Home/ Intrebari/Q 85482
Urmator
In Process

AniDeȘcoală.ro Latest Intrebari

Marius963
Pe: 2014-02-17T16:01:22+02:00 In: In:

matrice

Pentru X apartine C se considera matricea A(x)= x+1 x^2-1
( ) ap M2(C)
1 x-1

Să se determine toate numerele complexe x pentru care (A(x))^4+(A(x))^2=O2

Similare

1 raspuns

  1. xor_NTG
    maestru (V)

        \[ \begin{array}{l} A\left( x \right) = \left( {\begin{array} {x + 1} & {x^2 - 1} \\ 1 & {x - 1} \\ \end{array}} \right) \\ A\left( x \right)^4 + A\left( x \right)^2 = O_2 \\ A\left( x \right)^2 = \left( {\begin{array} {x + 1} & {x^2 - 1} \\ 1 & {x - 1} \\ \end{array}} \right)\left( {\begin{array} {x + 1} & {x^2 - 1} \\ 1 & {x - 1} \\ \end{array}} \right) = \left( {\begin{array} {\left( {x + 1} \right)^2 + x^2 - 1} & {\left( {x + 1} \right)\left( {x^2 - 1} \right) + \left( {x^2 - 1} \right)\left( {x - 1} \right)} \\ {x + 1 + x - 1} & {x^2 - 1 + \left( {x - 1} \right)^2 } \\ \end{array}} \right) \\ = \left( {\begin{array} {x^2 + 2x + 1 + x^2 - 1} & {\left( {x^2 - 1} \right)\left( {x + 1 + x - 1} \right)} \\ {2x} & {x^2 - 1 + x^2 - 2x + 1} \\ \end{array}} \right) = \left( {\begin{array} {2x^2 + 2x} & {2x\left( {x^2 - 1} \right)} \\ {2x} & {2x^2 - 2x} \\ \end{array}} \right) = 2x \cdot A\left( x \right) \\ A\left( x \right)^3 = A\left( x \right)^2 \cdot A\left( x \right) = 2x \cdot A\left( x \right) \cdot A\left( x \right) = 2x \cdot A\left( x \right)^2 = 2x \cdot 2x \cdot A\left( x \right) = 4x^2 A\left( x \right) \\ A\left( x \right)^4 = A\left( x \right)^3 \cdot A\left( x \right) = 4x^2 A\left( x \right)A\left( x \right) = 4x^2 A\left( x \right)^2 = 4x^2 2x \cdot A\left( x \right) = 8x^3 A\left( x \right) \\ \end{array} \]

    Acum revenim expresia initială:

        \[ \begin{array}{l} A\left( x \right)^4 + A\left( x \right)^2 = O_2 \Leftrightarrow 8x^3 A\left( x \right) + 2x \cdot A\left( x \right) = O_2 \Leftrightarrow 2x\left( {4x^2 + 1} \right)A\left( x \right) = O_2 \Leftrightarrow \\ \Leftrightarrow 2x\left( {4x^2 + 1} \right)\left( {\begin{array} {x + 1} & {x^2 - 1} \\ 1 & {x - 1} \\ \end{array}} \right) = \left( {\begin{array} 0 & 0 \\ 0 & 0 \\ \end{array}} \right) \Rightarrow \left\{ \begin{array}{l} 2x\left( {4x^2 + 1} \right)\left( {x + 1} \right) = 0 \\ 2x\left( {4x^2 + 1} \right)\left( {x - 1} \right)\left( {x + 1} \right) = 0 \\ 2x\left( {4x^2 + 1} \right) = 0 \\ 2x\left( {4x^2 + 1} \right)\left( {x - 1} \right) = 0 \\ \end{array} \right. \\ \end{array} \]

    Din a doua ecuatie, avem că:

        \[ 2x\left( {4x^2 + 1} \right)\left( {x - 1} \right)\left( {x + 1} \right) = 0 \Leftrightarrow \left\{ \begin{array}{l} 2x = 0\,\,\,\, \vee \\ 4x^2 + 1 = 0\,\,\, \vee \\ x - 1 = 0\,\,\, \vee \\ x + 1 = 0\, \\ \end{array} \right. \Leftrightarrow x \in \left\{ { - 1,0,1,\frac{1}{4}i} \right\} \]

Raspunde

Raspunde