Question

1º Escreva o sistema associado a cada equação matricial e resolva-o

a) ( 1 1 ) ( x ) ( 10 )         ( 1 1 1 ) ( x ) ( 6 )
( 0 1 ) . ( y ) = ( 8 )      b) ( 0 1 1 ) ( y ) ( 5 )
( 0 0 1) . ( z ) = ( 3 )

2º Calcule K para que (1,2,k) seja a solução de:

( 2 -1 -1 ) ( x ) ( -k )
( 1 1 2 ) ( y ) ( k² + 3 )
( 3 2 -5 ) . ( z ) = ( 7 )

Answer 1

Bom dia!

Solução!

Para resolver esse exercício você pode substituir os pontos nos sistema e igualar a zero ,ou resolves o sistema.

20)

$\begin{pmatrix} 1 & 1&1 \\ 1&-1&2 \\ \end{pmatrix}.\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix}=\begin{pmatrix} 0 \\ 0 \\ \end{pmatrix}\\\\\\\\\ \begin{cases} x+y+z=0\\ x-y+2z=0 \end{cases}\\\\\\\\\ \begin{cases} 0+0+0=0\\ 0-0+2.(0)=0 \end{cases}\\\\\\\\\ \boxed{Resposta:~~Terno~~ordenado~~(0,0,0)~~\boxed{Alternativa B}}$



21)

$Sendo~~o~~ponto~~~~(1+m,2),perceba~~que~~e~~uma~~coordenada,\\\\\ basta~~substituir~~no ~~sistema~~para~~determinar~~o valor~~de~~m.\\\\\\\\\\\ \begin{cases} x+y=5\\ 2x-3y=0 \end{cases}\\\\\\\ \begin{cases} 1+m+2=5\\ 2(1+m)-3(2)=0 \end{cases}\\\\\\\ \begin{cases} 1+m+2=5\\ 2+2m-6=0 \end{cases}\\\\\\\ m+2m+1+2-6+2=5\\\\\ 3m-6+5=5\\\\ 3m-1=5\\\\\ 3m=5+1\\\\\ 3m=6\\\\\ m= \dfrac{6}{3}\\\\\\ \boxed{m=2}$


22)

$A)\\\\\ \begin{pmatrix} 1 & 1 \\ 0 & 1 \\ \end{pmatrix}.\begin{bmatrix} x \\ y \end{bmatrix}=\begin{pmatrix} 10 \\ 8 \end{pmatrix}\\\\\\\ Sistema~~associado!\\\\\\\\ \begin{cases} x+y=10\\ 0x+y=8 \end{cases}\\\\\\ \boxed{y=8}\\\\\ x+8=10\\\\\ x=10-8\\\\\ \boxed{x=2}\\\\\\ \boxed{Resposta:~~S=\{2,8\}}$


$b)\\\\\ \begin{pmatrix} 1 & 1&1 \\ 0 & 1&1 \\ 0&0&1 \end{pmatrix}.\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix}= \begin{pmatrix} 6 \\ 5 \\ 3\\ \end{pmatrix}\\\\\\\ Sistema~~associado!\\\\\\ \begin{cases} x+y+z=6\\ ~~~~~y+z=5\\ ~~~~~~~~~~z=3 \end{cases}\\\\\\\ Veja,z~~e~~igual~~a~~tres,basta~~ir~~fazendo~~a~~substituic\~ao\\\\\ para~~determinar~~as~~outras~~variaveis.\\\\\ z=3\\\\\\\\ y+z=5\\\\ y+3=5\\\\\ y=5-3\\\\\ \boxed{y=2}$

$x+y+z=6\\\\\ x+2+3=6\\\\\ x+5=6\\\\\ x=6-5\\\\ \boxed{x=1}\\\\\\\ \boxed{Resposta:~~S=\{1,2,3\}}$

23)
$Ponto~~(1,2,k)~~so~~substiuir~~e~~determinar~~k.$

$\begin{pmatrix} 2 & -1&-1 \\ 1 & 1&2 \\ 3&2&-5 \end{pmatrix}.\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix}= \begin{pmatrix} -k \\ k^{2}+3 \\ 7 \end{pmatrix}$


$Sistema~~associado!\\\\\\\\\ \begin{cases} 2x-y-z=-k\\ x+y+2z=k^{2}+3\\ 3x+2y-5z=7 \end{cases}\\\\\\\\ Com~~o~~sistema~~montado~~substituir~~o~~ponto\\\\\\\ P(1,2,k)\\\\\ II)\\\\\\ x+y+2z=k^{2}+3\\\\\\ 1+2+2k=k^{2}+3\\\\\ 3+2k=k^{2}+3\\\\\\ k^{2}-2k=0\\\\ k(k-2)=0\\\\ k=0\\\\\\ k=2\\\\\\\boxed{k=0}\\\\\\\ \boxed{Resposta:~~k=0}$

Bom dia!
Bons estudos!