Ejercicio resuelto de Matrices · Matemáticas II · 2020

Problema 4.

Se dan las matrices $A=\left(\begin{array}{cc}1& 2\\ b& 0\\ -1& 2\end{array}\right)$ y $B=\left(\begin{array}{ccc}-1& 0& 2\\ -1& b& -1\end{array}\right)$, que dependen del parámetro real $b$. Obtener razonadamente, escribiendo todos los pasos del razonamiento utilizado:

a) Los valores de $b$ para que cada una de las matrices $AB$ y $BA$ tenga inversa. (3 puntos)
b) Los valores de $b$ para que la matriz $A^{T}A$ tenga inversa, siendo $A^T$ la matriz traspuesta de $A$. (3 puntos)
c) La inversa de $A^{T}A$, cuando dicha inversa exista. (4 puntos)

Calculamos $AB$,

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

$\mathrm{\exists }(AB{)}^{-1}⟺\mathrm{\mid }AB\mathrm{\mid }\mathrm{\ne }0$

$\mathrm{\mid }AB\mathrm{\mid }=\mid \begin{array}{ccc}-3& 2b& 0\\ -b& 0& 2b\\ -1& 2b& -4\end{array}\mid =-4b^2+12b^2-8b^2=0\Rightarrow \text{no }\mathrm{\exists }(AB{)}^{-1}$

Calculamos $BA$,

$BA=\left(\begin{array}{ccc}-1& 0& 2\\ -1& b& -1\end{array}\right)\left(\begin{array}{cc}1& 2\\ b& 0\\ -1& 2\end{array}\right)=\left(\begin{array}{cc}-3& 2\\ b^2& -4\end{array}\right)$

$\mathrm{\exists }(BA{)}^{-1}⟺\mathrm{\mid }BA\mathrm{\mid }\mathrm{\ne }0$

$\mathrm{\mid }BA\mathrm{\mid }=\mid \begin{array}{cc}-3& 2\\ b^2& -4\end{array}\mid =12-2b^2$

$12-2b^2=0\Rightarrow b^2=6\Rightarrow b=\pm \sqrt{6}\Rightarrow \mathrm{\exists }(BA{)}^{-1}\text{ si }b\mathrm{\ne }\pm \sqrt{6}$


Calculamos $A^{T}A$,

$A^{T}A=\left(\begin{array}{ccc}1& b& -1\\ 2& 0& 2\end{array}\right)\left(\begin{array}{cc}1& 2\\ b& 0\\ -1& 2\end{array}\right)=\left(\begin{array}{cc}{b}^2+2& 0\\ 0& 8\end{array}\right)$

$\mathrm{\exists }(A^{T}A{)}^{-1}⟺\mathrm{\mid }{A}^{T}A\mathrm{\mid }\mathrm{\ne }0$

$\mathrm{\mid }{A}^{T}A\mathrm{\mid }=\mid \begin{array}{cc}{b}^2+2& 0\\ 0& 8\end{array}\mid =8(b^2+2)$

$8(b^2+2)\mathrm{\ne }0\Rightarrow b^2+2\mathrm{\ne }0\Rightarrow b^2=-2\Rightarrow b=\pm \sqrt{-2}\text{ no existe}$



Sabemos de b) que $\mathrm{\mid }{A}^{T}A\mathrm{\mid }=8(b^2+2)$

$A^{T}A=\left(\begin{array}{cc}{b}^2+2& 0\\ 0& 8\end{array}\right)$ $\stackrel{\text{menores}}{\to }$ $\left(\begin{array}{cc}8& 0\\ 0& b^2+2\end{array}\right)$ $\stackrel{\text{adjunta}}{\to }$$\left(\begin{array}{cc}8& 0\\ 0& b^2+2\end{array}\right)$$\stackrel{\text{transpuesta}}{\to }$$\left(\begin{array}{cc}8& 0\\ 0& b^2+2\end{array}\right)$

$Y(A^{T}A{)}^{-1}=\frac{1}{8(b^2+2)}\left(\begin{array}{cc}8& 0\\ 0& b^2+2\end{array}\right)=\left(\begin{array}{cc}\frac{1}{b^2+2}& 0\\ 0& \frac{1}{8}\end{array}\right)$

Por tanto, $\mathrm{\forall },b\in \mathbb{R},(A^{T}A{)}^{-1}=\left(\begin{array}{cc}\frac{1}{b^2+2}& 0\\ 0& \frac{1}{8}\end{array}\right)$.