Question

Seja A=...(IMAGEM do EXERCÍCIO) então 2detA é:

Answer 1

Resposta:

1

Explicação passo-a-passo:

$det(a) = \begin{vmatrix}</p><p> log_{3}(2) & log_{3}(4) & log_{3}(8) & log_{3}(2) & log_{3}(4) \\ log_{3}(16) & log_{3}(32) & log_{3}(64) & log_{3}(16) & log_{3}(32)\\ log_{3}(128) & log_{3}(256) & log_{3}(512) & log_{3}(128) & log_{3}(256)</p><p>\end{vmatrix} \\ det(a) = log_{3}(2) . log_{3}(32) . log_{3}(512) + log_{3}(4) . log_{3}(64) . log_{3}(128) + log_{3}(8) . log_{3}(16) . log_{3}(256) - log_{3}(128) . log_{3}(32) . log_{3}(8) - log_{3}(256) . log_{3}(64) . log_{3}(2) - log_{3}(512) . log_{3}(16) . log_{3}(4) \\ det(a) = log_{3}(2) . log_{3}( {2}^{5} ) . log_{3}( {2}^{9} ) + log_{3}( {2}^{2} ) . log_{3}( {2}^{6} ) . log_{3}( {2}^{7} ) + log_{3}( {2}^{3} ) . log_{3}( {2}^{4} ) . log_{3}( {2}^{8} ) - log_{3}( {2}^{7} ) . log_{3}( {2}^{5} ) . log_{3}( {2}^{3} ) - log_{3}( {2}^{8} ) . log_{3}( {2}^{6} ) . log_{3}(2) - log_{3}( {2}^{9} ) . log_{3}( {2}^{4} ) . log_{3}( {2}^{2} ) \\ det(a) = log_{3}(2) .5 log_{3}(2) .9 log_{3}(2) + 2 log_{3}(2) .6 log_{3}(2) .7 log_{3}(2) + 3 log_{3}(2) .4 log_{3}(2) .8 log_{3}(2) - 7 log_{3}(2) .5 log_{3}(2) .3 log_{3}(2) - 8 log_{3}(2) .6 log_{3}(2) . log_{3}(2) - 9 log_{3}(2) .4 log_{3}(2) .2 log_{3}(2) \\ det(a) = 45. {( log_{3}(2) )}^{3} + 84 {( log_{3}(2)) }^{3} + 96{( log_{3}(2)) }^{3} - 105{( log_{3}(2)) }^{3} - 48{( log_{3}(2)) }^{3} - 72{( log_{3}(2)) }^{3} \\ det(a) = {( log_{3}(2)) }^{3}(45 + 84 + 96 - 105 - 48 - 72) \\ det(a) = 0.({( log_{3}(2)) }^{3}) \\ det(a) = 0$

Fazendo 2^det(a), temos:

${2}^{0} = 1$