# How to solve this linear system?

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.

• In the last equation, should it be $4x - y ...$? Nov 4, 2018 at 19:22
• @hjpotter92 my bad. I edited it. Nov 4, 2018 at 19:26
• You're supposed to arrive first at an upper triangular matrix. Nov 4, 2018 at 19:35
• Hint: Find the row reduced echelon form of the augmented matrix. Nov 4, 2018 at 19:38
• Yes. It can happen. In that case, you'd get a row in the row reduced echelon matrix, consisting of 0, except for the entry corresponding to the column for constant terms. Nov 4, 2018 at 20:00

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}
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.