MATRIZES:
Se o determinante da matriz abaixo
( 2 1 0 )
(k k k)
(1 2 -2)
[tex] \: \: \binom{2 \: \: 1 \: \: 0 \: }{k \: \: k \: \: k \: } \: \\ \binom{1 \: \: 2 \: - 2}{}[/tex]
É igual a 10 ,então qual o determinante da matriz :
.. ..( 2 ) . . . ( 1 ) . . . ( 0 )
( k + 4 ) = ( k + 3 ) = ( k - 1 )
. ...( 1 ) . . . . ( 2 ) . . . ( -2 )
[tex] \binom{2}{k \: + \: 4} = \binom{1}{k \: + \: 3} = \binom{0}{k \: - \: 1 \: } \ \: \\ \: \: \: \ \: \: \: : \: \: \: \: \: \: \: \binom{ \: \: \: 1 \: \: \: }{} \: \: \: \: \: \: \: \binom{ \: \: 2 \: \: \:}{} \: \: \: \: \: \: \: \: \: \: \: \ \binom{ - \: 2}{} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: [/tex]
Se o determinante da matriz abaixo
( 2 1 0 )
(k k k)
(1 2 -2)
[tex] \: \: \binom{2 \: \: 1 \: \: 0 \: }{k \: \: k \: \: k \: } \: \\ \binom{1 \: \: 2 \: - 2}{}[/tex]
É igual a 10 ,então qual o determinante da matriz :
.. ..( 2 ) . . . ( 1 ) . . . ( 0 )
( k + 4 ) = ( k + 3 ) = ( k - 1 )
. ...( 1 ) . . . . ( 2 ) . . . ( -2 )
[tex] \binom{2}{k \: + \: 4} = \binom{1}{k \: + \: 3} = \binom{0}{k \: - \: 1 \: } \ \: \\ \: \: \: \ \: \: \: : \: \: \: \: \: \: \: \binom{ \: \: \: 1 \: \: \: }{} \: \: \: \: \: \: \: \binom{ \: \: 2 \: \: \:}{} \: \: \: \: \: \: \: \: \: \: \: \ \binom{ - \: 2}{} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: [/tex]
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Resposta:
De acordo com o enunciado:
[tex]\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[/tex]
[tex]\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[/tex]
[tex]\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[/tex]
Pela regra de Sarrus:
[tex]\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[/tex]
[tex]\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[/tex]
[tex]\sf-\,4k+k+0-(-\,2k+4k+0)=10[/tex]
[tex]\sf-\,3k-2k=10[/tex]
[tex]\sf-\,5k=10[/tex]
[tex]\sf k=-\dfrac{10}{5}[/tex]
[tex]\sf k=-\,2[/tex]
Assim:
[tex]\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)[/tex]
[tex]\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|[/tex]
[tex]\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|[/tex]
[tex]\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|[/tex]
[tex]\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}[/tex]
[tex]\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)[/tex]
[tex]\sf det\,B=-\,4-3+0-(-\,4-12+0)[/tex]
[tex]\sf det\,B=-\,7+16[/tex]
[tex]\red{\sf det\,B=9}[/tex]
O determinante desta matriz é igual a 9.
Verified answer
Calculando o determinante da primeira matriz:
[tex]\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] [/tex]
Temos que:
[tex]\sf det\,A=-4k+k+0-0-4k+2k[/tex]
[tex]\sf det\,A=-5k [/tex]
Como det A = 10, temos que:
[tex]\sf -5k=10 [/tex]
[tex]\sf k=\dfrac{10}{-5} [/tex]
[tex]\sf k=-2[/tex]
Sabendo que k = -2, então agora é possível calcular o determinante da segunda matriz, substituindo k por -2.
[tex]\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][/tex]
[tex]\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] [/tex]
[tex]\sf B=\left[\begin{array}{c c c}\sf2&\sf1&\sf0\\\sf2&\sf1&\sf-3\\\sf1&\sf2&\sf-2\end{array}\right] [/tex]
Calculando det B, temos que:
[tex]\sf det\,B=-4-3+0+0+12+4 [/tex]
[tex]\red{\sf det\,B=9}\leftarrow\sf resposta [/tex]