2014 – Juny – Opció B – Exercici 3

Considereu les matrius $$A=\left(\begin{array}{ccc} 0 & 1 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right)\qquad i\qquad B=\left(\begin{array}{ccc} 1 & -1 & 1 \\ 1 & -1 & 0 \\ -1 & 2 & 3 \end{array}\right)$$

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

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

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

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

Calculem la inversa d’$A$, utilitzem el mètode de Gauss:

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

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

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

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

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