Question

usando o escalonamento ​

Answer 1

b)

$L2\leftarrow L2-L1\\\\L3\leftarrow L3+L1\\\\L4\leftarrow L4-L1\\\\\\\left\{\begin{array}{ccccccccc}x&+&y&-&z&+&t&=&0\\0&-&2y&+&2z&-&2t&=&2\\0&+&2y&+&0&+&0&=&-4\\0&-&2y&+&0&-&2t&=&-4\end{array}\right$

$L3\leftarrow L3+L2\\\\L4\leftarrow L4-L2\\\\\\\left\{\begin{array}{ccccccccc}x&+&y&-&z&+&t&=&0\\0&-&2y&+&2z&-&2t&=&2\\0&~&0&+&2z&-&2t&=&-2\\0&~&0&-&2z&+&0&=&-6\end{array}\right$

$L4\leftarrow L4+L3\\\\\\\left\{\begin{array}{ccccccccc}x&+&y&-&z&+&t&=&0\\0&-&2y&+&2z&-&2t&=&2\\0&~&0&+&2z&-&2t&=&-2\\0&~&0&~&0&-&2t&=&-8\end{array}\right$

Pela L4:

$-2t~=~-8\\\\\\t~=~\frac{-8}{-2}\\\\\\\boxed{t~=~4}$

Pela L3:

$2z~-~2~.~(4)~=~-2\\\\\\2z-8~=~-2\\\\\\2z~=~-2+8\\\\\\z~=~\frac{6}{2}\\\\\\\boxed{z~=~3}$

Pela L2:

$-2y~+~2~.~(3)~-~2~.~(4)~=~2\\\\\\-2y+6-8~=~2\\\\\\-2y~=~2+2\\\\\\y~=~\frac{4}{-2}\\\\\\\boxed{y~=~-2}$

Pela L1:

$x~+~(-2)~-~(3)~+~(4)~=~0\\\\\\x-2-3+4~=~0\\\\\\\boxed{x~=~1}$

c)

Sistema inderterminado

$L1\leftrightarrow L3\\\\\left\{\begin{array}{ccccccc}x&+&4y&+&2z&=&7\\x&-&y&+&2z&=&2\\2x&+&3y&+&4z&=&9\end{array}\right\\\\\\\\L2\leftarrow L2-L1\\\\L3\leftarrow L3-2L1\\\\\left\{\begin{array}{ccccccc}x&+&4y&+&2z&=&7\\0&-&5y&+&0&=&-5\\0&-&5y&+&0&=&-5\end{array}\right$

Pela L3 (ou L2):

$-5y~=~-5\\\\\\y~=~\frac{-5}{-5}\\\\\\\boxed{y~=~1}$

Pela L1:

$x+4y+2z~=~7\\\\\\x~+~4~.~(1)~+~2z~=~7\\\\\\x+2z~=~3\\\\\\x~=~3-2z\\\\\\Utilizando~"t"~como~parametro:\\\\\\\boxed{z~=~t}\\\\\boxed{x~=~3-2t}$