RREF of a matrix when the first column is all zeros

Question

I am given the following matrix and asked to convert to row reduced echelon form. $$ \begin{bmatrix} 0 & 2 & 3 & 0\\ 0 & 3 & 10 & 0\\ 0 & 0 & -1 & 0\\ \end{bmatrix} $$

The answer key states that $$ \begin{bmatrix} 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\\ \end{bmatrix} $$

Is the correct solution. My first instinct was to row reduce so I would get the following $$ \begin{bmatrix} 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ \end{bmatrix} $$ Why is the above matrix incorrect? Is it because that if there is any row with all zeros it must be at the bottom for the matrix to be in RREF?

Answer 1

All zero rows of a matrix must be at the bottom for a matrix to be considered in Row Reduced Echelon Form.

Answer 2

In short

In this RREF:

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

1. There is no relationship between columns with only 0s and rows with only 0s.

2. Rows with only 0s in the RREF are due to the fact you have more polynomials than basic variables, it is the result of a redundancy in the polynomials. You have 3 polynomials/rows, but only two basic variables/basic columns (the two other variables are free variables).

3. Columns with only 0s in the RREF, that is without pivot, corresponds to free variables, variables which can take any values independently of the basic variables. The values are not constrained by the polynomials used.

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More details

Your matrix:

\begin{bmatrix} 0 & 2 & 3 & 0 \\ 0 & 3 & 10 & 0 \\ 0 & 0 & -1 & 0 \\ \end{bmatrix}

corresponds to a system of 3 polynomials (equations if you use an augmented matrix) with 4 variables. But in this system the first and last variables have always a null coefficient, (e.g. $\small 0x+2y+3z+0w$) meaning the polynomials don't tell anything about them. So their value can't be obtained this way, they must be either considered free variables or more simply removed from your matrix:

\begin{bmatrix} 2 & 3 \\ 3 & 10 \\ 0 & -1 \\ \end{bmatrix}

This leads to the RREF:

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

The last row is null because you have more polynomials than variables, there is one redundancy. Keeping columns with 0s gives a similar RREF:

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

A column with a pivot is a basic column, it determines a basic variable. A column without pivot determines a free variable, an independent variable which can take any value. More.

Why is the above matrix incorrect?

It is not incorrect.

Is it because that if there is any row with all zeros it must be at the bottom for the matrix to be in RREF?

Rows with all 0s must be placed at the bottom to get the canonical RREF, but usually this has no impact. These rows are ignored and correspond to a redundancy in the polynomials, e.g. $\small 6x + 3y = 12$ and $\small 2x + y = 4$ will lead to a RREF with one row of 0s, which can be ignored. You could as well eliminate one of the equations upfront, this will lead to a RREF with only one row.