50 pontos para quem resolver essa questão (+25 pela melhor resposta)!Dadas as matrizes: [tex]A = \le.... Pergunta de ideia deAnniex

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A)

O produto é possível, pois o número de colunas de A (2) é igual ao número de linhas de B (2)

O produto AB será uma matriz 2 (nº linhas de A) x 1 (nº colunas de B)

$A\cdot B=\left[\begin{array}{ccc}log_{2}(x)&log_{2}(2x)\\y&\frac{y}{2}\end{array}\right]\cdot\left[\begin{array}{ccc}4\\4\end{array}\right]\\\\\\A\cdot B=\left[\begin{array}{ccc}log_{2}(x)\cdot4+log_{2}(2x)\cdot4\\y\cdot4+\frac{y}{2}\cdot4\end{array}\right]\\\\\\A\cdot B=\left[\begin{array}{ccc}4\cdot log_{2}(x)+4\cdot log_{2}(2x)\\4y+2y\end{array}\right]\\\\\\A\cdot B=\left[\begin{array}{ccc}log_{2}(x^{4})+log_{2}(2x)^{4}\\6y\end{array}\right]$

Resolvendo a potência e juntando os logaritmos pela propriedade:

$log_{b}(a\cdot c)\rightleftharpoons log_{b}(a)+lob_{b}(c)$

$A\cdot B=\left[\begin{array}{ccc}log_{2}(x^{4})+log_{2}(16x^{4})\\6y\end{array}\right]\\\\\\A\cdot B=\left[\begin{array}{ccc}log_{2}(x^{4}\cdot16x^{4})\\6y\end{array}\right]\\\\\\\boxed{\boxed{A\cdot B=\left[\begin{array}{ccc}log_{2}(16x^{8})\\6y\end{array}\right]}}$

B)

$A\cdot B=C\\\\\\\left[\begin{array}{ccc}log_{2}(16x^{8})\\6y\end{array}\right]=\left[\begin{array}{ccc}28\\10\end{array}\right]$

As matrizes serão iguais se suas entradas correspondentes forem iguais:

$log_{2}(16x^{8})=28\\2^{28}=16x^{8}\\2^{4}x^{8}=2^{28}\\x^{8}=2^{28}/2^{4}\\x^{8}=2^{28-4}\\x^{8}=2^{24}\\x^{8/8}=2^{24/8}\\x^{1}=2^{3}\\\\\boxed{\boxed{x=8}}$

e:

$6y=10\\y=10/6\\\\\\\boxed{\boxed{y=\dfrac{5}{3}}}$

Anniex Obrigada! Consegui compreender!

Niiya Disponha, e obrigado pela melhor resposta! :)

marcosnobre5

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Vamos lá..

a)
$\left[\begin{array}{ccc}log_{2}x&log_{2}2x\\y& \frac{y}{2}\end{array}\right]* \left[\begin{array}{ccc}4\\4\end{array}\right]= \left[\begin{array}{ccc}4log_{2}x+4log_{2}(2x)\\4y+ \frac{4y}{2}\end{array}\right] \\ \\ \\ = \left[\begin{array}{ccc}4log_{2}x+4(log_{2}2+log_{2}x)\\4y+2y\end{array}\right] = \left[\begin{array}{ccc}4log_{2}x+4(1+log_{2}x)\\6y\end{array}\right] \\ \\ \\ = \left[\begin{array}{ccc}4log_{2}x+4+4log_{2}x\\6y\end{array}\right]$

$= \left[\begin{array}{ccc}8log_{2}x+4\\6y\end{array}\right]$ << Esse é o produto A.B

b)
$\left[\begin{array}{ccc}8log_{2}x+4\\6y\end{array}\right]= \left[\begin{array}{ccc}28\\10\end{array}\right]$

Isso implica que:

$\left \{ {{8log_{2}x+4=28} \atop {6y=10}} \right. \\ \\ 6y=10 \\ \\ y= \frac{10}{6}= \frac{5}{3} \\ \\ 8log_{2}x+4=28 \\ \\ 8log_{2}x=24 \\ \\ log_{2}x= \frac{24}{8} \\ \\ log_{2}x=3 \\ \\ 2^[3}=x \\ \\ x=8$

Resp.: y = 5/3 ; x = 8

Abraço!

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