Matematică, întrebare adresată de mcmc, 9 ani în urmă

Rog ajutor la exercitii!!!!!!!!! exercitii cu matrice.

Anexe:

albatran: prea multe si prea grele ultimele; dacerau mai putine te ajutam

albatran: daca erau putin mai multe ti le stergeam pt probleme multiple...asa le las..cum ti-o fi norocul...dar pt ca te-ai aruncat la multe, pe ajutorul meu nu te baza! poate alti amatori

mcmc: am scris pentru cei ce stiu

mcmc: ;)

albatran: bvo, au stiut si sa redacteze frumos...acum vei sti si tu, pt.ca ai de unde invăţa...:)))))

albatran: fara gluma ,chiar daca doar vei copia cu manuta ta pe caietulde teme cata munca e acolo si tot vei invata.Raman la aprecierea, confirmata de munca altuia, ca au fost multe si grele (in fine apreciere asta as pute-o face daca as sti...) ...tot respectul meu pt Atsuko.Maximum de stele si un multumesc.Vad ca ale mele au fost primele.

albatran: adica multumescul acela de la inimioara...ia vezi, poate faci si tu o matrice linie de stelute si o inmultesti la dreapta cu numarul real "multiumesc" reprezentat algebric de o inimioara....

Răspunsuri la întrebare

Răspuns de Utilizator anonim

$\displaystyle \mathtt{1)A= \left(\begin{array}{ccc}\mathtt1&\mathtt i&\mathtt{-1}\\\mathtt0&\mathtt2&\mathtt{3i}\end{array}\right);~B=\left(\begin{array}{ccc}\mathtt i&\mathtt 1&\mathtt0\\\mathtt1&\mathtt i&\mathtt{i+1}\end{array}\right)~~~~~~~~~~~~~~~~~~~~~iA+2B=?}$

$\displaystyle \mathtt{i A=i \cdot \left(\begin{array}{ccc}\mathtt1&\mathtt i&\mathtt{-1}\\\mathtt0&\mathtt2&\mathtt{3i}\end{array}\right) = \left(\begin{array}{ccc}\mathtt i&\mathtt i^2&\mathtt{-i}\\\mathtt0&\mathtt{2i}&\mathtt{3i^2}\end{array}\right)= \left(\begin{array}{ccc}\mathtt i&\mathtt {-1}&\mathtt{-i}\\\mathtt0&\mathtt{2i}&\mathtt{-3}\end{array}\right) }\\ \\ \mathtt{iA= \left(\begin{array}{ccc}\mathtt i&\mathtt {-1}&\mathtt{-i}\\\mathtt0&\mathtt{2i}&\mathtt{-3}\end{array}\right) }$

$\displaystyle \mathtt{2B=2\cdot \left(\begin{array}{ccc}\mathtt i&\mathtt 1&\mathtt0\\\mathtt1&\mathtt i&\mathtt{i+1}\end{array}\right)=\left(\begin{array}{ccc}\mathtt {2 i}&\mathtt {2 \cdot 1}&\mathtt{2 \cdot 0}\\\mathtt{2 \cdot 1}&\mathtt {2i}&\mathtt{2(i+1)}\end{array}\right)=\left(\begin{array}{ccc}\mathtt {2i}&\mathtt 2&\mathtt0\\\mathtt2&\mathtt {2i}&\mathtt{2i+2}\end{array}\right) }$

$\displaystyle \mathtt{2B=\left(\begin{array}{ccc}\mathtt {2i}&\mathtt 2&\mathtt0\\\mathtt2&\mathtt {2i}&\mathtt{2i+2}\end{array}\right) }$

$\displaystyle \mathtt{iA+2B=\left(\begin{array}{ccc}\mathtt i&\mathtt{-1}&\mathtt{-i}\\\mathtt0&\mathtt{2i}&\mathtt{-3}\end{array}\right)+\left(\begin{array}{ccc}\mathtt {2i}&\mathtt 2&\mathtt0\\\mathtt2&\mathtt {2i}&\mathtt{2i+2}\end{array}\right)=}\\ \\ \mathtt{=\left(\begin{array}{ccc}\mathtt {i+2i}&\mathtt {-1+2}&\mathtt{-i+0}\\\mathtt{0+2}&\mathtt {2i+2i}&\mathtt{-3+2i+2}\end{array}\right)= \left(\begin{array}{ccc}\mathtt{3i}&\mathtt1&\mathtt{-i}\\\mathtt2&\mathtt{4i}&\mathtt{-1+2i}\end{array}\right)}$

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

$\displaystyle \mathtt{2) \left(\begin{array}{ccc}\mathtt{y+3x}&\mathtt{-1}\\\mathtt3&\mathtt{y-x}\end{array}\right)= \left(\begin{array}{ccc}\mathtt{x^2+1}&\mathtt{-1}\\\mathtt3&\mathtt6\end{array}\right)~~~~~~~~~~~~~~~~~~~~~~~~x,y=?}$

$\displaystyle \mathtt{ \left \{ {{y+3x=x^2+1} \atop {y-x=6}\Rightarrow y=6+x} \right. }\\ \\ \mathtt{y+3x=x^2+1 \Rightarrow 6+x+3x=x^2+1 \Rightarrow 6+x+3x-x^2-1=0 \Rightarrow } \\ \\ \mathtt{-x^2+4x+5=0 \Rightarrow x^2-4x-5=0}\\ \\ \mathtt{\Delta=(-4)^2-4 \cdot 1 \cdot (-5)=16+20=36\ \textgreater \ 0}\\ \\ \mathtt{x_1= \frac{4-6}{2}= \frac{-2}{2}=-1;~~~~x_2= \frac{4+6}{2}= \frac{10}{2}=5}$

$\displaystyle \mathtt{ \left \{ {{x=x_1} \atop {y=6+x_1}} \right. ~~~sau~~~ \left \{ {{x=x_2} \atop {y=6+x_2}} \right. }\\ \\ \mathtt{ \left \{ {{x=-1} \atop {y=6+(-1)}} \right.~~~sau~~~ \left \{ {{x=5} \atop {y=6+5}} \right. }\\ \\ \mathtt{ \left \{ {{x=-1} \atop {y=5}} \right. ~~~sau~~~ \left \{ {{x=5} \atop {y=11}} \right. }$

$\displaystyle \mathtt{ \left \{ {{x=-1} \atop {y=5}} \right. \Rightarrow \left(\begin{array}{ccc}\mathtt{5+3\cdot(-1)}&\mathtt{-1}\\\mathtt3&\mathtt{5-(-1)}\end{array}\right)=\left(\begin{array}{ccc}\mathtt{(-1)^2+1}&\mathtt{-1}\\\mathtt3&\mathtt6\end{array}\right) \Rightarrow}\\ \\ \mathtt{~~~~~~~~~~~~~\Rightarrow\left(\begin{array}{ccc}\mathtt2&\mathtt{-1}\\\mathtt3&\mathtt6\end{array}\right) =\left(\begin{array}{ccc}\mathtt2&\mathtt{-1}\\\mathtt3&\mathtt6\end{array}\right) }$

$\displaystyle \mathtt{ \left \{ {{x=5} \atop {y=11}} \right. \Rightarrow \left(\begin{array}{ccc}\mathtt{11+3\cdot 5}&\mathtt{-1}\\\mathtt3&\mathtt{11-5}\end{array}\right)=\left(\begin{array}{ccc}\mathtt{5^2+1}&\mathtt{-1}\\\mathtt3&\mathtt6\end{array}\right) \Rightarrow }\\ \\ \mathtt{~~~~~~~~~~~~\Rightarrow \left(\begin{array}{ccc}\mathtt{26}&\mathtt{-1}\\\mathtt3&\mathtt{6}\end{array}\right) = \left(\begin{array}{ccc}\mathtt{26}&\mathtt{-1}\\\mathtt3&\mathtt{6}\end{array}\right) }$

$\displaystyle \mathtt{3a)A=\left(\begin{array}{ccc}\mathtt{\sqrt{3}}&\mathtt1\\\mathtt{-3}&\mathtt{-\sqrt{3}}\end{array}\right)~~~~~~~~~~~~~~~~~~~~det(A)=?}\\ \\ \mathtt{det(A)=\left|\begin{array}{ccc}\mathtt{\sqrt{3}}&\mathtt1\\\mathtt{-3}&\mathtt{-\sqrt{3}}\end{array}\right|=\sqrt{3}\cdot \left(-\sqrt{3}\right)-1 \cdot(-3)=-3+3=0}\\ \\ \mathtt{det(A)=0}$

$\displaystyle \mathtt{b)\left(\begin{array}{ccc}\mathtt1&\mathtt{-1}&\mathtt{2i}\\\mathtt2&\mathtt1&\mathtt{3i}\\\mathtt4&\mathtt5&\mathtt{-2i}\end{array}\right)~~~~~~~~~~~~~~~~~~~~~det(A)=?}~\\ \\ \mathtt{det(A)=\left|\begin{array}{ccc}\mathtt1&\mathtt{-1}&\mathtt{2i}\\\mathtt2&\mathtt1&\mathtt{3i}\\\mathtt4&\mathtt5&\mathtt{-2i}\end{array}\right|=1\cdot1\cdot(-2i)+2i\cdot2\cdot5+(-1)\cdot3i\cdot4-}\\\\\mathtt{-2i\cdot1\cdot4-(-1)\cdot2\cdot(-2i)-1\cdot3i\cdot5=}\\\\\mathtt{=-2i+20i-12i-8i-4i-15i=-21i}\\ \\ \mathtt{det(A)=-21i}$