Question

matriz inversa de A= [ 1 2] [1 3 ]

Answer 1

$\mathrm{A=\left[\begin{array}{ccc}1&2\\1&3\end{array}\right]\ \ \bigg\| \ \ A^{-1}=\ ?}\\\\\\ \mathbf{Sendo\ A^{-1}=\left[\begin{array}{ccc}a&b\\c&d\end{array}\right],\ teremos\ que:} \\\\\\ \mathrm{A.A^{-1}=I\ \to\ \left[\begin{array}{ccc}1&2\\1&3\end{array}\right].\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]=\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]}\\\\ \mathrm{\left[\begin{array}{ccc}a+2c&b+2d\\a+3c&b+3d\end{array}\right]=\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]}$

$\begin{cases} \mathrm{a+2c=1\ \mathbf{(I)}} \\ \mathrm{a+3c=0\ \mathbf{(II)}} \\ \mathrm{b+2d=0\ \mathbf{(III)}} \\ \mathrm{b+3d=1\ \mathbf{(IV)}} \end{cases}$

$\mathrm{Somando\ II\ com\ I.(-1),\ vem:}\\ \mathrm{a+3c-a-2c=0-1\ \to\ \boxed{\mathrm{c=-1}}}\\\\ \mathrm{Substituindo\ c\ em\ I,\ vem:}\\ \mathrm{a+2.(-1)=1\ \to\ a-2=1\ \to\ \boxed{\mathrm{a=3}}}\\\\ \mathrm{Somando\ IV\ com\ III.(-1),\ vem:}\\ \mathrm{b+3d-b-2d=1-0\ \to\ \boxed{\mathrm{d=1}}}\\\\ \mathrm{Substituindo\ d\ em\ III,\ vem:}\\ \mathrm{b+2.1=0\ \to\ \boxed{\mathrm{b=-2}}}$

$\mathrm{A^{-1}=\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]\ \to\ \mathbf{A^{-1}=\left[\begin{array}{ccc}3&-2\\-1&1\end{array}\right]}}$

Answer 2

Resposta:

tem uma forma mais fácil

Explicação passo-a-passo:

1º) você divide a matriz pelo determinante;

2º) Troca os números da diagonal principal de posição.

3º) Inverte o sinal da diagonal secundária.