Question

ME AJUDEM PFVR! 4. (UECE – 2006) O determinante da matriz é nulo para um valor de x situado no intervalo:
$\left[\begin{array}{ccc}2x&3&1\\1&0&4\\0&1&x-1\end{array}\right]$

Answer 1

Pelo Teorema de Laplace, temos 

$\left|\begin{array}{ccc}2x&3&1\\1&0&4\\0&1&x-1\end{array}\right|=\\\\\\=\ell_{11}a_{11}\left|\begin{array}{ccc}0&4\\1&x-1\end{array}\right|+\ell_{12}a_{12}\left|\begin{array}{ccc}3&1\\1&x-1\end{array}\right|+\ell_{13}a_{13}\left|\begin{array}{ccc}3&1\\0&4\end{array}\right|$

Como $a_{13}=0$, resumimos o cálculo em

$\left|\begin{array}{ccc}2x&3&1\\1&0&4\\0&1&x-1\end{array}\right|=\ell_{11}a_{11}\left|\begin{array}{ccc}0&4\\1&x-1\end{array}\right|+\ell_{12}a_{12}\left|\begin{array}{ccc}3&1\\1&x-1\end{array}\right|$

Onde $\ell_{ij}=(-1)^{i+j}$ (coeficientes de Laplace), e $a_{11}=2x,~a_{12}=1$

Portanto:

$\left|\begin{array}{ccc}2x&3&1\\1&0&4\\0&1&x-1\end{array}\right|=(-1)^{1+1}2x\left|\begin{array}{ccc}0&4\\1&x-1\end{array}\right|+(-1)^{1+2}\left|\begin{array}{ccc}3&1\\1&x-1\end{array}\right|\\\\\\=2x\left|\begin{array}{ccc}0&4\\1&x-1\end{array}\right|-\left|\begin{array}{ccc}3&1\\1&x-1\end{array}\right|\\\\\\=2x\cdot[0(x-1)-4(1)]-[3(x-1)-1(1)]\\\\=2x\cdot[0-4]-[3x-3-1]\\\\=2x\cdot(-4)-(3x-4)\\\\=-8x-3x+4\\\\=-11x+4$

Portanto, teremos que

$\left|\begin{array}{ccc}2x&3&1\\1&0&4\\0&1&x-1\end{array}\right|=-11x+4=0~\Leftrightarrow~\boxed{\boxed{x=\dfrac{4}{11}}}$

Basta verificar qual intervalo das alternativas contém esse valor de $x$