Question

encontre a matriz inversa de A= (1 1 )
(0 2)

Answer 1

$\left[\begin{array}{cc}1&1\\0&2\end{array}\right]=A$
a matriz inversa é dada por:
$\displaystyle A^{-1}=\frac{1}{\det A}\bar{A}$
onde o A com barra é a matriz adjunta e det A o determinante de A.

1 - calcular adjunta de A:
    i) encontrar matriz dos cofatores:
   $A=\left[\begin{array}{cc}1&1\\0&2\end{array}\right]\implies\\\\ A_{11}=(-1)^{2}\cdot\det\left|2\right|=2\\A_{12}=(-1)^3\cdot \det\left|0\right|=0\\A_{21}=(-1)^3\cdot\det\left|1\right|=-1\\A_{22}=(-1)^4\cdot\det\left|1\right|=1\\\\ \boxed{\bar{A}= \left[\begin{array}{cc}2&0\\-1&1\end{array}\right]}$
  
2 - encontrar determinante de A:
$\det A= \left[\begin{array}{cc}1&1\\0&2\end{array}\right] =(1\cdot2)-(1\cdot0)=\boxed{2}$

3 - encontrar matriz inversa:
$\displaystyle A^{-1}=\frac{1}{\det A}\cdot\bar{A}^{t}\implies A^{-1}=\frac{1}{2}\cdot\left[\begin{array}{cc}2&-1\\0&1\end{array}\right] \implies \left[\begin{array}{cc}\frac{2}{2}&\frac{0}{2}\\-\frac{1}{2}&\frac{1}{2}\end{array}\right] =\\\\ \boxed{\left[\begin{array}{cc}1&-0,5\\0&0,5\end{array}\right] =A^{-1}}$

prova:
a multiplicação de uma matriz pela sua inversa é igual a uma matriz identidade:
$\left[\begin{array}{cc}1&-0,5\\0&0,5\end{array}\right]\cdot \left[\begin{array}{cc}1&1\\0&2\end{array}\right]= \left[\begin{array}{cc}(1.1+-0,5\cdot0)&(1.1+(-0,5\cdot2))\\(0.1+(0,5\cdot0))&(0.1+0,5\cdot2)\end{array}\right] \\\\= \left[\begin{array}{cc}(1+0)&(1-1)\\(0+0)&(0+1)\end{array}\right] = \left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$