Question

MATRIZES:

Se o determinante da matriz abaixo
( 2 1 0 )
(k k k)
(1 2 -2)
$\: \: \binom{2 \: \: 1 \: \: 0 \: }{k \: \: k \: \: k \: } \: \\ \binom{1 \: \: 2 \: - 2}{}$
É igual a 10 ,então qual o determinante da matriz :

.. ..( 2 ) . . . ( 1 ) . . . ( 0 )
( k + 4 ) = ( k + 3 ) = ( k - 1 )
. ...( 1 ) . . . . ( 2 ) . . . ( -2 )

$\binom{2}{k \: + \: 4} = \binom{1}{k \: + \: 3} = \binom{0}{k \: - \: 1 \: } \ \: \\ \: \: \: \ \: \: \: : \: \: \: \: \: \: \: \binom{ \: \: \: 1 \: \: \: }{} \: \: \: \: \: \: \: \binom{ \: \: 2 \: \: \:}{} \: \: \: \: \: \: \: \: \: \: \: \ \binom{ - \: 2}{} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

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Answer 1

Resposta:

De acordo com o enunciado:

$\sf \left(\begin{array}{ccc}\sf 2&\sf 1&\sf 0\\\sf k&\sf k&\sf k\\\sf 1&\sf 2&\sf \!\!\!-2\end{array}\right)=A$

$\sf \left|\begin{array}{ccc}\sf 2&\sf 1&\sf 0\\\sf k&\sf k&\sf k\\\sf 1&\sf 2&\sf \!\!\!-2\end{array}\right|=det\,A$

$\sf \left|\begin{array}{ccc}\sf 2&\sf 1&\sf 0\\\sf k&\sf k&\sf k\\\sf 1&\sf 2&\sf \!\!\!-2\end{array}\right|=10$

Pela regra de Sarrus:

$\sf \left|\begin{array}{ccc}\sf 2&\sf 1&\sf 0\\\sf k&\sf k&\sf k\\\sf 1&\sf 2&\sf \!\!\!-2\end{array}\right|\begin{matrix}\sf 2&\sf 1\\\sf k&\sf k\\\sf 1&\sf 2\end{matrix}~=10$

$\sf2\cdot k\cdot(-2)+1\cdot k\cdot1+0\cdot k\cdot2-(1\cdot k\cdot(-2)+2\cdot k\cdot2+0\cdot k\cdot1)=10$

$\sf-\,4k+k+0-(-\,2k+4k+0)=10$

$\sf-\,3k-2k=10$

$\sf-\,5k=10$

$\sf k=-\dfrac{10}{5}$

$\sf k=-\,2$

Assim:

$\sf B=\left(\begin{array}{ccc}\sf 2&\sf 1&\sf 0\\\sf k+4&\sf k+3&\sf k-1\\\sf 1&\sf 2&\sf \!\!\!-2\end{array}\right)$

$\sf det\,B=\left|\begin{array}{ccc}\sf 2&\sf 1&\sf 0\\\sf k+4&\sf k+3&\sf k-1\\\sf 1&\sf 2&\sf \!\!\!-2\end{array}\right|$

$\sf det\,B=\left|\begin{array}{ccc}\sf 2&\sf 1&\sf 0\\\sf -\,2+4&\sf -\,2+3&\sf -\,2-1\\\sf 1&\sf 2&\sf \!\!\!-2\end{array}\right|$

$\sf det\,B=\left|\begin{array}{ccc}\sf 2&\sf 1&\sf 0\\\sf 2&\sf 1&\sf \!\!\!-\,3\\\sf 1&\sf 2&\sf \!\!\!-2\end{array}\right|$

$\sf det\,B=\left|\begin{array}{ccc}\sf 2&\sf 1&\sf 0\\\sf 2&\sf 1&\sf \!\!\!-\,3\\\sf 1&\sf 2&\sf \!\!\!-2\end{array}\right|\begin{matrix}\sf 2&\sf 1\\\sf 2&\sf 1\\\sf 1&\sf 2\end{matrix}$

$\sf det\,B=2\cdot1\cdot(-2)+1\cdot(-3)\cdot1+0\cdot2\cdot2-(1\cdot2\cdot(-2)+2\cdot(-3)\cdot2+0\cdot1\cdot1)$

$\sf det\,B=-\,4-3+0-(-\,4-12+0)$

$\sf det\,B=-\,7+16$

$\red{\sf det\,B=9}$

O determinante desta matriz é igual a 9.

Answer 2

Calculando o determinante da primeira matriz:

$\sf A=\left[\begin{array}{c c c}\sf2&\sf1&\sf0\\\sf k&\sf k &\sf k\\\sf 1&\sf2&\sf-2\end{array}\right]$

Temos que:

$\sf det\,A=-4k+k+0-0-4k+2k$

$\sf det\,A=-5k$

Como det A = 10, temos que:

$\sf -5k=10$

$\sf k=\dfrac{10}{-5}$

$\sf k=-2$

Sabendo que k = -2, então agora é possível calcular o determinante da segunda matriz, substituindo k por -2.

$\sf B=\left[\begin{array}{c c c}\sf 2&\sf1&\sf0\\\sf k+4&\sf k+3&\sf k-1\\\sf 1&\sf2&\sf-2\end{array}\right]$

$\sf B=\left[\begin{array}{c c c}\sf2&\sf1&\sf0\\\sf-2+4&\sf-2+3&\sf-2-1\\\sf1&\sf2&\sf-2\end{array}\right]$

$\sf B=\left[\begin{array}{c c c}\sf2&\sf1&\sf0\\\sf2&\sf1&\sf-3\\\sf1&\sf2&\sf-2\end{array}\right]$

Calculando det B, temos que:

$\sf det\,B=-4-3+0+0+12+4$

$\red{\sf det\,B=9}\leftarrow\sf resposta$