Matriu no invertible

Sigueu la matriu $$A=\left(\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right)$$ i $I$ la matriu identitat d’ordre $3$.
a) Troba els valors de $m$ perquè la matriu $A-MI$ no tingui inversa.
b) Troba $x$, diferent de zero, perquè $A-xI$ sigui la inversa de la matriu $\dfrac{1}{x}\left(A-I\right)$.

a) $$A-mI=\left(\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right)-m\cdot\left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right)=$$
$$=\left(\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right)-\left(\begin{array}{ccc} m & 0 & 0 \\ 0 & m & 0 \\ 0 & 0 & m \end{array}\right)=\left(\begin{array}{ccc} 1-m & 1 & 1 \\ 1 & 1-m & 1 \\ 1 & 1 & 1-m \end{array}\right)$$
$A-mI$ no tindrà inversa si $|A-mI|=0$

$$|A-mI|=\left|\begin{array}{ccc} 1-m & 1 & 1 \\ 1 & 1-m & 1 \\ 1 & 1 & 1-m \end{array}\right| =$$

$$=(1-m)^3+1+1-(1-m)-(1-m)-(1-m)$$

$$|A-mI|=(1-m)^3+3m-1=1-3m+3m^2-m^3+3m-1=$$

$$=-m^3+3m^2=m^2(3-m)$$

$|A-mI|=0 \Leftrightarrow m^2(3-m)=0 \Leftrightarrow \left\{\begin{array}{l} m=0 \ m=3 \end{array}\right. \Rightarrow A-mI$ no té inversa si $\boxed{m=0}\ \text{ o } \boxed{m=3}$

b)

• $$A-xI=\left(\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right)-x\cdot\left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right)=\left(\begin{array}{ccc} 1-x & 1 & 1 \\ 1 & 1-x & 1 \\ 1 & 1 & 1-x \end{array}\right)$$
• $$\dfrac{1}{x}\left(A-I\right)=\dfrac{1}{x}\left(\left(\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right)-\left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right)\right)=\dfrac{1}{x}\left(\begin{array}{ccc} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}\right)$$
  

Perquè siguin inverses, el producte ha de ser igual a la matriu identitat.

$$\dfrac{1}{x}\left(A-I\right)\cdot\left(A-xI\right)= \dfrac{1}{x}\left(\begin{array}{ccc} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}\right) \cdot \left(\begin{array}{ccc} 1-x & 1 & 1 \\ 1 & 1-x & 1 \\ 1 & 1 & 1-x \end{array}\right)=$$

$$=\dfrac{1}{x}\left(\begin{array}{ccc} 2 & 2-x & 2-x \\ 2-x & 2 & 2-x \\ 2-x & 2-x & 2 \end{array}\right)=\left(\begin{array}{ccc} \frac{2}{x} & \frac{2-x}{x} & \frac{2-x}{x } \\ \frac{2-x}{x} & \frac{2}{x} & \frac{2-x}{x} \\ \frac{2-x}{x} & \frac{2 -x}{x} & \frac{2}{x} \end{array}\right)$$

$$\left(\begin{array}{ccc} \frac{2}{x} & \frac{2-x}{x} & \frac{2-x}{x} \\ \frac{2-x} {x} & \frac{2}{x} & \frac{2-x}{x} \\ \frac{2-x}{x} & \frac{2-x}{x} & \frac{ 2}{x} \end{array}\right)=I=\left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right) \Leftrightarrow \left\{\begin{array}{l} \dfrac{2}{x}=1 \ \dfrac{2-x}{x}=0 \end{array}\right. \Leftrightarrow \boxed{x=2}$$