Question

achar a matriz inversa da matriz

ps: pfv me ajudem ​

Answer 1

Resposta:

$A^-^1=\left(\begin{array}{ccc}\frac{1}{5}~~~~-\frac{1}{10} \\-\frac{1}{5}~~~~~~ \frac{3}{10} \end{array}\right)$

Explicação passo a passo:

$Seja A^-^1 =\left(\begin{array}{ccc}a~~~~b\\c~~~~d\end{array}\right)\\\\A*A^-^1=I\\\\\left(\begin{array}{ccc}3~~~~1\\2~~~~4\end{array}\left(\begin{array}{ccc}a~~~~b\\c~~~~d\end{array}\right)=\left(\begin{array}{ccc}1~~~~0\\0~~~~1\end{array}\right)$

$\left(\begin{array}{}3a~~+~~c~~~~~~3b~~+~~d\\2a~~+~~4c~~~~~~2b~~+~~4d\end{array}\right)=\left(\begin{array}{ccc}1~~~~0\\0~~~~1\end{array}\right)$

$\left\{\begin{array}{cccc}3a+c=1\\2a+4c=0\\\end{array}\right~~e~~\left\{\begin{array}{cccc}3b+d=0\\2b+4d=1\\\end{array}\right$

$\left\{\begin{array}{cccc}-12a-4c=-4\\2a+4c=0\\\end{array}\right\\\\-10a=-4\\\\10a=4\\\\a=\frac{2}{5} \\\\2(\frac{2}{5})+4c=0\\\\\frac{4}{5} +4c=0\\\\4c=-\frac{4}{5} \\\\c=-\frac{1}{5}$

$\left\{\begin{array}{cccc}-12b-4d=0\\2b+4d=1\\\end{array}\right\\\\-10b=1\\\\b=-\frac{1}{10} \\\\3(-\frac{1}{10})+d=0\\\\d=\frac{3}{10}$