0
$\begingroup$

\begin{cases} 5x_1 + 3x_2 + 7x_3 -x_4 =8 \\ -2x_1 -x_2-3x_3+x_4=-2 \\ 2x_1+2x_2+2x_3+2x_4=8 \end{cases}

Could anyone help me figure out a solution to this question? Thanks a lot. I tried to transform the augmented matrix to REF, and get those xs' relations with each other. However, it seemed that there are several free variables which prevent me from figuring out a true numerical solution to this system of linear equations.

$\endgroup$
  • $\begingroup$ So, what is the RREF that you ended up with? $\endgroup$ – amd Oct 13 '17 at 5:41
0
$\begingroup$

COMMENT.-Your third equation $III$ is not independant of the other $I$ and $II$; in fact you have $$I+2(II)=III$$. Consequently you have only two independant equations $$\begin{cases} 5x_1 + 3x_2 + 7x_3 -x_4 =8 \\ -2x_1 -x_2-3x_3+x_4=-2 \end{cases}$$

which gives by elimination of $x_4$ $$3x_1+2x_2+4x_3=6$$ This is the equation of a plan of which it is known a parametrization.

$\endgroup$
1
$\begingroup$

This system can be written as the matrix equation $$\begin{bmatrix}5&3&7&-1\\-2&-1&-3&1\\2&2&2&2\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}=\begin{bmatrix}8\\-2\\8\end{bmatrix}$$ Row reduce the augmented matrix $$\begin{bmatrix}5&3&7&-1&|&8\\-2&-1&-3&1&|&-2\\2&2&2&2&|&8\end{bmatrix}$$ to obtain the solution system.

$\endgroup$
  • $\begingroup$ The OP says that he’s already gotten the RREF, but doesn’t know what to do next. $\endgroup$ – amd Oct 13 '17 at 5:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.