Question

URGENTE!!! JÁ TENTEI VÁRIAS VEZES MAS NN ACHO A RESPOSTA. PFV ME AJUDEM.

Answer 1

Sendo

$A=\left[\begin{array}{cc}1&3\\2&-5 \end{array}\right ],\;\;B=\left[\begin{array}{cc}5\\2 \end{array}\right ]~~\text{ e }~~AX=2B,$

Encontrar a matrix $X.$

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$AX=2B\\\\ \left[\begin{array}{cc}1&3\\2&-5 \end{array}\right ]\cdot X=2\left[\begin{array}{cc}5\\2 \end{array}\right ]\\\\\\ \left[\begin{array}{cc}1&3\\2&-5 \end{array}\right ]\cdot X=\left[\begin{array}{cc}10\\4 \end{array}\right ]$


Para que o produto das matrizes $A_{2\times2}$ e $X$ resulte em uma matriz $2\times 1,\,$ $X$ deve ser uma matriz $2\times 1:$

( duas linhas e uma coluna )

$X=\left[\begin{array}{c}x_1\\x_2 \end{array} \right ]$


Sendo assim,

$\left[\begin{array}{cc}1&3\\2&-5 \end{array}\right ]\cdot \left[\begin{array}{c}x_1\\x_2\end{array}\right ]=\left[\begin{array}{c}10\\4\end{array}\right ]\\\\\\ \left\{ \begin{array}{ccr} 1x_1+3x_2&\!\!=\!\!&10\\\\ 2x_1+(-5)x_2&\!\!=\!\!&4 \end{array} \right.\\\\\\\\ \left\{ \begin{array}{ccrc} x_1+3x_2&\!\!=\!\!&10&~~~~\mathbf{(i)}\\\\ 2x_1-5x_2&\!\!=\!\!&4&~~~~\mathbf{(ii)} \end{array} \right.$


Resolvendo o sistema acima:

$x_1=10-3x_2~~~~~\big(\text{da equa\c{c}\~ao }\mathbf{(i)}\big)$


Substituindo na equação $\mathbf{(ii)}:$

$2\,(10-3x_2)-5x_2=4\\\\ 20-6x_2-5x_2=4\\\\ 20-11x_2=4\\\\ -11x_2=4-20\\\\ -11x_2=-16\\\\ x_2=\dfrac{-16}{-11}\\\\\\ \boxed{\begin{array}{c} x_2=\dfrac{16}{11} \end{array}}$


Encontrando $x_1:$

$x_1=10-3x_2\\\\ x_1=10-3\cdot \dfrac{16}{11}\\\\\\ x_1=10-\dfrac{48}{11}\\\\\\ x_1=\dfrac{110}{11}-\dfrac{48}{11}\\\\\\ x_1=\dfrac{110-48}{11}\\\\\\ \boxed{\begin{array}{c}x_1=\dfrac{62}{11} \end{array}}$

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Logo, a matrix $X$ é

$X=\left[\begin{array}{c}\frac{62}{11}\\\\ \frac{16}{11} \end{array} \right ]$


Resposta: alternativa $\text{c) }\left[\begin{array}{c}\frac{62}{11}\\\\ \frac{16}{11} \end{array} \right ].$