Question

sa se determine numerele intregi x pentru care matricea M(x) este egala cu inversa sa​

Answer 1

$\displaystyle 1.~~A=\left(\begin{array}{ccc}-1&-2\\1&2\end{array}\right),~I_2=\left(\begin{array}{ccc}1&0\\0&1\end{array}\right),~M(x)=I_2+xA,~x\in\mathbb{R}\\\\ c)~M(x)=M^{-1}(x),~x=?$

$xA=x \cdot \left(\begin{array}{ccc}-1&-2\\1&2\end{array}\right)=\left(\begin{array}{ccc}-x&-2x\\x&2x\end{array}\right)\\\\ I_2+xA=\left(\begin{array}{ccc}1&0\\0&1\end{array}\right)+\left(\begin{array}{ccc}-x&-2x\\x&2x\end{array}\right)=\left(\begin{array}{ccc}1-x&-2x\\x&1+2x\end{array}\right)=M(x)$

$det(M(x))=\left|\begin{array}{ccc}1-x&-2x\\x&1+2x\end{array}\right|=(1-x)(1+2x)-x \cdot (-2x)=\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=1+x-2x^2+2x^2=1+x\\\\1+x \not= 0 \Rightarrow x\not = -1\\\\ Transpusa~matricei~M(x):~^tM(x)=\left(\begin{array}{ccc}1-x&x\\-2x&1+2x\end{array}\right)\\\\ Construim~matricea~adjuncta~a~matricei~M(x):$

$a_{11}=(-1)^{1+1}\cdot (1+2x)=1 \cdot (1+2x)=1+2x\\\\ a_{12}=(-1)^{1+2}\cdot (-2x)=(-1) \cdot (-2x)=2x\\\\ a_{21}=(-1)^{2+1}\cdot x=(-1) \cdot x=-x\\\\ a_{22}=(-1)^{2+2}\cdot (1-x)=1 \cdot (1-x)=1-x$

$\displaystyle Matricea~adjuncta:~M^*(x)=\left(\begin{array}{ccc}1+2x&2x\\-x&1-x\end{array}\right)\\\\ Aplicam~formula:~M^{-1}(x)=\cfrac{1}{det(M(x))}\cdot M^*(x) \\\\ M^{-1}(x)=\frac{1}{1+x}\cdot \left(\begin{array}{ccc}1+2x&2x\\-x&1-x\end{array}\right)=\left(\begin{array}{ccc}\cfrac{1+2x}{1+x} &\cfrac{2x}{1+x} \\-\cfrac{x}{1+x} &\cfrac{1-x}{1+x} \end{array}\right)$

$\displaystyle M(x)=M^{-1}(x) \Rightarrow \left(\begin{array}{ccc}1-x&-2x\\x&1+2x\end{array}\right)=\left(\begin{array}{ccc}\cfrac{1+2x}{1+x} &\cfrac{2x}{1+x} \\-\cfrac{x}{1+x} &\cfrac{1-x}{1+x} \end{array}\right)\\\\ 1-x=\frac{1+2x}{1+x} \Rightarrow (1-x)(1+x)=1+2x \Rightarrow 1-x^2=1+2x\Rightarrow \\\\ \Rightarrow -x^2-2x+1-1=0 \Rightarrow -x^2-2x=0 \Rightarrow x(-x-2)=0 \Rightarrow x=0\\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow x=-2$

$\displaystyle -2x=\frac{2x}{1+x} \Rightarrow -2x(1+x)=2x \Rightarrow -2x-2x^2-2x=0\Rightarrow \\\\ \Rightarrow -2x^2-4x=0 \Rightarrow 2x(-x-2)=0 \Rightarrow x=0\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow x=-2$

$\displaystyle x=-\frac{x}{1+x} \Rightarrow x(1+x)=-x \Rightarrow x+x^2+x=0 \Rightarrow x^2+2x=0 \Rightarrow \\\\ \Rightarrow x(x+2)=0 \Rightarrow x=0 \\\\~~~~~~~~~~~~~~~~~~~~~\Rightarrow x=-2$

$\displaystyle 1+2x=\frac{1-x}{1+x} \Rightarrow (1+2x)(1+x)=1-x \Rightarrow 1+3x+2x^2-1+x=0\Rightarrow \\\\ \Rightarrow 2x^2+4x=0\Rightarrow 2x(x+2)=0 \Rightarrow x=0 \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow x=-2\\\\ x\in\{0;-2\}$