Question

dau coroană. va rog ajutor la punctul a) ex E2

Answer 1

$\displaystyle \mathtt{A=\left(\begin{array}{ccc}\mathtt1&\mathtt3\\\mathtt3&\mathtt1\\\end{array}\right),~~~B=\left(\begin{array}{ccc}\mathtt{\displaystyle- \frac{1}{2} }&\mathtt1\\\\\mathtt1&\mathtt{\displaystyle - \frac{1}{2} }\\\end{array}\right)}\\ \\ \mathtt{AB=?}$

$\displaystyle \mathtt{AB=\left(\begin{array}{ccc}\mathtt1&\mathtt3\\\mathtt3&\mathtt1\\\end{array}\right)\cdot\left(\begin{array}{ccc}\mathtt{\displaystyle- \frac{1}{2} }&\mathtt1\\\\\mathtt1&\mathtt{\displaystyle - \frac{1}{2} }\\\end{array}\right)=}$

$\displaystyle\mathtt{=\left(\begin{array}{ccc}\mathtt{\displaystyle1\cdot\left(- \frac{1}{2}\right)+3\cdot1}&\mathtt{\displaystyle1\cdot1+3\cdot\left(-\frac{1}{2}\right) }\\\\\mathtt{\displaystyle3\cdot\left(-\frac{1}{2}\right)+1\cdot 1 }&\mathtt{\displaystyle3\cdot1+1\cdot\left(-\frac{1}{2}\right) }\\\end{array}\right)=}$

$\displaystyle \mathtt{=\left(\begin{array}{ccc}\mathtt{\displaystyle-\frac{1}{2}+3 }&\mathtt{\displaystyle1-\frac{3}{2}}\\\\\mathtt{\displaystyle \left(-\frac{3}{2}\right)+1}&\mathtt{\displaystyle3-\frac{1}{2} }\\\end{array}\right)=\left(\begin{array}{ccc}\mathtt{\displaystyle \frac{5}{2} }&\mathtt{\displaystyle- \frac{1}{2} }\\\\\mathtt{\displaystyle -\frac{1}{2} }&\mathtt{\displaystyle\frac{5}{2} }\\\end{array}\right)}$

$\displaystyle \mathtt{AB=\left(\begin{array}{ccc}\mathtt{\displaystyle \frac{5}{2} }&\mathtt{\displaystyle- \frac{1}{2} }\\\\\mathtt{\displaystyle -\frac{1}{2} }&\mathtt{\displaystyle\frac{5}{2} }\\\end{array}\right)}$

$\displaystyle\mathtt{BA=?}\\\\\mathtt{BA=\left(\begin{array}{ccc}\mathtt{\displaystyle-\frac{1}{2} }&\mathtt1\\\\\mathtt1&\mathtt{\displaystyle-\frac{1}{2} }\\\end{array}\right)\cdot\left(\begin{array}{ccc}\mathtt1&\mathtt3\\\mathtt3&\mathtt1\\\end{array}\right)=}$

$\displaystyle \mathtt{=\left(\begin{array}{ccc}\mathtt{\displaystyle\left(-\frac{1}{2}\right)\cdot1+1\cdot3}&\mathtt{\displaystyle \left(- \frac{1}{2}\right)\cdot3+1\cdot1 }\\\\\mathtt{\displaystyle 1\cdot1+\left(- \frac{1}{2\right)\cdot3} }&\mathtt{\displaystyle1\cdot3+\left(- \frac{1}{2}\right)\cdot1 }\\\end{array}\right)=}$

$\displaystyle\mathtt{=\left(\begin{array}{ccc}\mathtt{\displaystyle \left(- \frac{1}{2}\right)+3 }&\mathtt{\displaystyle \left(- \frac{3}{2}\right)+1 }\\\\\mathtt{\displaystyle1- \frac{3}{2} }&\mathtt{\displaystyle3-\frac{1}{2} }\\\end{array}\right)=\left(\begin{array}{ccc}\mathtt{\displaystyle \frac{5}{2} }&\mathtt{\displaystyle - \frac{1}{2} }\\\\\mathtt{\displaystyle- \frac{1}{2} }&\mathtt{\displaystyle\frac{5}{2} }\\\end{array}\right)}$

$\displaystyle \mathtt{BA=\left(\begin{array}{ccc}\mathtt{\displaystyle \frac{5}{2} }&\mathtt{\displaystyle - \frac{1}{2} }\\\\\mathtt{\displaystyle- \frac{1}{2} }&\mathtt{\displaystyle\frac{5}{2} }\\\end{array}\right)}$

$\displaystyle \mathtt{^tA^tB=?}\\ \\ \mathtt{A=\left(\begin{array}{ccc}\mathtt1&\mathtt3\\\mathtt3&\mathtt1\\\end{array}\right)\Rightarrow ^tA=\left(\begin{array}{ccc}\mathtt1&\mathtt3\\\mathtt3&\mathtt1\\\end{array}\right)}$

$\displaystyle \mathtt{B=\left(\begin{array}{ccc}\mathtt{\displaystyle- \frac{1}{2} }&\mathtt1\\\\\mathtt1&\mathtt{\displaystyle - \frac{1}{2} }\\\end{array}\right)\Rightarrow ^tB=\left(\begin{array}{ccc}\mathtt{\displaystyle- \frac{1}{2} }&\mathtt1\\\\\mathtt1&\mathtt{\displaystyle - \frac{1}{2} }\\\end{array}\right)}$

$\displaystyle \mathtt{^tA^tB=\left(\begin{array}{ccc}\mathtt1&\mathtt3\\\mathtt3&\mathtt1\\\end{array}\right)\cdot \left(\begin{array}{ccc}\mathtt{\displaystyle- \frac{1}{2} }&\mathtt1\\\\\mathtt1&\mathtt{\displaystyle - \frac{1}{2} }\\\end{array}\right)=\left(\begin{array}{ccc}\mathtt{\displaystyle \frac{5}{2} }&\mathtt{\displaystyle- \frac{1}{2} }\\\\\mathtt{\displaystyle -\frac{1}{2} }&\mathtt{\displaystyle\frac{5}{2} }\\\end{array}\right)=AB}$

$\displaystyle \mathtt{^tA^tB=\left(\begin{array}{ccc}\mathtt{\displaystyle \frac{5}{2} }&\mathtt{\displaystyle- \frac{1}{2} }\\\\\mathtt{\displaystyle -\frac{1}{2} }&\mathtt{\displaystyle\frac{5}{2} }\\\end{array}\right)~~~~~~~~~~~~~~~~~~~~~~~~~^tA^tB=AB}$

$\displaystyle \mathtt{^tB^tA=\left(\begin{array}{ccc}\mathtt{\displaystyle- \frac{1}{2} }&\mathtt1\\\\\mathtt1&\mathtt{\displaystyle - \frac{1}{2} }\\\end{array}\right)\cdot \left(\begin{array}{ccc}\mathtt1&\mathtt3\\\mathtt3&\mathtt1\\\end{array}\right)=\left(\begin{array}{ccc}\mathtt{\displaystyle \frac{5}{2} }&\mathtt{\displaystyle - \frac{1}{2} }\\\\\mathtt{\displaystyle- \frac{1}{2} }&\mathtt{\displaystyle\frac{5}{2} }\\\end{array}\right)=BA}$

$\displaystyle \mathtt{^tB^tA=\left(\begin{array}{ccc}\mathtt{\displaystyle \frac{5}{2} }&\mathtt{\displaystyle - \frac{1}{2} }\\\\\mathtt{\displaystyle- \frac{1}{2} }&\mathtt{\displaystyle\frac{5}{2} }\\\end{array}\right)~~~~~~~~~~~~~~~~~~~~~~~~~^ tB^tA=BA}$