Question

Îmi poate explica cineva de ce rezulta matricea respectivă?

Answer 1

   
[tex]\displaystyle\\ \text{Regula adunarii matricilor:}\\\\ \left(\begin{array}{cc}a_{11}&a_{12}\\a_{21}&a_{22}\end{array}\right)+ \left(\begin{array}{cc}b_{11}&b_{12}\\b_{21}&b_{22}\end{array}\right)= \left(\begin{array}{cc}a_{11}+b_{11}&a_{12}+b_{12}\\a_{21}+b_{21}&a_{22}+b_{22}\end{array}\right)\\\\ \text{unde }a_{ij} \text{ sunt numere sau expresii.}\\ \text{Practic se aduna elementele de pe aceeasi pozitie ale matricilor.}\\ [/tex]


[tex]\displaystyle\\ \text{Rezolvare:}\\\\ \left(\begin{array}{cc} \dfrac{1+1}{2} &\dfrac{1-1}{2} \\\\\dfrac{1-1}{2}&\dfrac{1+1}{2}\end{array}\right)+ \left(\begin{array}{cc} \dfrac{2+1}{2} &\dfrac{2-1}{2} \\\\\dfrac{2-1}{2}&\dfrac{2+1}{2}\end{array}\right)+...+ \left(\begin{array}{cc} \dfrac{n+1}{2} &\dfrac{n-1}{2} \\\\\dfrac{n-1}{2}&\dfrac{n+1}{2}\end{array}\right) [/tex]


[tex]\displaystyle\\ \text{Adunam termenii din stanga sus de la toate matricile.}\\ a_{11}+b_{11}+c_{11} + ....\\\\ \frac{1+1}{2}+\frac{2+1}{2}+...+\frac{n+1}{2}= \frac{2}{2}+\frac{3}{2}+...+\frac{n+1}{2}\\\\ \text{Calculam numarul de termeni:}\\ Nr. = n+1 - 2 + 1 = \boxed{\bf n~termeni}\\ \text{Rezulta ca avem n matrice care se aduna.}\\\\ \text{Aplicam formula lui Gauss}\\\\ \frac{2}{2}+\frac{3}{2}+...+\frac{n+1}{2}=\\\\ =\frac{1}{2}(2+3+...+n+1)=\frac{1}{2}\times \frac{n(n+1+2)}{2}=\frac{n(n+3)}{4} [/tex]


[tex]\displaystyle\\ \text{Adunam termenii din dreapta sus de la toate matricile.}\\ a_{12}+b_{12}+c_{12} + ....\\\\ \frac{1-1}{2}+\frac{2-1}{2}+...+\frac{n-1}{2}= \frac{0}{2}+\frac{1}{2}+...+\frac{n-1}{2}\\\\ \text{Nu mai alculam numarul de termeni deoarece stim ca sunt n matrici.}\\ \text{Aplicam formula lui Gauss}\\\\ \frac{0}{2}+\frac{1}{2}+...+\frac{n-1}{2}=\\\\ =\frac{1}{2}(0+1+...+n-1)=\frac{1}{2}\times \frac{n(n-1+0)}{2}=\frac{n(n-1)}{4} [/tex]


[tex]\displaystyle\\ \text{Suma termenilor din stanga jos este egala cu suma termenilor }\\ \text{din dreapta sus, deoarece se insumeaza aceeiasi termeni.} \\\\ \text{Suma termenilor din dreapta jos este egala cu suma termenilor }\\ \text{din stanga sus, deoarece se insumeaza aceeiasi termeni.} \\\\ \text{Rezulta ca matricea rezultanta este: }\\\\ \left(\begin{array}{cc} \dfrac{n(n+3)}{4} &\dfrac{n(n-1)}{4} \\\\\dfrac{n(n-1)}{4}&\dfrac{n(n+3)}{4}\end{array}\right)[/tex]