mathrm{3}−\mathrm{3}^{\mathrm{2}{n}} \right)−\mathrm{33}^{\mathrm{2}{n}} \:+\mathrm{3}}{\mathrm{2}}\:{A} \\ $$$$−\mathrm{3}\frac{\mathrm{3}−\mathrm{3}^{\mathrm{2}{n}} }{\mathrm{2}}\:{I} \\ $$$$=\mathrm{8}\left(\mathrm{3}^{\mathrm{2}{n}} −\mathrm{1}\right){A}\:−\mathrm{6}\left(\mathrm{3}^{\mathrm{2}{n}} −\mathrm{1}\right){I}\:\:+\:\frac{\mathrm{15}−\mathrm{7}\:\mathrm{3}^{\mathrm{2}{n}} }{\mathrm{2}}\:{A} \\ $$$$−\mathrm{3}\:\frac{\mathrm{3}−\mathrm{3}^{\mathrm{2}{n}} }{\mathrm{2}}\:{I} \\ $$$$=\:\frac{\mathrm{16}\left(\mathrm{3}^{\mathrm{2}{n}} −\mathrm{1}\right)+\mathrm{15}−\mathrm{73}^{\mathrm{2}{n}} }{\mathrm{2}}\:{A}\:\:−\frac{\mathrm{12}\left(\mathrm{3}^{\mathrm{2}{n}} −\mathrm{1}\right)+\mathrm{3}\left(\mathrm{3}−\mathrm{3}^{\mathrm{2}{n}} \right)}{\mathrm{2}}\:{A} \\ $$$$=\frac{\mathrm{9}.\mathrm{3}^{\mathrm{2}{n}} \:−\mathrm{1}}{\mathrm{2}}\:{A}\:−\frac{\mathrm{9}\:\mathrm{3}^{\mathrm{2}{n}} \:−\mathrm{3}}{\mathrm{2}}\:{I} \\ $$$$=\frac{\mathrm{3}^{\mathrm{2}{n}+\mathrm{2}} −\mathrm{1}}{\mathrm{2}}\:{A}+\frac{\mathrm{3}−\mathrm{3}^{\mathrm{2}{n}+\mathrm{2}} }{\mathrm{2}}\:{I}\:\:{the}\:{relation}\:{is}\:{true} \\ $$$${at}\:{term}\:\left({n}+\mathrm{1}\right)\:{also}\:{we}\:{must}\:{prove}\:{by}\:{recurrence} \\ $$$${that} \\ $$$${A}^{\mathrm{2}{n}+\mathrm{1}} \:\:=\frac{\mathrm{3}^{\mathrm{2}{n}+\mathrm{1}} \:−\mathrm{1}}{\mathrm{2}}\:{A}\:\:+\frac{\mathrm{3}−\mathrm{3}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}}\:{I}\:. \\ $$

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