How to solve this linear system?

Question

I've tried solving this system of equations using Gauss-Jordan( I don't know how could I solve it any other way)

\begin{array}{c} x-y+2z-2t=-5 \\ 2x+3y-z+t=5 \\ 3x+y+2z-2t=-3 \\ 4x-y+5z-5t=-11 \end{array}

And using Gauss-Jordan, I've arrived to this matrix:

\begin{array}{rrrr|r} 1 & -1 & 2 & -2 & -5 \\ 2 & 3 & -1 & 1 & 5 \\ 1 & 0 & 1 & -1 & -2 \\ 0 & -1 & 1 & 1 & -3 \\ \end{array}

After this set of operations R3+R4, R3:7, R4-2R2, R4:7.

Answer 1

Here's how to solve it in $5$ steps. I leave it to you to find the row operations that were performed:

\begin{align}&\begin{bmatrix} \begin{array}{rrrr|r} 1&-1&2&-2&-5 \\ 2&3&-1&1&5 \\3&1&2&-2&-3\\4&-1&5&-5&-11\end{array} \end{bmatrix}\rightsquigarrow \begin{bmatrix}\begin{array}{rrrr|r} 1&-1&2&-2&-5 \\ 0&5&-5&5&15 \\0 & 4 & -4 & 4 & 12\\ 0&3&-3&3&9 \end{array}\end{bmatrix} \\ \rightsquigarrow &\begin{bmatrix}\begin{array}{rrrr|r} 1&-1&2&-2&-5 \\ 0&1&-1&1&3 \\0 & 1 & -1 & 1 & 3\\ 0&1&-1&1&3 \end{array}\end{bmatrix} \rightsquigarrow \begin{bmatrix}\begin{array}{rrrr|r} 1&-1&2&-2&-5\\ 0&1&-1&1&3 \\0 & 0&0&0&0\\ 0&0&0&0&0 \end{array}\end{bmatrix} \\[1ex] \rightsquigarrow & \begin{bmatrix}\begin{array}{rrrr|r} 1&0&1&-1&-2\\ 0&1&-1&1&3 \\0 & 0&0&0&0\\ 0&0&0&0&0 \end{array}\end{bmatrix} \ \end{align}

Answer 2

COMMENT.-The given equations are not independent. You proved it if you look carefully to get what equation is a linear combination of the other ones. Or if you want, calculating the determinant of the matrix to get that $$\det\begin{pmatrix} 1 & -1 & 2&-2 \\ 2 & 3 & -1&1 \\ 3 & 1 & 2&-2\\4&-1&5&-5 \end{pmatrix}=0$$ You have then an infinity of solutions(or maybe none).

You can get $x+y=1$, in particular.