Ejercicio resuelto de Matrices y Determinantes · Matemáticas CCSS

Problema 2.

Consideramos las matrices $A=\left(\begin{array}{cc}1& 0\\ 0& 1\\ -1& 3\end{array}\right)$, $B=\left(\begin{array}{ccc}-1& 4& 0\\ 0& 2& 3\end{array}\right)$ y $C=\left(\begin{array}{cc}3& 1\\ 1& 0\end{array}\right)$.

a) Analiza si la matriz $AB-2I$ es invertible, siendo $I$ la matriz identidad de orden 3. (3 puntos)
b) Determina la matriz $X$ que es solución de la ecuación $A+2XC=B^t$, siendo $B^t$ la traspuesta de la matriz $B$. (4 puntos)
c) Calcula para qué valores de $z$ la matriz $D=\left(\begin{array}{cc}1& -1\\ -1& z\end{array}\right)$ cumple la condición $CD=DC$. (3 puntos)

$A\cdot B=\left(\begin{array}{cc}1& 0\\ 0& 1\\ -1& 3\end{array}\right)\cdot \left(\begin{array}{ccc}-1& 4& 0\\ 0& 2& 3\end{array}\right)=\left(\begin{array}{ccc}-1& 4& 0\\ 0& 2& 3\\ 1& 2& 9\end{array}\right)$

$AB-2I=\left(\begin{array}{ccc}-1& 4& 0\\ 0& 2& 3\\ 1& 2& 9\end{array}\right)-2\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)=\left(\begin{array}{ccc}-3& 4& 0\\ 0& 0& 3\\ 1& 2& 7\end{array}\right)$

La matriz $(AB-2I)$ será invertible si $\det (AB-2I)\mathrm{\ne }0$

$\det (AB-2I)=\mid \begin{array}{ccc}-3& 4& 0\\ 0& 0& 3\\ 1& 2& 7\end{array}\mid =30\mathrm{\ne }0\Rightarrow \mathrm{\exists }(AB-2I{)}^{-1}$



$A+2XC=B^t\Rightarrow 2XC=B^t-A\Rightarrow X=\frac{1}{2}(B^t-A)C^{-1}$

$C^{-1}=\frac{1}{\det (C)}[\mathrm{Adj}(C){]}^t,\det (C)=\mid \begin{array}{cc}3& 1\\ 1& 0\end{array}\mid =-1\mathrm{\ne }0\Rightarrow \mathrm{\exists }{C}^{-1}$

$\mathrm{Adj}(C)=\left(\begin{array}{cc}0& -1\\ -1& 3\end{array}\right)\Rightarrow [\mathrm{Adj}(C){]}^t=\left(\begin{array}{cc}0& -1\\ -1& 3\end{array}\right)\Rightarrow C^{-1}=\left(\begin{array}{cc}0& 1\\ 1& -3\end{array}\right)$

$B^t-A=\left(\begin{array}{cc}-1& 0\\ 4& 2\\ 0& 3\end{array}\right)-\left(\begin{array}{cc}1& 0\\ 0& 1\\ -1& 3\end{array}\right)=\left(\begin{array}{cc}-2& 0\\ 4& 1\\ 1& 0\end{array}\right)$

$X=\frac{1}{2}(B^t-A)C^{-1}=\frac{1}{2}\left(\begin{array}{cc}-2& 0\\ 4& 1\\ 1& 0\end{array}\right)\left(\begin{array}{cc}0& 1\\ 1& -3\end{array}\right)=\frac{1}{2}\left(\begin{array}{cc}0& -2\\ 1& 1\\ 0& 1\end{array}\right)\Rightarrow X=\left(\begin{array}{cc}0& -1\\ \frac{1}{2}& \frac{1}{2}\\ 0& \frac{1}{2}\end{array}\right)$



$C\cdot D=D\cdot C\Rightarrow \left(\begin{array}{cc}3& 1\\ 1& 0\end{array}\right)\left(\begin{array}{cc}1& -1\\ -1& z\end{array}\right)=\left(\begin{array}{cc}1& -1\\ -1& z\end{array}\right)\left(\begin{array}{cc}3& 1\\ 1& 0\end{array}\right)$

$\Rightarrow \left(\begin{array}{cc}2& -3+z\\ 1& -1\end{array}\right)=\left(\begin{array}{cc}2& 1\\ -3+z& -1\end{array}\right)\Rightarrow -3+z=1\Rightarrow z=4$