Question

Ma jutati va rog la subiectul 2 exercitiul 1 subpunctul c)
Ati putea sa imi si explicati cum se face .
Mersii

Answer 1

$\displaystyle \mathtt{1.~c)~A=\left(\begin{array}{ccc}\mathtt{-1}&\mathtt1\\\mathtt0&\mathtt0\end{array}\right);~I_2=\left(\begin{array}{ccc}\mathtt1&\mathtt0\\\mathtt0&\mathtt1\end{array}\right)~~~~~~~~~~~~~~~~~~det~B=0,~m=?}\\ \\ \mathtt{B=A \cdot A+mI_2}$

$\displaystyle \mathtt{A \cdot A=\left(\begin{array}{ccc}\mathtt{-1}&\mathtt1\\\mathtt0&\mathtt0\end{array}\right)\cdot \left(\begin{array}{ccc}\mathtt{-1}&\mathtt1\\\mathtt0&\mathtt0\end{array}\right)=}\\ \\ \mathtt{=\left(\begin{array}{ccc}\mathtt{(-1)\cdot(-1)+1\cdot0 }&\mathtt{(-1) \cdot 1+1 \cdot 0}\\\mathtt{0 \cdot (-1)+0 \cdot 0}&\mathtt{0 \cdot 1+0 \cdot 0}\end{array}\right)=\left(\begin{array}{ccc}\mathtt{1}&\mathtt{-1}\\\mathtt{0}&\mathtt{0}\end{array}\right)}$

$\displaystyle \mathtt{A \cdot A=\left(\begin{array}{ccc}\mathtt{1}&\mathtt{-1}\\\mathtt{0}&\mathtt{0}\end{array}\right)}$

$\displaystyle \mathtt{mI_2=m \cdot \left(\begin{array}{ccc}\mathtt1&\mathtt0\\\mathtt0&\mathtt1\end{array}\right)=\left(\begin{array}{ccc}\mathtt{m \cdot 1}&\mathtt{m \cdot 0}\\\mathtt{m\cdot0}&\mathtt {m\cdot1}\end{array}\right) =\left(\begin{array}{ccc}\mathtt m&\mathtt0\\\mathtt0&\mathtt m\end{array}\right)}\\ \\ \mathtt{mI_2=\left(\begin{array}{ccc}\mathtt m&\mathtt0\\\mathtt0&\mathtt m\end{array}\right) }$

$\displaystyle \mathtt{A \cdot A+mI_2=\left(\begin{array}{ccc}\mathtt{1}&\mathtt{-1}\\\mathtt{0}&\mathtt{0}\end{array}\right)+\left(\begin{array}{ccc}\mathtt m&\mathtt0\\\mathtt0&\mathtt m\end{array}\right)=\left(\begin{array}{ccc}\mathtt {1+m}&\mathtt{(-1)+0}\\\mathtt{0+0}&\mathtt{0+m}\end{array}\right)=}\\ \\ \mathtt{=\left(\begin{array}{ccc}\mathtt{1+m}&\mathtt{-1}\\\mathtt0&\mathtt m\end{array}\right)}$

$\displaystyle \mathtt{A \cdot A+mI_2=\left(\begin{array}{ccc}\mathtt{1+m}&\mathtt{-1}\\\mathtt0&\mathtt m\end{array}\right) \Rightarrow B=\left(\begin{array}{ccc}\mathtt{1+m}&\mathtt{-1}\\\mathtt0&\mathtt m\end{array}\right)}$

$\mathtt{det~B=0 \Rightarrow \left|\begin{array}{ccc}\mathtt{1+m}&\mathtt{-1}\\\mathtt0&\mathtt m\end{array}\right|=0}\\ \\ \mathtt{det~B=\left|\begin{array}{ccc}\mathtt{1+m}&\mathtt{-1}\\\mathtt0&\mathtt m\end{array}\right|=(1+m) \cdot m-(-1) \cdot 0=m^2+m}\\ \\ \mathtt{det~B=0 \Rightarrow m^2+m=0\Rightarrow m(m+1)=0 \Rightarrow \underline{m=0}}\\ \\ \mathtt{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow \underline{m=-1}}$