Respuestas
La inversa de una matriz
Para resolver el ejercicio planteado se hará uso del siguiente Teorema
Teorema: Sea A una matriz de nxn, entonces A es invertible si y solo si el determinante de A ≠0
A^{-1}=\frac{1}{det(A)}(Adj(A))
Sea A=\left[\begin{array}{ccc}2&1&-2\\3&2&2\\-1&2&3\end{array}\right]
Determinante de A
\left[\begin{array}{ccc}2&1&-2\\3&2&2\\-1&2&3\end{array}\right]=2\left[\begin{array}{ccc}2&2&\\2&3&\end{array}\right]-\left[\begin{array}{ccc}3&2&\\-1&3&\end{array}\right]-2\left[\begin{array}{ccc}3&2&\\-1&2&\end{array}\right]=-23
Adjunta de A
A_{11}=\left[\begin{array}{ccc}2&2&\\2&3&\end{array}\right]=6-4=2
A_{12}=-\left[\begin{array}{ccc}3&2&\\-1&3&\end{array}\right]=-(9+2)=-11
A_{13}=\left[\begin{array}{ccc}3&2&\\-1&2&\end{array}\right]=6+2=8
A_{21}=-\left[\begin{array}{ccc}1&-2&\\2&3&\end{array}\right]=-(3+4)=-7
A_{22}=\left[\begin{array}{ccc}2&-2&\\-1&3&\end{array}\right]=6-2=4
A_{23}=\left[\begin{array}{ccc}2&1&\\-1&2&\end{array}\right]=4+1=5
A_{31}=\left[\begin{array}{ccc}1&-2&\\2&2&\end{array}\right]=2+4=6
A_{32}=-\left[\begin{array}{ccc}2&-2&\\3&2&\end{array}\right]=-(4+6)=-10
A_{33}=\left[\begin{array}{ccc}2&1&\\3&2&\end{array}\right]=4-3=1
A^{-1}=\frac{1}{det(A)}(Adj(A))
A^{-1}=\frac{1}{-23}(\left[\begin{array}{ccc}2&-7&6\\-11&4&-10\\8&5&1\end{array}\right])
A^{-1}=\left[\begin{array}{ccc}-2/23&7/23&-6/23\\11/23&-4/23&10/23\\-8/23&-5/23&-1/23\end{array}\right]
2 A^{-1}=3B entonces B=2/3 A^{-1}
B=2/3(\left[\begin{array}{ccc}-2/23&7/23&-6/23\\11/23&-4/23&10/23\\-8/23&-5/23&-1/23\end{array}\right])
B=-2/69(\left[\begin{array}{ccc}2&-7&6\\-11&4&-10\\8&5&1\end{array}\right])
B^{-1}=\frac{1}{det(B)}(Adj(B))
Determinante de B
\left[\begin{array}{ccc}2&-7&6\\-11&4&-10\\8&5&1\end{array}\right]=2\left[\begin{array}{ccc}4&-10&\\5&1&\end{array}\right]-7\left[\begin{array}{ccc}-11&-10&\\8&1&\end{array}\right]+6\left[\begin{array}{ccc}-11&4&\\8&5&\end{array}\right]=2(4+50)-7(-11+80)+6(-55-32)= -897
Adjunta de B
B_{11}=\left[\begin{array}{ccc}4&-10&\\5&1&\end{array}\right]=4+50=54
B_{12}=-\left[\begin{array}{ccc}-11&-10&\\8&1&\end{array}\right]=-(-11+80)=-69
B_{13}=\left[\begin{array}{ccc}-11&4&\\8&5&\end{array}\right]=-55-32=-87
B_{21}=-\left[\begin{array}{ccc}-7&6&\\5&1&\end{array}\right]=-(-7-30)=-37
B_{22}=\left[\begin{array}{ccc}2&6&\\8&1&\end{array}\right]=2-48=-46
B_{23}=\left[\begin{array}{ccc}2&-7&\\8&5&\end{array}\right]=10+56=66
B_{31}=\left[\begin{array}{ccc}-7&6&\\4&-10&\end{array}\right]=70-24=46
B_{32}=-\left[\begin{array}{ccc}2&6&\\-11&-10&\end{array}\right]=-(-20+66)=-46
B_{33}=\left[\begin{array}{ccc}2&-7&\\-11&4&\end{array}\right]=8-77=-69
B^{-1}=-2/69(1/ -897)(\left[\begin{array}{ccc}54&-37&46\\-69&-46&46\\-87&66&-69\end{array}\right])
B^{-1}=2/61893(\left[\begin{array}{ccc}54&-37&46\\-69&-46&46\\-87&66&-69\end{array}\right])