Resolució de l’equació matricial

Considera les matrius $$A=\left(\begin{array}{ccc} 0 & 1 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right)\text{ i } B=\left(\begin{array}{ccc} 1 & -1 & 1 \\ 1 & -1 & 0 \\ -1 & 2 & 3 \end{array}\right)$$ Determina, si existeix, la matriu $X$ que verifica $AX+B=A^2$.

Determina, si existe, la matriz $X$ que verifica $AX+B=A^2$.

Como queremos que $AX+B=A^2\Rightarrow AX=A^2-B\Rightarrow X=A^{-1}\left(A^2-B\right)$

$$A^2=\left(\begin{array}{ccc} 0 & 1 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right)\cdot\left(\begin{array}{ccc} 0 & 1 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right)=\left(\begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right)$$

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

Calculamos la inversa de $A$, utilizamos el método de Gauss:

$$\left(\begin{array}{ccc|ccc} 0 & 1 & 1 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \end{array}\right)\rightarrow\left(\begin{array}{ccc|ccc} 1 & 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \end{array}\right)\rightarrow$$

$$\rightarrow\left(\begin{array}{ccc|ccc} 1 & 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 0 & 0 & 1 \end{array}\right)\Rightarrow A^{-1}=\left(\begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & -1 \\ 0 & 0 & 1 \end{array}\right)$$

$X=A^{-1}\left(A^2-B\right)\Rightarrow$

$$\Rightarrow X=\left(\begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & -1 \\ 0 & 0 & 1 \end{array}\right)\cdot\left(\begin{array}{ccc} 0 & 1 & 0 \\ -1 & 2 & 1 \\ 1 & -2 & -2 \end{array}\right)=\left(\begin{array}{ccc} -1 & 2 & 1 \\ -1 & 3 & 2 \\ 1 & -2 & -2 \end{array}\right)$$

$$\boxed{X=\left(\begin{array}{ccc} -1 & 2 & 1 \\ -1 & 3 & 2 \\ 1 & -2 & -2 \end{array}\right)}$$