Question

Encontre a matriz inversa

Answer 1

$\boxed{A^{-1}=\dfrac{1}{\det{(A)}}\bigg[\text{adj}{(A)}\bigg]}$

$\boxed{\text{adj}(A)=[A_{ij}]^t}\ \to\ \text{Transposta da matriz dos cofatores de A.}$

$\boxed{A_{ij}=(-1)^{i+j}|M_{ij}|}\ \to\ \text{Matriz dos cofatores de A.}$

A matriz A é:

$A=\left[\begin{array}{ccc}2&0&1\\6&0&2\\4&2&-1\end{array}\right]$

O determinante de A será:

$\det{(A)}=1(6)(2)-2(2)(2)=12-8\ \therefore\ \boxed{\det{(A)}=4}$

A matriz dos cofatores de A será:

$a_{11}=(-1)^{1+1}|M_{11}|=\left|\begin{array}{cc}0&2\\2&-1\end{array}\right|\ \therefore\ a_{11}=-4$

$a_{12}=(-1)^{1+2}|M_{12}|=-\left|\begin{array}{cc}6&2\\4&-1\end{array}\right|\ \therefore\ a_{12}=14$

$a_{13}=(-1)^{1+3}|M_{13}|=\left|\begin{array}{cc}6&0\\4&2\end{array}\right|\ \therefore\ a_{13}=12$

$a_{21}=(-1)^{2+1}|M_{21}|=-\left|\begin{array}{cc}0&1\\2&-1\end{array}\right|\ \therefore\ a_{21}=2$

$a_{22}=(-1)^{2+2}|M_{22}|=\left|\begin{array}{cc}2&4\\1&-1\end{array}\right|\ \therefore\ a_{22}=-6$

$a_{23}=(-1)^{2+3}|M_{23}|=-\left|\begin{array}{cc}2&0\\4&2\end{array}\right|\ \therefore\ a_{23}=-4$

$a_{31}=(-1)^{3+1}|M_{31}|=\left|\begin{array}{cc}0&1\\0&2\end{array}\right|\ \therefore\ a_{31}=0$

$a_{32}=(-1)^{3+2}|M_{32}|=-\left|\begin{array}{cc}2&1\\6&2\end{array}\right|\ \therefore\ a_{32}=2$

$a_{33}=(-1)^{3+3}|M_{33}|=\left|\begin{array}{cc}2&0\\6&0\end{array}\right|\ \therefore\ a_{33}=0$

$A_{ij}=\left[\begin{array}{ccc}-4&14&12\\2&-6&-4\\0&2&0\end{array}\right]$

A adjunta de A será:

$\text{adj}(A)=[A_{ij}]^t\ \therefore\ \text{adj}(A)=\left[\begin{array}{ccc}-4&2&0\\14&-6&2\\12&-4&0\end{array}\right]$

Dessa forma, a matriz inversa de A será:

$A^{-1}=\dfrac{1}{4}\left[\begin{array}{ccc}-4&2&0\\14&-6&2\\12&-4&0\end{array}\right]\ \therefore\ \boxed{A^{-1}=\left[\begin{array}{ccc}-1&\frac{1}{2}&0\\\frac{7}{2}&-\frac{3}{2}&\frac{1}{2}\\3&-1&0\end{array}\right]}$