Question

rezolvati cel putin 2 exercitii va rog

Answer 1

$\displaystyle \mathtt{1.~~~A(a)= \left(\begin{array}{ccc}\mathtt1&\mathtt{-1}\\\mathtt a&\mathtt3\end{array}\right)}\\ \\ \mathtt{a)~A(-1)+A(1)=2A(0)}\\ \\ \mathtt{A(-1)= \left(\begin{array}{ccc}\mathtt1&\mathtt{-1}\\\mathtt{-1}&\mathtt3\end{array}\right);~~~A(1)= \left(\begin{array}{ccc}\mathtt1&\mathtt{-1}\\\mathtt1&\mathtt3\end{array}\right)}$

$\displaystyle \mathtt{A(-1)+A(1)= \left(\begin{array}{ccc}\mathtt1&\mathtt{-1}\\\mathtt{-1}&\mathtt3\end{array}\right)+\left(\begin{array}{ccc}\mathtt1&\mathtt{-1}\\\mathtt1&\mathtt3\end{array}\right)=}\\ \\ \mathtt{=\left(\begin{array}{ccc}\mathtt{1+1}&\mathtt{-1+(-1)}\\\mathtt{-1+1}&\mathtt{3+3}\end{array}\right)=\left(\begin{array}{ccc}\mathtt2&\mathtt{-2}\\\mathtt0&\mathtt6\end{array}\right)=2\left(\begin{array}{ccc}\mathtt1&\mathtt{-1}\\\mathtt0&\mathtt3\end{array}\right)=2A(0)}$

$\displaystyle \mathtt{b)~A(a)\cdot \left(\begin{array}{ccc}\mathtt3&\mathtt1\\\mathtt{-2}&\mathtt1\end{array}\right)=5I_2} \\ \\ \mathtt{\left(\begin{array}{ccc}\mathtt1&\mathtt{-1}\\\mathtt a&\mathtt3\end{array}\right) \left(\begin{array}{ccc}\mathtt3&\mathtt1\\\mathtt{-2}&\mathtt1\end{array}\right)=5\left(\begin{array}{ccc}\mathtt1&\mathtt0\\\mathtt0&\mathtt1\end{array}\right)}$

$\displaystyle \mathtt{\left(\begin{array}{ccc}\mathtt{1\cdot3+(-1)\cdot(-2)}&\mathtt{1\cdot1+(-1)\cdot1}\\\mathtt{a\cdot3+3\cdot(-2)}&\mathtt{a\cdot1+3\cdot1}\end{array}\right)=\left(\begin{array}{ccc}\mathtt5&\mathtt0\\\mathtt0&\mathtt5\end{array}\right)}\\ \\ \mathtt{\left(\begin{array}{ccc}\mathtt{3+2}&\mathtt{1-1}\\\mathtt{3a-6}&\mathtt{a+3}\end{array}\right)=\left(\begin{array}{ccc}\mathtt5&\mathtt0\\\mathtt0&\mathtt5\end{array}\right)}$

$\displaystyle \mathtt{\left(\begin{array}{ccc}\mathtt5&\mathtt0\\\mathtt{3a-6}&\mathtt{a+3}\end{array}\right)=\left(\begin{array}{ccc}\mathtt5&\mathtt0\\\mathtt0&\mathtt5\end{array}\right)}\\ \\ \mathtt{3a-6=0\Rightarrow 3a=0+6\Rightarrow 3a=6\Rightarrow a= \frac{6}{3} \Rightarrow a=2}\\ \\ \mathtt{a+3=5\Rightarrow a=5-3\Rightarrow a=2}\\ \\ ~~~~~~~~~~~\mathtt{\mathbf{a=2}}$

$\displaystyle\mathtt{c)~A(1)\cdot X=4A(2)}\\ \\ \mathtt{A(1)=\left(\begin{array}{ccc}\mathtt1&\mathtt{-1}\\\mathtt1&\mathtt3\end{array}\right);~~~A(2)=\left(\begin{array}{ccc}\mathtt1&\mathtt{-1}\\\mathtt2&\mathtt3\end{array}\right)}\\ \\ \mathtt{\left(\begin{array}{ccc}\mathtt1&\mathtt{-1}\\\mathtt1&\mathtt3\end{array}\right)\cdot X=4\left(\begin{array}{ccc}\mathtt1&\mathtt{-1}\\\mathtt2&\mathtt3\end{array}\right)}$

$\displaystyle\underbrace{\mathtt{\left(\begin{array}{ccc}\mathtt1&\mathtt{-1}\\\mathtt1&\mathtt3\end{array}\right)}}\cdot X=\underbrace{\left(\begin{array}{ccc}\mathtt4&\mathtt{-4}\\\mathtt8&\mathtt{12}\end{array}\right)}~~~~~~~~~~~~~~~~~~~\mathtt{AX=B\Rightarrow X=A^{-1}B}\\ ~~~~~~~~\mathtt{A~~~~~~~~~~~~~~~~~~~~~~~B}$

$\displaystyle\mathtt{\mathtt{X=\left(\begin{array}{ccc}\mathtt1&\mathtt{-1}\\\mathtt1&\mathtt3\end{array}\right)^{-1}\left(\begin{array}{ccc}\mathtt4&\mathtt{-4}\\\mathtt8&\mathtt{12}\end{array}\right)~~~~~~~~~~~~~~~~~~~~A^{-1}=\frac{1}{det(A)}\cdot A^*}}\\\\ \mathtt{det(A)=\left|\begin{array}{ccc}\mathtt1&\mathtt{-1}\\\mathtt1&\mathtt3\end{array}\right|=1\cdot3-1\cdot(-1)=3+1=4}\\\\\mathtt{det(A)=4\ne0\Rightarrow inversabil\u{a}}$

$\displaystyle\mathtt{A=\left(\begin{array}{ccc}\mathtt1&\mathtt{-1}\\\mathtt1&\mathtt3\end{array}\right)\Rightarrow A^{tr}=\left(\begin{array}{ccc}\mathtt1&\mathtt1\\\mathtt{-1}&\mathtt3\end{array}\right)}\\\\ \mathtt{\delta_{11}=(-1)^{1+1}\cdot3=3;~~~~~~~~~~~~~~~~~\delta_{12}=(-1)^{1+2}\cdot (-1)=1}\\\\\mathtt{\delta_{21}=(-1)^{2+1}\cdot1=-1;~~~~~~~~~~~~~~~\delta_{22}=(-1)^{2+2}\cdot1=1}\\\\ \mathtt{A^*=\left(\begin{array}{ccc}\mathtt3&\mathtt1\\\mathtt{-1}&\mathtt1\end{array}\right)}$

$\displaystyle\mathtt{A^{-1}=\frac{1}{4}\cdot\left(\begin{array}{ccc}\mathtt3&\mathtt1\\\mathtt{-1}&\mathtt1\end{array}\right)=\left(\begin{array}{ccc}\mathtt{\displaystyle\frac{3}{4} }&\mathtt{\displaystyle\frac{1}{4} }\\\\\mathtt{\displaystyle-\frac{1}{4}}&\mathtt{\displaystyle \frac{1}{4} }\end{array}\right)}$

$\displaystyle\mathtt{X=\left(\begin{array}{ccc}\mathtt{\displaystyle\frac{3}{4}}&\mathtt{\displaystyle\frac{1}{4}}\\\\\mathtt{\displaystyle-\frac{1}{4} }&\mathtt{\displaystyle\frac{1}{4}}\end{array}\right)\left(\begin{array}{ccc}\mathtt4&\mathtt{-4}\\\mathtt8&\mathtt{12}\end{array}\right)}$

$\displaystyle\mathtt{X=\left(\begin{array}{ccc}\mathtt{\displaystyle\frac{3}{4}\cdot4+\frac{1}{4}\cdot8 }&\mathtt{\displaystyle\frac{3}{4}\cdot(-4)+\frac{1}{4}\cdot12}\\\\\mathtt{\displaystyle \left(-\frac{1}{4}\right)\cdot4+\frac{1}{4}\cdot 8 }&\mathtt{\displaystyle \left(-\frac{1}{4}\right)\cdot(-4)+\frac{1}{4}\cdot12}\end{array}\right)}$

$\displaystyle \mathtt{X=\left(\begin{array}{ccc}\mathtt{3+2}&\mathtt{-3+3}\\\mathtt{-1+2}&\mathtt{1+3}\end{array}\right)}\\\\\mathtt{X=\left(\begin{array}{ccc}\mathtt5&\mathtt0\\\mathtt1&\mathtt4\end{array}\right)}$