Question

Va rog putin ajutoor....am lasat poza!!

Answer 1

$\mathtt{A(x)=\left(\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt0\\\mathtt x&\mathtt1&\mathtt1\\\mathtt1&\mathtt{-1}&\mathtt1\end{array}\right)}$

$\displaystyle \mathtt{a)~A(2)+A(6)=2A(4)}$

$\displaystyle \mathtt{A(2)=\left(\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt0\\\mathtt 2&\mathtt1&\mathtt1\\\mathtt1&\mathtt{-1}&\mathtt1\end{array}\right),~~~A(6)=\left(\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt0\\\mathtt 6&\mathtt1&\mathtt1\\\mathtt1&\mathtt{-1}&\mathtt1\end{array}\right)}\\ \\ \mathtt{~~~~~~~~~~~~~~~~~~A(4)=\left(\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt0\\\mathtt 4&\mathtt1&\mathtt1\\\mathtt1&\mathtt{-1}&\mathtt1\end{array}\right)}$

$\displaystyle \mathtt{A(2)+A(6)=\left(\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt0\\\mathtt 2&\mathtt1&\mathtt1\\\mathtt1&\mathtt{-1}&\mathtt1\end{array}\right)+\left(\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt0\\\mathtt 6&\mathtt1&\mathtt1\\\mathtt1&\mathtt{-1}&\mathtt1\end{array}\right)=}$

$\displaystyle \mathtt{=\left(\begin{array}{ccc}\mathtt{1+1}&\mathtt{1+1}&\mathtt{0+0}\\\mathtt{2+6}&\mathtt{1+1}&\mathtt{1+1}\\\mathtt{1+1}&\mathtt{(-1)+(-1)}&\mathtt{1+1}\end{array}\right)=\left(\begin{array}{ccc}\mathtt2&\mathtt2&\mathtt0\\\mathtt8&\mathtt2&\mathtt2\\\mathtt2&\mathtt{-2}&\mathtt1\end{array}\right)=}\\ \\ \\ \mathtt{=2\left(\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt0\\\mathtt 4&\mathtt1&\mathtt1\\\mathtt1&\mathtt{-1}&\mathtt1\end{array}\right)=2A(4)}$

$\displaystyle \mathtt{b)~\big(A(2)\big)^{-1}=?}\\ \\ \mathtt{det\big(A(2)\big)=\left(\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt0\\\mathtt 2&\mathtt1&\mathtt1\\\mathtt1&\mathtt{-1}&\mathtt1\end{array}\right)=1\cdot1\cdot1+0\cdot2\cdot(-1)+1\cdot1\cdot1-}\\ \\ \mathtt{-0\cdot1\cdot1-1\cdot2\cdot1-1\cdot1\cdot(-1)=1+0+1-0-2+1=1}\\ \\ \mathtt{det\big(A(2)\big)=1\ne 0\Rightarrow Matricea~A(2)~este~inversabil\u{a}}$

$\displaystyle \mathtt{A(2)=\left(\begin{array}{ccc}\mathtt1&\mathtt1&\mathtt0\\\mathtt 2&\mathtt1&\mathtt1\\\mathtt1&\mathtt{-1}&\mathtt1\end{array}\right)\Rightarrow \big(A(2)\big)^{tr}=\left(\begin{array}{ccc}\mathtt1&\mathtt2&\mathtt1\\\mathtt 1&\mathtt1&\mathtt{-1}\\\mathtt0&\mathtt1&\mathtt1\end{array}\right)}$

$\displaystyle \mathtt{\delta_{11}=(-1)^{1+1}\cdot \left|\begin{array}{ccc}\mathtt1&\mathtt{-1}\\\mathtt1&\mathtt1\end{array}\right|=1\cdot2=2} \\ \\ \mathtt{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\delta_{11}=2}$

$\displaystyle \mathtt{\delta_{12}=(-1)^{1+2}\cdot \left|\begin{array}{ccc}\mathtt1&\mathtt{-1}\\\mathtt0&\mathtt1\end{array}\right|=(-1)\cdot1=-1}\\ \\ \mathtt{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\delta_{12}=-1}$

$\displaystyle \\ \mathtt{\delta_{13}=(-1)^{1+3}\cdot \left|\begin{array}{ccc}\mathtt1&\mathtt1\\\mathtt0&\mathtt1\end{array}\right|=1\cdot1=1}\\ \\ \mathtt{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\delta_{13}=1}$

$\displaystyle \mathtt{\delta_{21}=(-1)^{2+1}\cdot \left|\begin{array}{ccc}\mathtt2&\mathtt1\\\mathtt1&\mathtt1\end{array}\right|=(-1)\cdot1=-1}\\ \\ \mathtt{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\delta_{21}=-1}$

$\displaystyle \mathtt{\delta_{22}=(-1)^{2+2}\cdot \left|\begin{array}{ccc}\mathtt1&\mathtt1\\\mathtt0&\mathtt1\end{array}\right|=1\cdot1=1}\\ \\ \mathtt{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\delta_{22}=1}$

$\displaystyle \mathtt{\delta_{23}=(-1)^{2+3}\cdot \left|\begin{array}{ccc}\mathtt1&\mathtt2\\\mathtt0&\mathtt1\end{array}\right|=(-1)\cdot1=-1}\\ \\ \mathtt{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\delta_{23}=-1}$

$\displaystyle \mathtt{\delta_{31}=(-1)^{3+1}\cdot \left|\begin{array}{ccc}\mathtt2&\mathtt1\\\mathtt1&\mathtt{-1}\end{array}\right|=1\cdot(-3)=-3}\\ \\ \mathtt{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\delta_{31}=-3}$

$\displaystyle \mathtt{\delta_{32}=(-1)^{3+2}\cdot \left|\begin{array}{ccc}\mathtt1&\mathtt1\\\mathtt1&\mathtt{-1}\end{array}\right|=(-1)\cdot(-2)=2}\\ \\\mathtt{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\delta_{32}=2}$

$\displaystyle \mathtt{\delta_{33}=(-1)^{3+3}\cdot \left|\begin{array}{ccc}\mathtt1&\mathtt2\\\mathtt1&\mathtt1\end{array}\right|=1\cdot(-1)=-1}\\ \\ \mathtt{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\delta_{33}=-1}$

$\displaystyle \mathtt{\big(A(2)\big)^*=\left(\begin{array}{ccc}\mathtt2&\mathtt{-1}&\mathtt1\\\mathtt{-1}&\mathtt1&\mathtt{-1}\\\mathtt{-3}&\mathtt2&\mathtt{-1}\end{array}\right)}$

$\displaystyle \mathtt{\big(A(2)\big)^{-1}= \frac{1}{1}\cdot \left(\begin{array}{ccc}\mathtt2&\mathtt{-1}&\mathtt1\\\mathtt{-1}&\mathtt1&\mathtt{-1}\\\mathtt{-3}&\mathtt2&\mathtt{-1}\end{array}\right)=\left(\begin{array}{ccc}\mathtt2&\mathtt{-1}&\mathtt1\\\mathtt{-1}&\mathtt1&\mathtt{-1}\\\mathtt{-3}&\mathtt2&\mathtt{-1}\end{array}\right)}$