Question

Subiectul al || -lea 1 C va rog !

Answer 1

$\displaystyle \mathtt{A=\left(\begin{array}{ccc}\mathtt0&\mathtt1&\mathtt0\\\mathtt1&\mathtt0&\mathtt1\\\mathtt0&\mathtt1&\mathtt0\end{array}\right),~~B=\left(\begin{array}{ccc}\mathtt0&\mathtt0&\mathtt1\\\mathtt0&\mathtt1&\mathtt0\\\mathtt1&\mathtt0&\mathtt0\end{array}\right)}\\ \\ \\ \mathtt{det(B+xA)=1}$

$\displaystyle \mathtt{xA=x\cdot\left(\begin{array}{ccc}\mathtt0&\mathtt1&\mathtt0\\\mathtt1&\mathtt0&\mathtt1\\\mathtt0&\mathtt1&\mathtt0\end{array}\right)=\left(\begin{array}{ccc}\mathtt{x\cdot0}&\mathtt{x\cdot1}&\mathtt{x\cdot0}\\\mathtt{x\cdot1}&\mathtt{x\cdot0}&\mathtt{x\cdot1}\\\mathtt{x\cdot0}&\mathtt{x\cdot1}&\mathtt{x\cdot0}\end{array}\right)=\left(\begin{array}{ccc}\mathtt0&\mathtt x&\mathtt0\\\mathtt x&\mathtt0&\mathtt x\\\mathtt0&\mathtt x&\mathtt0\end{array}\right)}$

$\displaystyle \mathtt{B+xA=\left(\begin{array}{ccc}\mathtt0&\mathtt0&\mathtt1\\\mathtt0&\mathtt1&\mathtt0\\\mathtt1&\mathtt0&\mathtt0\end{array}\right)+\left(\begin{array}{ccc}\mathtt0&\mathtt x&\mathtt0\\\mathtt x&\mathtt0&\mathtt x\\\mathtt0&\mathtt x&\mathtt0\end{array}\right)=}$

$\displaystyle \mathtt{=\left(\begin{array}{ccc}\mathtt{0+0}&\mathtt{0+x}&\mathtt{1+0}\\\mathtt{0+x}&\mathtt{1+0}&\mathtt{0+x}\\\mathtt{1+0}&\mathtt{0+x}&\mathtt{0+0}\end{array}\right)=\left(\begin{array}{ccc}\mathtt0&\mathtt x&\mathtt1\\\mathtt x&\mathtt1&\mathtt x\\\mathtt1&\mathtt x&\mathtt0\end{array}\right)}$

$\displaystyle \mathtt{det(B+xA)=1}\\ \\ \mathtt{\left|\begin{array}{ccc}\mathtt0&\mathtt x&\mathtt1\\\mathtt x&\mathtt1&\mathtt x\\\mathtt1&\mathtt x&\mathtt0\end{array}\right|=1}\\ \\\\ \mathtt{0\cdot1\cdot0+1\cdot x\cdot x+x\cdot x\cdot1-1\cdot1\cdot1-x\cdot x \cdot0-0\cdot x\cdot x=1}\\\\\mathtt{0+x^2+x^2-1-0-0=1}\\ \\ \mathtt{2x^2-1=1}\\ \\ \mathtt{2x^2-1-1=0}\\\\\mathtt{2x^2-2=0}\\ \\ \mathtt{2(x^2-1)=0}\\ \\ \mathtt{2(x-1)(x+1)=0}\\\\\mathtt{x-1=0\Rightarrow x=1}\\ \\ \mathtt{x+1=0\Rightarrow x=-1}$