Question

va implor ajutatima plsssss

Answer 1

$\displaystyle \mathtt{1.~~~~O(0,0),~A(0,2)~B(3,5),~C(6,8)}\\\\\mathtt{a)~ \left|\begin{array}{ccc}\mathtt x&\mathtt y&\mathtt1\\\mathtt0&\mathtt2&\mathtt1\\\mathtt6&\mathtt8&\mathtt1\end{array}\right|=0}\\\\\mathtt{x\cdot2\cdot1+1\cdot 0\cdot8+y\cdot1\cdot6-1\cdot 2 \cdot6-y\cdot0\cdot1-x\cdot1\cdot8=0}\\\\\mathtt{2x+6y-12-8x=0}\\\\ \mathtt{-6x+6y-12=0}\\\\\mathtt{x-y+2=0}$

$\displaystyle\mathtt{b)~A(0,2),~B(3,5),~C(6,8)}\\ \\ \mathtt{\left|\begin{array}{ccc}\mathtt{0}&\mathtt{2}&\mathtt{1}\\\mathtt{3}&\mathtt{5}&\mathtt{1}\\\mathtt{6}&\mathtt{8}&\mathtt{1}\end{array}\right|=0}\\\\\mathtt{0\cdot5\cdot1+1 \cdot3\cdot8+2\cdot1\cdot6-1\cdot5\cdot6-2\cdot3\cdot1-0\cdot1\cdot 8=0}\\\\\mathtt{0+24+12-30-6+0=0}\\\\ \mathtt{0=0~adev\u{a}rat\Rightarrow punctele~A(0,2),~B(3,~5),~C(6,8)~sunt~coliniare.}$

$\displaystyle\mathtt{c)~A_{AOB}=A_{ BOC}}\\\\\mathtt{A_{ AOB}= \frac{1}{2}|D|}\\\\\mathtt{D=\left|\begin{array}{ccc}\mathtt0&\mathtt2&\mathtt1\\\mathtt0&\mathtt0&\mathtt1\\\mathtt3&\mathtt5&\mathtt1\end{array}\right|=0\cdot 0\cdot1+1\cdot0\cdot5+2\cdot1\cdot 3-1\cdot0\cdot3-2\cdot 0\cdot 1-}\\ \\ \mathtt{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-0 \cdot1\cdot5=6}\\ \\ \mathtt{A_{AOB}=\frac{1}{2}|6|=\frac{6}{2}=3}$

$\displaystyle \mathtt{A_{BOC}=\frac{1}{2}|D|}\\ \\ \mathtt{D=\left|\begin{array}{ccc}\mathtt3&\mathtt5&\mathtt1\\\mathtt0&\mathtt0&\mathtt1\\\mathtt6&\mathtt8&\mathtt1\end{array}\right|=3\cdot0\cdot1+1\cdot0\cdot8+5\cdot1\cdot 6-1\cdot0\cdot6-5\cdot0\cdot 1-}\\ \\ \mathtt{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-~3 \cdot1\cdot 8=6}\\\\ \mathtt{A_{BOC}=\frac{1}{2}|6|=\frac{6}{2}=3}\\\\\mathtt{\left\begin{array}{ccc}\mathtt{A_{AOB}=3}\\\mathtt{A_{BOC}=3}\end{array}\right]\Rightarrow A_{AOB}=A_{BOC}}$

$\displaystyle \mathtt{2.~~~~~~A= \left(\begin{array}{ccc}\mathtt1&\mathtt2\\\mathtt3&\mathtt4\end{array}\right);~B=\left(\begin{array}{ccc}\mathtt4&\mathtt3\\\mathtt2&\mathtt1\end{array}\right)}\\ \\ \mathtt{a)~2A+2B=?}$

$\displaystyle\mathtt{2A=2\cdot \left(\begin{array}{ccc}\mathtt1&\mathtt2\\\mathtt3&\mathtt4\end{array}\right) =\left(\begin{array}{ccc}\mathtt{2\cdot1}&\mathtt{2 \cdot 2}\\\mathtt{2\cdot 3}&\mathtt{2\cdot 4}\end{array}\right)=\left(\begin{array}{ccc}\mathtt2&\mathtt4\\\mathtt6&\mathtt8\end{array}\right)}$

$\displaystyle \mathtt{2B=2 \cdot \left(\begin{array}{ccc}\mathtt4&\mathtt3\\\mathtt2&\mathtt1\end{array}\right)=\left(\begin{array}{ccc}\mathtt{2 \cdot 4}&\mathtt{2 \cdot 3}\\\mathtt{2 \cdot 2}&\mathtt{2 \cdot 1}\end{array}\right)=\left(\begin{array}{ccc}\mathtt8&\mathtt6\\\mathtt4&\mathtt2\end{array}\right)}$

$\displaystyle\mathtt{2A+2B=\left(\begin{array}{ccc}\mathtt2&\mathtt4\\\mathtt6&\mathtt8\end{array}\right)+\left(\begin{array}{ccc}\mathtt8&\mathtt6\\\mathtt4&\mathtt2\end{array}\right)=\left(\begin{array}{ccc}\mathtt{2+8}&\mathtt{4+6}\\\mathtt{6+4}&\mathtt{8+2}\end{array}\right)=\left(\begin{array}{ccc}\mathtt{10}&\mathtt{10}\\\mathtt{10}&\mathtt{10}\end{array}\right)}$

$\displaystyle \mathtt{b)~(A-B)\cdot(B-A)=-8I_2}\\ \\ \mathtt{A-B=\left(\begin{array}{ccc}\mathtt1&\mathtt2\\\mathtt3&\mathtt4\end{array}\right)-\left(\begin{array}{ccc}\mathtt4&\mathtt3\\\mathtt2&\mathtt1\end{array}\right)=\left(\begin{array}{ccc}\mathtt{1-4}&\mathtt{2-3}\\\mathtt{3-2}&\mathtt{4-1}\end{array}\right)=\left(\begin{array}{ccc}\mathtt{-3}&\mathtt{-1}\\\mathtt1&\mathtt3\end{array}\right)}$

$\displaystyle \mathtt{B-A=\left(\begin{array}{ccc}\mathtt4&\mathtt3\\\mathtt2&\mathtt1\end{array}\right)-\left(\begin{array}{ccc}\mathtt1&\mathtt2\\\mathtt3&\mathtt4\end{array}\right)=\left(\begin{array}{ccc}\mathtt{4-1}&\mathtt{3-2}\\\mathtt{2-3}&\mathtt{1-4}\end{array}\right)=\left(\begin{array}{ccc}\mathtt3&\mathtt1\\\mathtt{-1}&\mathtt{-3}\end{array}\right)}$

$\displaystyle \mathtt{(A-B)\cdot(B-A)=\left(\begin{array}{ccc}\mathtt{-3}&\mathtt{-1}\\\mathtt1&\mathtt3\end{array}\right)\cdot \left(\begin{array}{ccc}\mathtt3&\mathtt1\\\mathtt{-1}&\mathtt{-3}\end{array}\right) =}\\ \\ \mathtt{=\left(\begin{array}{ccc}\mathtt{(-3)\cdot3+(-1)\cdot (-1)}&\mathtt{(-3)\cdot1+(-1)\cdot(-3)}\\\mathtt{1\cdot 3+3\cdot (-1)}&\mathtt{1\cdot1+3\cdot(-3)}\end{array}\right)=}$

$\displaystyle \mathtt{=\left(\begin{array}{ccc}\mathtt{-9+1}&\mathtt{-3+3}\\\mathtt{3-3}&\mathtt{1-9}\end{array}\right)=\left(\begin{array}{ccc}\mathtt{-8}&\mathtt0\\\mathtt{0}&\mathtt{-8}\end{array}\right)=-8\cdot \left(\begin{array}{ccc}\mathtt1&\mathtt0\\\mathtt0&\mathtt1\end{array}\right)=-8I_2}$