Question

determine a matriz inversa das seguintes matrizes​

Answer 1

Sua questão:

Determine a matriz inversa das seguintes matrizes:

$\large \sf A=\left[\begin{array}{ccc}0&1\\1&2\end{array}\right]$

$\large \sf B=\left[\begin{array}{ccc}3&2\\7&5\end{array}\right]$

$\large \sf C=\left[\begin{array}{ccc}5&3\\3&2\end{array}\right]$

$\large \sf D=\left[\begin{array}{ccc}2&-1\\3&5\end{array}\right]$

Resolução:

$\large \sf A=\left[\begin{array}{ccc}0&1\\1&2\end{array}\right]$

$\large \sf \left[\begin{array}{ccc}0&1\\1&2\end{array}\right] \cdot \large \sf \left[\begin{array}{ccc}a&b\\c&d \end{array}\right] = \large \sf \left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

$\large \sf \left[\begin{array}{ccc}\not{0*a} +1*c&\not{0*b}+1*d\\1*a+2*c&1*b+2*d\end{array}\right]= \large \sf \left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

$\large\begin{array}{lr} \sf 1c = 1\\\\\sf \underline{\boxed{\red{\sf c=1}}}\\\\\\ \sf 1d = 0\\\\\sf \underline{\boxed{\red{\sf d=0}}}\\\\\\\sf 1*a+2*c=0\\\\\sf 1*a + 2*1=0\\\\\sf a+2=0\\\\\sf \underline{\boxed{\red{\sf a=-2}}}\\\\\\\sf 1*b + 2*d = 1\\\\\sf 1*b + 2*0 = 1\\\\\sf 1b = 1\\\\\sf \underline{\boxed{\red{\sf b=1}}}\end{array}$

Escrevendo a matriz inversa de A:

$\large \sf \left[\begin{array}{ccc}-2&1\\1&0\end{array}\right] \leftarrow \sf Resposta\ do\ item\ A.$

__________#___________

$\large \sf B=\left[\begin{array}{ccc}3&2\\7&5\end{array}\right]$

Irei fazer pela fórmula pois odeio fazer sistemas ... mas por definição dará o mesmo resultado rsrs.

$\large\begin{array}{lr} \sf B^{-1} = \dfrac{1}{ad - bc} \cdot \left[\begin{array}{ccc}d&-b\\-c&a\end{array}\right] \end{array}$

a => 3

b => 2

c => 7

d => 5

$\large\begin{array}{lr} \sf B^{-1} = \dfrac{1}{3*5 - 2*7} \cdot \left[\begin{array}{ccc}5&-2\\-7&3\end{array}\right] \end{array}$

$\large\begin{array}{lr} \sf B^{-1} = \dfrac{1}{15 - 14} \cdot \left[\begin{array}{ccc}5&-2\\-7&3\end{array}\right] \end{array}$

$\large\begin{array}{lr} \sf B^{-1} = \dfrac{1}{1} \cdot \left[\begin{array}{ccc}5&-2\\-7&3\end{array}\right] \end{array}$

$\large\begin{array}{lr} \sf B^{-1} = 1 \cdot \left[\begin{array}{ccc}5&-2\\-7&3\end{array}\right] \end{array}$

$\large\begin{array}{lr} \sf B^{-1} = \left[\begin{array}{ccc}5&-2\\-7&3\end{array}\right] \end{array}$

Escrevendo a matriz inversa de B:

$\large\begin{array}{lr} \sf \left[\begin{array}{ccc}5&-2\\-7&3\end{array}\right] \end{array} \leftarrow \sf Resposta\ do\ item\ B.$

__________#___________

$\large \sf C=\left[\begin{array}{ccc}5&3\\3&2\end{array}\right]$

Mesma coisa ... não gosto de sistemas rs.

$\large\begin{array}{lr} \sf C^{-1} = \dfrac{1}{ad - bc} \cdot \left[\begin{array}{ccc}d&-b\\-c&a\end{array}\right] \end{array}$

a => 5

b => 3

c => 3

d => 2

$\large\begin{array}{lr} \sf C^{-1} = \dfrac{1}{5*2 - 3*3} \cdot \left[\begin{array}{ccc}2&-3\\-3&5\end{array}\right] \end{array}$

$\large\begin{array}{lr} \sf C^{-1} = \dfrac{1}{10 - 9} \cdot \left[\begin{array}{ccc}2&-3\\-3&5\end{array}\right] \end{array}$

$\large\begin{array}{lr} \sf C^{-1} = \dfrac{1}{1} \cdot \left[\begin{array}{ccc}2&-3\\-3&5\end{array}\right] \end{array}$

$\large\begin{array}{lr} \sf C^{-1} =1\cdot \left[\begin{array}{ccc}2&-3\\-3&5\end{array}\right] \end{array}$

$\large\begin{array}{lr} \sf C^{-1} = \left[\begin{array}{ccc}2&-3\\-3&5\end{array}\right] \end{array}$

Escrevendo a matriz inversa de C:

$\large\left[\begin{array}{ccc}2&-3\\-3&5\end{array}\right] \end{array} \leftarrow \sf Resposta\ do\ item\ C.$

__________#___________

$\large \sf D=\left[\begin{array}{ccc}2&-1\\3&5\end{array}\right]$

$\large\begin{array}{lr} \sf D^{-1} = \dfrac{1}{ad - bc} \cdot \left[\begin{array}{ccc}d&-b\\-c&a\end{array}\right] \end{array}$

a => 2

b => -1

c => 3

d => 5

$\large\begin{array}{lr} \sf D^{-1} = \dfrac{1}{2*5 - (-1)*3} \cdot \left[\begin{array}{ccc}5&-(-1)\\-3&2\end{array}\right] \end{array}$

$\large\begin{array}{lr} \sf D^{-1} = \dfrac{1}{13} \cdot \left[\begin{array}{ccc}5&-(-1)\\-3&2\end{array}\right] \end{array}$

$\large\begin{array}{lr} \sf D^{-1} = \dfrac{1}{13} \cdot \left[\begin{array}{ccc}5&1\\-3&2\end{array}\right] \end{array}$

$\large\begin{array}{lr} \sf D^{-1} = \cdot \left[\begin{array}{ccc}5/13&1/13\\-3/13&2/13\end{array}\right] \end{array}$

Escrevendo a matriz inversa de D:

$\large\begin{array}{lr} \sf \left[\begin{array}{ccc}5/13&1/13\\-3/13&2/13\end{array}\right] \end{array}\leftarrow \sf Resposta\ do\ item\ D.$

Espero ter ajudado.

Bons estudos.

• Att. FireClassis.

Answer 2

Come Together <33

SORRY