Question

Resolver a equação na variável x:
$\left[\begin{array}{cccc}x&a&a&a\\a&x&a&a\\a&a&x&a\\a&a&a&x\end{array}\right] = 0$

Answer 1

Sendo $\mathsf{A = \left[\begin{array}{cccc}x&a&a&a\\a&x&a&a\\a&a&x&a\\a&a&a&x\end{array}\right] }$
Podemos usar o teorema de Laplace:
$\mathsf{Det \ A = a_{11} A_{11} + a_{12} A_{12} + a_{13} A_{13} + a_{14} A_{14}}$ 
Vamos fazer tudo separadamente:
$\mathsf{a_{11} A_{11} = x \cdot (-1)^{1 + 1} \cdot \left|\begin{array}{ccc}x&a&a\\a&x&a\\a&a&x\end{array}\right| = x( x^{3} + 2a^{3} - 3a^{2}x) = } \\ \mathsf{x^{4} + 2a^{3}x - 3a^{2} x^{2} }$
$\mathsf{a_{12}A_{12} = a \cdot (-1)^{1 +2} \cdot \left|\begin{array}{ccc}a&a&a\\a&x&a\\a&a&x\end{array}\right| = (-a)( ax^{2} + a^{3} - 2a^{2}x)= } \\ \mathsf{ (-a)^{2} x^{2} - a^{4} + 2a^{3}x }$
 $\mathsf{a_{13}A_{13} = a \cdot (-1)^{1+ 3} \cdot \left|\begin{array}{ccc}a&x&a\\a&a&a\\a&a&x\end{array}\right| = a(2a^{2}x - a x^{2} - a^{3}) = } \\ \mathsf{2a^{3}x - a^{2} x^{2} - a^{4} }$
$\mathsf{a_{14}A_{14} = a \cdot(-1)^{1 + 4} \cdot \left|\begin{array}{ccc}a&a&a\\a&x&a\\a&a&x\end{array}\right| = (-a)(a x^{2} + a^{3} - 2a^{2}x) } \\ \mathsf{-a^{2}x^{2} - a^{4} + 2a^{3} x }$
Unindo tudo temos a seguinte solucao : 
x = a ou x = -3a