Problema 163

Se sabe que el determinante de la matriz $A=\begin{pmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{pmatrix}$ es -3. Calcula, indicando las propiedades que utilices, los siguientes determinantes:

a) |-2A| y |A⁻¹|.

b) $\begin{vmatrix}a_{21}&a_{22}&a_{23}\\7a_{11}&7a_{12}&7a_{13}\\2a_{31}&2a_{32}&2a_{33}\end{vmatrix}$ y $\begin{vmatrix}a_{11}&a_{21}+2a_{31}&5a_{31}\\a_{12}&a_{22}+2a_{32}&5a_{32}\\a_{13}&a_{23}+2a_{33}&5a_{33}\end{vmatrix}$

Solución:

Para calcular estos determinantes utilizaremos las siguientes propiedades de los determinantes.

$\displaystyle \mbox{a) }|-2A|\overset{P.6}=(-2)^3|A|=-8\cdot(-3)=24\\\\|A^{-1}|\overset{P.4}=\frac 1{|A|}=\frac 1{-3}$

$\mbox{b) }\begin{vmatrix}a_{21}&a_{22}&a_{23}\\7a_{11}&7a_{12}&7a_{13}\\2a_{31}&2a_{32}&2a_{33}\end{vmatrix}\overset{P.6}=7\cdot 2\cdot\begin{vmatrix}a_{21}&a_{22}&a_{23}\\a_{11}&a_{12}&a_{13}\\a_{31}&a_{32}&a_{33}\end{vmatrix}=\\\\\overset{P.5}=-14\begin{vmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{vmatrix}=-14\cdot(-3)=42$

$\begin{vmatrix}a_{11}&a_{21}+2a_{31}&5a_{31}\\a_{12}&a_{22}+2a_{32}&5a_{32}\\a_{13}&a_{23}+2a_{33}&5a_{33}\end{vmatrix}\overset{P.7}=\begin{vmatrix}a_{11}&a_{21}&5a_{31}\\a_{12}&a_{22}&5a_{32}\\a_{13}&a_{23}&5a_{33}\end{vmatrix}+\begin{vmatrix}a_{11}&2a_{31}&5a_{31}\\a_{12}&2a_{32}&5a_{32}\\a_{13}&2a_{33}&5a_{33}\end{vmatrix}=\\\\\overset{P.6}=5\cdot\begin{vmatrix}a_{11}&a_{21}&a_{31}\\a_{12}&a_{22}&a_{32}\\a_{13}&a_{23}&a_{33}\end{vmatrix}+2\cdot 5\cdot\begin{vmatrix}a_{11}&a_{31}&a_{31}\\a_{12}&a_{32}&a_{32}\\a_{13}&a_{33}&a_{33}\end{vmatrix}=\\\\\overset{P.1}=5\cdot\begin{vmatrix}a_{11}&a_{21}&a_{31}\\a_{12}&a_{22}&a_{32}\\a_{13}&a_{23}&a_{33}\end{vmatrix}+2\cdot 5\cdot0=\\\\\overset{P.2}=5\cdot\begin{vmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{vmatrix}=5\cdot(-3)=-15$

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eMatecs

Exámenes resueltos de Selectividad, EVAU, EBAU

Problema 163

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