Răspuns :
[tex]\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=?}[/tex]
[tex]\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)=}[/tex]
[tex]\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)=} [/tex]
[tex]\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)} [/tex]
[tex]\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)}[/tex]
[tex]\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)=}[/tex]
[tex]\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)=}[/tex]
[tex]\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)} [/tex]
[tex]\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)}[/tex]
[tex]\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)} [/tex]
[tex]\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)}[/tex]
[tex]\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}[/tex]
[tex]\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}[/tex]
[tex]\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}[/tex]
[tex]\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}[/tex]
[tex]\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)=}[/tex]
[tex]\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)=} [/tex]
[tex]\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)} [/tex]
[tex]\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)}[/tex]
[tex]\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)=}[/tex]
[tex]\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)=}[/tex]
[tex]\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)} [/tex]
[tex]\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)}[/tex]
[tex]\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)} [/tex]
[tex]\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)}[/tex]
[tex]\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}[/tex]
[tex]\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}[/tex]
[tex]\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}[/tex]
[tex]\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}[/tex]
Vă mulțumim pentru vizita pe site-ul nostru dedicat Matematică. Sperăm că informațiile disponibile v-au fost utile și inspiraționale. Dacă aveți întrebări sau aveți nevoie de suport suplimentar, suntem aici pentru a vă ajuta. Ne face plăcere să vă revedem și vă invităm să adăugați site-ul nostru la favorite pentru acces rapid!