Solve the linear system by Gauss - Jordan elimination

Question

$$ \begin{align} x& - y + 2z - w &= -1\\ 2x& + y - 2z - 2w &= -2\\ -x& + 2y - 4z + w &= 1\\ 3x& -3w &= -3 \end{align} $$

\begin{bmatrix}1&-1&2&-1&-1\\2&1&-2&-2&-2\\-1&2&4&1&1\\3&0&0&-3&-3\end{bmatrix}

I did -3p1 + p4. I did (1/3) * p4. I did 2 * (p3) + p2. I did (1/5) * p2. I did p1 + p3.

\begin{bmatrix}1&-1&2&-1&-1\\0&1&-2&-0&0\\0&0&0&0&0\\0&0&0&0&0\end{bmatrix}.

Answer: First : w -x + 2y -z = 1.

Second: x - 2y = 0.

Answer 1

You've got the right idea. Remember, we want to make sure that: $\\$

1. The leftmost nonzero entry of a row is equal to 1. $\\$
2. The leftmost nonzero entry of a row is the only nonzero entry in its column. $\\$
3. If there is a row where every entry is zero, then that row lies below all other rows that contain a nonzero entry $\\$

\begin{bmatrix} 1 & -1 & 2 & -1 & -1 \\[0.3em] 2 & 1 & -2 & -2 & -2 \\[0.3em] -1 & 2 & 4 & 1 & 1 \\[0.3em] 3 & 0 & 0 & -3 & -3 \end{bmatrix} $\\$ $R_{2}$ - $2R_{1}$ | $R_{3}$ + $R_{1}$ | $R_{4}$ - $3R_{1}$ \begin{bmatrix} 1 & -1 & 2 & -1 & -1 \\[0.3em] 0 & 3 & -6 & 0 & 0 \\[0.3em] 0 & 1 & 6 & 0 & 0 \\[0.3em] 0 & 3 & -6 & 0 & 0 \end{bmatrix} $\\$ $\frac{1}{3}$$R_{2}$ | swap $R_{3}$,$R_{4}$ | $\frac{1}{3}$$R_{3}$ \begin{bmatrix} 1 & -1 & 2 & -1 & -1 \\[0.3em] 0 & 1 & -2 & 0 & 0 \\[0.3em] 0 & 1 & -2 & 0 & 0 \\[0.3em] 0 & 1 & 6 & 0 & 0 \end{bmatrix} $\\$ $R_{1}$ + $R_{2}$ | $R_{3}$ - $R_{2}$ | $R_{4}$ - $R_{2}$ \begin{bmatrix} 1 & 0 & 0 & -1 & -1 \\[0.3em] 0 & 1 & -2 & 0 & 0 \\[0.3em] 0 & 0 & 0 & 0 & 0 \\[0.3em] 0 & 0 & 8 & 0 & 0 \end{bmatrix} $\\$ $\frac{1}{8}$$R_{4}$ | $R_{2}$ + $2R_{4}$ \begin{bmatrix} 1 & 0 & 0 & -1 & -1 \\[0.3em] 0 & 1 & 0 & 0 & 0 \\[0.3em] 0 & 0 & 0 & 0 & 0 \\[0.3em] 0 & 0 & 1 & 0 & 0 \end{bmatrix} $\\$ swap $R_{3}$, $R_{4}$ \begin{bmatrix} 1 & 0 & 0 & -1 & -1 \\[0.3em] 0 & 1 & 0 & 0 & 0 \\[0.3em] 0 & 0 & 1 & 0 & 0 \\[0.3em] 0 & 0 & 0 & 0 & 0 \end{bmatrix} $\\$

The system is consistent since there is at least one solution. In fact, there are infinite solutions since we have a free variable $x_{4}$. The solution to this linear system can be written as

$\begin{pmatrix} -1 + x_{4}\\ 0\\ 0\\ x_{4}\end{pmatrix} | x_{4} \in\ S\\$