Checking a Linear Algebra result

[Char. Limit]

Homework Statement

Yeah, so I'm just starting LinAlg, and I wanted to check my answer on a certain question. The question reads:

"Describe the possible echelon forms of a nonzero 3x2 matrix. Use the symbols \diamond, *, and 0, as in the first part of example 1."

Now, here, the \diamond indicates a nonzero entry, the * indicates an entry that can be any value, and the 0 indicates, well, a zero entry.

The Attempt at a Solution

So I got four matrices:

\left[ \begin{array}{cc}<br /> \diamond &amp; 0 \\<br /> 0 &amp; 0 \\<br /> 0 &amp; 0 \\<br /> \end{array} \right]

\left[ \begin{array}{cc}<br /> \diamond &amp; * \\<br /> 0 &amp; \diamond \\<br /> 0 &amp; 0 \\<br /> \end{array} \right]

\left[ \begin{array}{cc}<br /> \diamond &amp; * \\<br /> 0 &amp; * \\<br /> 0 &amp; \diamond \\<br /> \end{array} \right]

\left[ \begin{array}{cc}<br /> \diamond &amp; * \\<br /> 0 &amp; * \\<br /> 0 &amp; * \\<br /> \end{array} \right]

However, I feel like I'm missing some. Can you help me with which ones?

 

[vela]

In echelon form, the leading entry in a row has to be strictly to the right of the leading entry of the row above it, so your third and fourth matrices wouldn't generally be in echelon form.

Another case you might consider is when the first column has no leading entry.

 

[Char. Limit]

Do you mean that the next leading entry can't be two rows down from the previous one? I... didn't honestly remember that.

And I'll take a look at the cases of "no leading entry in the first row". Thanks, I knew I had forgotten something.

 

[vela]

No, it's more that you can't have two rows where the first non-zero element is in the same column. The matrix

\begin{bmatrix} 1 &amp; 1 \\ 0 &amp; 2 \\ 0 &amp; 3 \end{bmatrix}

is not in echelon form though the matrix would qualify as your fourth type.

 

[Char. Limit]

So, I have this as my answer now:

\left[ \begin{array}{cc}<br /> \diamond &amp; 0 \\<br /> 0 &amp; 0 \\<br /> 0 &amp; 0 \\<br /> \end{array} \right]

\left[ \begin{array}{cc}<br /> \diamond &amp; * \\<br /> 0 &amp; \diamond \\<br /> 0 &amp; 0 \\<br /> \end{array} \right]

\left[ \begin{array}{cc}<br /> 0 &amp; \diamond \\<br /> 0 &amp; 0 \\<br /> 0 &amp; 0 \\<br /> \end{array} \right]

I do wonder though, is an all-zero matrix in echelon form?

 

[vela]

Looks good.

An all-zero matrix is in echelon form. It satisfies all the requirements. There are no leading entries, so there's nothing to worry about there. And all the zero rows are at the bottom.

 

[Char. Limit]

I'd better add it then. Thanks for the help.

 

[vela]

The problem specified non-zero matrices, so you don't need the 0 matrix.

 

[Char. Limit]

Oh, of course. How foolish of me.

I guess I need to read the problem more carefully...