# Help regarding row echelon and reduced row echelon form

I have been given three matrices and I must identify which are in row echelon or reduced row echelon form and mark the leading 1's in each row. The matrices are:

\begin{bmatrix} \bf1 & 2 & 1\\ 0 & \bf1 & 0\\ 0 & 0 & 0\\ \end{bmatrix}

\begin{bmatrix} \bf1 & -5 & 0 & 5\\ 0 & \bf1 & 3 & 2\\ \end{bmatrix}

\begin{bmatrix} 1 & 0 & 0 & 5\\ 0 & 0 & 1 & 7\\ 0 & 1 & 0 & 4\\ \end{bmatrix}

From my understanding, would I be right in saying that the first two are in reduced row echelon form and the bottom one is neither? I have highlighted what I think are the leading 1's in the top two matrices. Is this correct?

The main problem that I have is that I then have to assume that these matrices are obtained by elementary row operations from the augmented matrix of some systems of linear equations and I must determine all solutions of the corresponding systems of equations. Now this is what I do not understand on how to do?

Any help is appreciated, thanks.

Row echelon form is:

1. Rows with only zeros are at the bottom
2. Leading coefficients are to the right of leading coefficients in rows above it
3. Elements below leading coefficients are zero

For reduced row echelon form we need, in addition to row echelon form, that the following is satisfied:

1. The leading coefficients are all 1 and the only elements in their respective columns

Pay particular attention to this last one, the condition for reduced row echelon form. Is this satisfied?

To solve a system by elementary row operations, you need to do the following. Let's say your system to solve is:

$$\mathbf{Ax}=\mathbf{b}$$ where $\mathbf{A}$ is one of the matrices above. Augment the $\mathbf{b}$ vector:

$$\big[\begin{array}{c|c}\mathbf{A}&\mathbf{b}\end{array}\big]$$ and perform elementary row operations on this augmented matrix.

I'll exemplify this with your first matrix. Suppose you have:

\begin{align} \mathbf{Ax}&=\mathbf{b}\\ \begin{pmatrix}1 & 2 & 1\\ 0 & 1 & 0\\ 0 & 0 & 0\end{pmatrix}\begin{pmatrix}x_1 \\ x_2 \\ x_3\end{pmatrix}&=\begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix} \end{align} then the augmented matrix is $$\left[\begin{array}{ccc|c}1 & 2 & 1 & b_1\\ 0 & 1 & 0 & b_2\\ 0 & 0 & 0&b_3\end{array}\right].$$ Subtract two times the middle row from the first row: $$\left[\begin{array}{ccc|c}1 & 0 & 1 & b_1-2b_2\\ 0 & 1 & 0 & b_2\\ 0 & 0 & 0&b_3\end{array}\right]$$ and hence you get that \begin{align} x_1+x_3&=b_1-2b_2\\ x_2&=b_2. \end{align}