Question

preciso encontrar a A⁻¹ da matriz $\left[\begin{array}{ccc}1 & 1\\0,1 & 0,35\end{array}\right]$ , sei que o resultado final fica $\left[\begin{array}{ccc}\frac{7}{5} & -4\\\frac{-2}{5} & 4\\\end{array}\right]$. Preciso do passo a passo

Answer 1

Para uma matriz de ordem 2, existe uma fórmula para a inversa

Seja A uma matriz com determinante não nulo da forma

$A=\left[\begin{array}{cc}a&b\\c&d\end{array}\right]$

Então

$\boxed{\boxed{A^{-1}=\dfrac{1}{\det A}\left[\begin{array}{cc}d&-b\\-c&a\end{array}\right]}}$
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Vamos achar a inversa de A utilizando a fórmula acima:

$A=\left[\begin{array}{cc}1&1\\0.1&0.35\end{array}\right]=\left[\begin{array}{cc}1&1\\1/10&35/100\end{array}\right]$

Achando o determinante de A:

$\det A=1\cdot0.35-1\cdot0.1=0.35-0.1=0.25=1/4=25/100$

Logo, a inversa de A é

$A^{-1}=\dfrac{1}{\det A}\left[\begin{array}{cc}35/100&-1\\-1/10&\,\,\,\,1\end{array}\right]\\\\\\A^{-1}=\dfrac{1}{25/100}\left[\begin{array}{cc}35/100&-1\\-1/10&\,\,\,\,1\end{array}\right]\\\\\\A^{-1}=\dfrac{100}{25}\left[\begin{array}{cc}35/100&-1\\-1/10&\,\,\,\,1\end{array}\right]\\\\\\A^{-1}=\left[\begin{array}{cc}\frac{100}{25}\cdot\frac{35}{100}&-\frac{100}{25}\\\\-\frac{100}{25}\cdot\frac{1}{10}&\,\,\,\,\frac{100}{25}\end{array}\right]$

$A^{-1}=\left[\begin{array}{cc}\frac{35}{25}&-4\\\\-4\cdot\frac{1}{10}&\,\,\,\,4\end{array}\right]\\\\\\A^{-1}=\left[\begin{array}{cc}\frac{7}{5}&-4\\-\frac{4}{10}&\,\,\,\,4\end{array}\right]\\\\\\A^{-1}=\left[\begin{array}{cc}\,\,\,7/5&-4\\-2/5&\,\,\,\,4\end{array}\right]$
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Outra forma seria usar a definição de inversa.

Se $B=\left[\begin{array}{cc}a&b\\c&d\end{array}\right]$ é a inversa de $\left[\begin{array}{cc}1&1\\0.1&0.35\end{array}\right]$, então

$\left[\begin{array}{cc}a&b\\c&d\end{array}\right]\cdot\left[\begin{array}{cc}1&1\\0.1&0.35\end{array}\right]=\left[\begin{array}{cc}1&0\\0&1\end{array}\right]=\mathsf{matriz\,identidade}$

Fazendo o produto de matrizes:

$\left[\begin{array}{cc}a+0.1b&a+0.35b\\c+0.1d&c+0.35d\end{array}\right]=\left[\begin{array}{cc}1&0\\0&1\end{array}\right]$

Com isso, obtemos dois sistemas lineares:

$\begin{cases}a+0.1b=1\\a+0.35b=0\end{cases}\,\,\,\,\mathsf{e}\,\,\,\,\begin{cases}c+0.1d=0\\c+0.35d=1\end{cases}$

Subtraindo as equações do primeiro:

$(a+0.1b)-(a+0.35b)=1-0\\\\0.1b-0.35=1\\\\-0.25b=1\\\\-b/4=1\\\\b=-4$

Substituindo na primeira:

$a+0.1b=1\\\\a+0.1(-4)=1\\\\a-0.4=1\\\\a=1+0.4\\\\a=1.4\\\\a=1.4\cdot5/5\\\\a=7/5$

Agora, vamos achar c e d. Subtraindo as equações do segundo sistema:

$(c+0.1d)-(c+0.35d)=0-1\\\\0.1d-0.35d=-1\\\\-0.25d=-1\\\\d/4=1\\\\d=4$

Substituindo na primeira:

$c+0.1d=0\\\\c+0.1\cdot4=0\\\\c+0.4=0\\\\c=-0.4\\\\c=-0.4\cdot5/5\\\\c=-2/5$

Logo,

$B=\left[\begin{array}{cc}a&b\\c&d\end{array}\right]=\left[\begin{array}{cc}\,\,\,\,\,7/5&-4\\-2/5&4\end{array}\right]$

Answer 2

1              1         |      1        0
0,1           0,35    |      0        1

L2=L2-0,1L1

1              1         |         1         0
0           0,25       |      -0,1        1

L2 =(1/0,25) * L2

1              1         |         1                0
0              1         |      -10/25        1/0,25

1              1         |         1                0
0              1         |       -2/5              4

L1=L1-L2

1              0         |        7/5              -4
0              1         |       -2/5               4

A⁻¹ =
 7/5              -4
-2/5               4