Question

DETERMINE A MATRIZ INVERSA DE A = ( 3 4)
2 3

Answer 1

Boa noite Mariana!


Solução!


$A= \begin{vmatrix} 3 & 4 \\ 2 & 3 \\ \end{vmatrix}$


$A^{-1}=\begin{vmatrix} a & b \\ c & d \\ \end{vmatrix}$

$\begin{pmatrix} 3 & 4 \\ 2 & 3 \\ \end{pmatrix} \times\begin{pmatrix} a & b \\ c & d\\ \end{pmatrix}=\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}\\\\\\\\ \begin{pmatrix} 3a+4c & 3b+4d \\ 2a+3c & 2b+3d \\ \end{pmatrix}=\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}\\\\\\ Temos ~~dois ~~sistemas!\\\\\\ I\begin{cases} 3a+4c=1\\ 2a+3c=0 \end{cases}~\\\\\\ II\begin{cases} 3b+4d=0\\ 2b+3d=1 \end{cases}$




$I\begin{cases} 3a+4c=1\\ 2a+3c=0 \end{cases}\\\\\\ \begin{cases} 3a+4c=1.(-2)\\ 2a+3c.=0.(-3) \end{cases}\\\\\\\ \end{cases}\\\\\begin{cases} -6a-8c=-2\\ 6a+9c=0 \end{cases}\\\\\\\\\\ -8c+9c=-2+0\\\\ c=-2\\\\ \boxed{c=-2}\\\\\\\ 3a+4(-2)=1\\\\ 3a-8=1\\\\ 3a=1+8\\\\\ 3a=9\\\\\ a=\dfrac{9}{3}\\\\\ \boxed{a= 3 }$




$II\begin{cases} 3b+4d=0\\ 2b+3d=1 \end{cases}\\\\\\\ \begin{cases} 3b+4d=0.-2)\\ 2b+3d=1.(3) \end{cases}\\\\\\\\ \begin{cases} -6b-8d=0\\ 6b+9d=3 \end{cases}\\\\\\\ -8d+9d=0+3\\\\\ \boxed{d=3}\\\\\\\\ 3b+4d=0 \\\\\\ 3b+4(3)=0\\\\\ 3b+12=0\\\\\ 3b=-12\\\\\ b= \dfrac{-12}{3} \\\\ b=-4$


$Resposta:~~A^{-1} =\begin{vmatrix} 3 & -4 \\ -2 & 3 \\ \end{vmatrix}$



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Bons estudos!