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JimmHEX
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Pe: 2020-11-05T12:02:51+02:00 In: In:

tema matematica det

tema matematica det

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    1.\det A=4\cdot(-3)-6\cdot(-2)=-12+12=0

    2.Din teorema lui Hamilton-Cayley, stim ca pentru orice matrice A de 2×2 are loc relatia:
    A^2=\text{Tr}A\cdot A-\det A\cdot I_2=(4+1)A-(4-2\cdot3)I_2=5A+2I_2
    Inlocuind, obtinem:
    \det(A^2-2A-I_2)=\det(5A+2I_2-2A-I_2)=\det(3A-I_2)=\det(\begin{pmatrix} 12&9\\6&3 \end{pmatrix}-I_2)=\det\begin{pmatrix}11&9\\6&2\end{pmatrix}=2*11-6*9=22-54=-32

    3.\det(A\cdot A^t)=\det(\begin{pmatrix} -1&2&2\\2&2&-1 \end{pmatrix}\cdot \begin{pmatrix}-1&2\\2&2\\2&-1\end{pmatrix})=\det(\begin{pmatrix}9&0\\0&9\end{pmatrix})=9\cdot9=81

    4.Adunam ultima linie peste prima, apoi dezvoltam in functie de prima linie:
    \begin{vmatrix}1&-1&-1\\-1&1&-1\\-1&-1&1\end{vmatrix}=\begin{vmatrix}0&-2&0\\-1&1&-1\\-1&-1&1\end{vmatrix}=(-2)\cdot(-1)^{1+2}\begin{vmatrix}-1&-1\\-1&1\end{vmatrix}=2(-1-1)=-4

    5.Dezvoltam dupa prima linie:
    1\cdot(-1)^{1+1}(x-1)+1\cdot(-1)^{1+2}(1-x)+x\cdot(-1)^{1+3}(1-x)=x-1-(1-x)+x(1-x)=x-1-1+x+x-x^2=-x^2+3x-2=-(x^2-3x+2)=-(x-1)(x-2)
    Produsul este 0 daca cel putin unul dintre termeni este egal cu 0, deci avem solutiile 1 si 2.

    6.2(x^2+2x)-5(x+3)=-6x+x+1
    2x^2+4x-5x-15=-5x+1
    2x^2+4x-16=0
    x^2+2x-8=0
    Solutiile sunt -4 si 2(din relatii lui viete, -4+2=-2 si -4*2=8).

    7.\det(A+B)+\det(AB)=\det(A+B)+\det(A)\cdot\det(B)=\begin{vmatrix}7&-2\\4&6\end{vmatrix}+(14+6)(-5-2)=42-(-8)+20\cdot-7=42+8-140=50-140=-90

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