Question

URGENTE POR FAVOR!!!!!!

Answer 1

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])