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Question Number 84068 by Power last updated on 09/Mar/20

Commented by Power last updated on 09/Mar/20

prove it please

$$\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{it}}\:\boldsymbol{\mathrm{please}} \\ $$

Commented by Power last updated on 09/Mar/20

please see

$$\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{see}} \\ $$

Commented by TANMAY PANACEA last updated on 09/Mar/20

Commented by arcana last updated on 20/Mar/20

para el caso n=4 determinant (((1+a_1 ),1,1,1),(1,(1+a_2 ),1,1),(1,1,(1+a_3 ),1),(1,1,1,(1+a_4 )))= determinant ((a_1 ,0,0,(−a_4 )),(0,a_2 ,0,(−a_4 )),(0,0,a_3 ,(−a_4 )),(1,1,1,(1+a_4 ))) =− determinant ((0,0,(−a_4 )),(a_2 ,0,(−a_4 )),(0,a_3 ,(−a_4 )))+ determinant ((a_1 ,0,(−a_4 )),(0,0,(−a_4 )),(0,a_3 ,(−a_4 )))− determinant ((a_1 ,0,(−a_4 )),(0,a_2 ,(−a_4 )),(0,0,(−a_4 )))+a_1 a_2 a_3 (1+a_4 ) =a_2 a_3 a_4 +a_1 a_3 a_4 +a_1 a_2 a_4 +a_1 a_2 a_3 +a_1 a_2 a_3 a_4 =a_1 a_2 a_3 a_4 (1+(1/a_1 )+(1/a_2 )+(1/a_3 )+(1/a_4 )) para el caso general puede usarse induccion matematica Algebra linear − Friedberg

$$\mathrm{para}\:\mathrm{el}\:\mathrm{caso}\:{n}=\mathrm{4} \\ $$$$ \\ $$$$\begin{vmatrix}{\mathrm{1}+{a}_{\mathrm{1}} }&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}+{a}_{\mathrm{2}} }&{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}+{a}_{\mathrm{3}} }&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}+{a}_{\mathrm{4}} }\end{vmatrix}=\begin{vmatrix}{{a}_{\mathrm{1}} }&{\mathrm{0}}&{\mathrm{0}}&{−{a}_{\mathrm{4}} }\\{\mathrm{0}}&{{a}_{\mathrm{2}} }&{\mathrm{0}}&{−{a}_{\mathrm{4}} }\\{\mathrm{0}}&{\mathrm{0}}&{{a}_{\mathrm{3}} }&{−{a}_{\mathrm{4}} }\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}+{a}_{\mathrm{4}} }\end{vmatrix} \\ $$$$=−\begin{vmatrix}{\mathrm{0}}&{\mathrm{0}}&{−{a}_{\mathrm{4}} }\\{{a}_{\mathrm{2}} }&{\mathrm{0}}&{−{a}_{\mathrm{4}} }\\{\mathrm{0}}&{{a}_{\mathrm{3}} }&{−{a}_{\mathrm{4}} }\end{vmatrix}+\begin{vmatrix}{{a}_{\mathrm{1}} }&{\mathrm{0}}&{−{a}_{\mathrm{4}} }\\{\mathrm{0}}&{\mathrm{0}}&{−{a}_{\mathrm{4}} }\\{\mathrm{0}}&{{a}_{\mathrm{3}} }&{−{a}_{\mathrm{4}} }\end{vmatrix}−\begin{vmatrix}{{a}_{\mathrm{1}} }&{\mathrm{0}}&{−{a}_{\mathrm{4}} }\\{\mathrm{0}}&{{a}_{\mathrm{2}} }&{−{a}_{\mathrm{4}} }\\{\mathrm{0}}&{\mathrm{0}}&{−{a}_{\mathrm{4}} }\end{vmatrix}+{a}_{\mathrm{1}} {a}_{\mathrm{2}} {a}_{\mathrm{3}} \left(\mathrm{1}+{a}_{\mathrm{4}} \right) \\ $$$$={a}_{\mathrm{2}} {a}_{\mathrm{3}} {a}_{\mathrm{4}} +{a}_{\mathrm{1}} {a}_{\mathrm{3}} {a}_{\mathrm{4}} +{a}_{\mathrm{1}} {a}_{\mathrm{2}} {a}_{\mathrm{4}} +{a}_{\mathrm{1}} {a}_{\mathrm{2}} {a}_{\mathrm{3}} +{a}_{\mathrm{1}} {a}_{\mathrm{2}} {a}_{\mathrm{3}} {a}_{\mathrm{4}} \\ $$$$={a}_{\mathrm{1}} {a}_{\mathrm{2}} {a}_{\mathrm{3}} {a}_{\mathrm{4}} \left(\mathrm{1}+\frac{\mathrm{1}}{{a}_{\mathrm{1}} }+\frac{\mathrm{1}}{{a}_{\mathrm{2}} }+\frac{\mathrm{1}}{{a}_{\mathrm{3}} }+\frac{\mathrm{1}}{{a}_{\mathrm{4}} }\right) \\ $$$$\mathrm{para}\:\mathrm{el}\:\mathrm{caso}\:\mathrm{general}\:\mathrm{puede}\:\mathrm{usarse}\:\mathrm{induccion} \\ $$$$\mathrm{matematica} \\ $$$$\mathrm{A}{lgebra}\:{linear}\:−\:\boldsymbol{\mathrm{Friedberg}} \\ $$

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