Si:
[tex]A = \left(\begin{array}{ccc}-2&5&-1\\3&0&-4\\3&1&-5\end{array}\right) [/tex]
Calcular: [tex]A^{-1}[/tex]
NOTA:
[tex]A^{-1} = \frac{1}{\|A\|} * Adj(A)[/tex]
• Solución:
i) Calculamos el determinante de la matriz A.( por el metodo de Jacobi: )
Nota: Solo poseen inversa , aquellas matrices cuadradas "no singulares" (determinante diferente de 0), además esta es única.
[tex]\|A\| = \left|\begin{array}{ccc}-2&5&-1\\3&0&-4\\3&1&-5\end{array}\right|\ ^{F_{12}(1)}_{F_{23}(-1)} = \left|\begin{array}{ccc}1&5&-5\\3&0&-4\\0&1&-1\end{array}\right|\ ^{F_{21}(-3)} [/tex]
[tex]\|A\| =\left|\begin{array}{ccc}1&5&-5\\0&-15&11\\0&1&-1\end{array}\right| = \left|\begin{array}{cc}-15&11\\1&-1\end{array}\right| = (-15)(-1) - (1)(11)
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\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \to \|A\| = 4[/tex]
ii) Calculamos la matriz de cofactores de la matriz A, denotado por: cof (A)
[tex]Cof(A) = (\alpha _{ij})_{nxn} \ \ \ \ \ / \alpha _{ij}(-1)^{i+j} M_{ij}[/tex]
⇒ [tex]\alpha _{11} = (-1)^{1+1} * \left|\begin{array}{cc}0&-4\\1&-5&\end{array}\right| = 4[/tex]
⇒ [tex]\alpha _{12} = (-1)^{1+2} * \left|\begin{array}{cc}3&-4\\3&-5&\end{array}\right| = 3[/tex]
⇒ [tex]\alpha _{13} = (-1)^{1+3} * \left|\begin{array}{cc}3&0\\3&1&\end{array}\right| = 3[/tex]
⇒ [tex]\alpha _{21} = (-1)^{2+1} * \left|\begin{array}{cc}5&-1\\1&-5&\end{array}\right| = 24[/tex]
⇒ [tex]\alpha _{22} = (-1)^{2+2} * \left|\begin{array}{cc}-2&-1\\3&-5&\end{array}\right| = 13[/tex]
⇒ [tex]\alpha _{23} = (-1)^{2+3} * \left|\begin{array}{cc}-2&5\\3&1&\end{array}\right| = 17[/tex]
⇒ [tex]\alpha _{31} = (-1)^{3+1} * \left|\begin{array}{cc}5&-1\\0&-4&\end{array}\right| = -20[/tex]
⇒ [tex]\alpha _{32} = (-1)^{3+2} * \left|\begin{array}{cc}-2&-1\\3&-4&\end{array}\right| = -11[/tex]
⇒ [tex]\alpha _{33} = (-1)^{3+3} * \left|\begin{array}{cc}-2&5\\3&0&\end{array}\right| = -15[/tex]
Por consiguiente:
[tex]cof (A) = \left(\begin{array}{ccc}4&3&3\\24&13&17\\-20&-11&-15\end{array}\right)[/tex]
iii) Calcular la matriz adjunta de A :
• Recuerda que: [tex]\{cof(A)\}^T = Adj(A)[/tex]
De tal modo:
[tex]Adj(A) = \left(\begin{array}{ccc}4&24&-20\\3&13&-11\\3&17&-15\end{array}\right) [/tex]
iv) Por último, calculamos la matriz inversa , denotado por: A⁻¹
[tex]A^{-1} = { \frac{1}{4}} * \left(\begin{array}{ccc}4&24&-20\\3&13&-11\\3&17&-15\end{array}\right)
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A^{-1} = \left(\begin{array}{ccc}1&6&-5\\ \\ \frac{3}{4} & \frac{13}{4} & -\frac{11}{4}\\ \\ \frac{3}{4} & \frac{17}{4} & -\frac{15}{4} \end{array}\right)[/tex]
Eso es todo!!! uff :)
