Problem solving a 4 by 4 linear system with infinite solutions

Question

I have a question here. The following is a 4 by 4 system of linear equations which has infinitely many solutions.

$$ \left\{ \begin{array}{c} x+y+3z+t=0 \\ x-y-z-t=0 \\ 3x+y+5z+3t=0 \\ x+5y+11z+8t=0 \end{array} \right. $$

An online calculator gave me these solutions:
$t = 0, x = - r_1, y = -2 r_1, z = r_1$, where $r_1$ is a parameter.

However, how did they solve this? It's kinda complicated for me, because what I'd do is I'd isolate "$x$" from the second equation $(x=y+z+t)$, then I would plug that into the first equation and so on...

Can someone help? Thank you and sorry for taking your time !

Answer 1

Use the matrix of the linear system and put it in reduced row echelon form: \begin{align} &\left[\begin{array}{rrrr} 1&1&3&1\\1&-1&-1&-1\\3&1&5&3\\1&5&11&8 \end{array}\right]\rightsquigarrow \left[\begin{array}{rrrr} 1&1&3&1\\0&-2&-4&-2\\0&-2&-4&0\\0&4&8&7 \end{array}\right]\rightsquigarrow \left[\begin{array}{rrrr} 1&1&3&1\\0&-2&-4&-2\\0&0&0&2\\0&0&0&3 \end{array}\right]\rightsquigarrow \left[\begin{array}{rrrr} 1&1&3&1\\0&1&2&1\\0&0&0&1\\0&0&0&3 \end{array}\right] \\[1ex] &{}\rightsquigarrow\left[\begin{array}{rrrr} 1&1&3&0\\0&1&2&0\\0&0&0&1\\0&0&0&0 \end{array}\right]\rightsquigarrow \left[\begin{array}{rrrr} 1&0&1&0\\0&1&2&0\\0&0&0&1\\0&0&0&0 \end{array}\right]. \end{align}

Answer 2

Substitute $x=y+z+t$ in the first equation and you have

$2y+4z+2t=0 \\ \Rightarrow y+2z=-t$

Substitute $x=y+z+t$ in the third equation and you have

$4y+8z+6t=0$

Then substitute $y+2z=-t$ and you get $t=0$. So $y=-2z$ and $x=y+z$.

Substitute $x=y+z$, $y=-2z$ and $t=0$ in the fourth equation and you have

$0=0$

So the fourth equation tells us nothing new (this is because the fourth equation is a linear combination of the other three equations).

So we have $y=-2z$ and $x=y+z=-z$.