Question

MATRIZ

me ajudem pfv

Answer 1

Resposta: x = 2

Explicação passo a passo: Seja A = A^(t)

$\text{Temos que: } A = A^{t} \,\, \text{multiplicando pela matriz inversa.} \\A \cdot A^{-1} = \mathcal{I} = A^t \cdot A^{-1}\,\, \text{Onde} \quad \mathcal{I} = \left[\begin{array}{cc}1&0\\0&1\end{array}\right] \quad \text{Vamos agora calcular } \,\, A^{-1}\\A^{-1} =\left[\begin{array}{cc}a&b\\c&d\end{array}\right]^{-1} = \dfrac{1}{(a.d - b.c)} \cdot \left[\begin{array}{cc}d&-b\\-c&a\end{array}\right]\\ A^{-1 } = \dfrac{1}{2(3-x^3)} \left[\begin{array}{cc}2&-x^2\\-2x&3\end{array}\right]$

Continuando.

$A^{-1} = \dfrac{1}{2(3-x^3)} \left[\begin{array}{cc}2&-x^2&-2x&3\\\end{array}\right] \text{multiplicado com } \,\, A^t= \left[\begin{array}{cc}3&2x&x^2&2\\\end{array}\right]\\\\A^t \cdot A^{-1} = \dfrac{1}{2(3-x^3)}\left[\begin{array}{cc} 6 -x^4&4x-2x^2&-6x+3x^2&-4x^2+6\\\end{array}\right] = \left[\begin{array}{cc}1&0&0&1\\\end{array}\right]$$\text{Fazendo:} \quad 4x -2x^2 = 0 \quad \to \quad x = 2$