# Finding rank and nullity of 3X4 matrix

I am given

\begin{bmatrix} 1 & 1 & 1 & 1 \\[0.3em] 1 & 2 & 4 & 2 \\[0.3em] 2 & 0 & -4 & 1 \end{bmatrix}

I know that I need to get it in reduced row echelon form so I first do $-R_1 + R_2$ to get.

\begin{bmatrix} 1 & 1 & 1 & 1 \\[0.3em] 0 & 1 & 3 & 1 \\[0.3em] 2 & 0 & -4 & 1 \end{bmatrix}

Now I know that I need to get the row 3 2 into a zero so I do $-2R+1 + R_3$

\begin{bmatrix} 1 & 1 & 1 & 1 \\[0.3em] 0 & 1 & 3 & 1 \\[0.3em] 0 & -2 & -6 & -1 \end{bmatrix}

Now I need the row 3 column 2 to be zero. so $2R_2 + R_3$ \begin{bmatrix} 1 & 1 & 1 & 1 \\[0.3em] 0 & 1 & 3 & 1 \\[0.3em] 0 & 0 & 0 & 1 \end{bmatrix}

This is inconsistent, was the step I did wrong or the system wrong? What went wrong? How do I find the rank and nullity on something that is inconsistent like this?

• What do you mean $2x3$? – Amzoti Sep 22 '13 at 20:31
• @Amzoti When talking about a matrix you call them by their rows and columns, and it is always rows and then columns so I mean 2 rows by 3 columns. Obviously that is a typo. Are you confused? – Paul the Pirate Sep 22 '13 at 20:40

We've shown that the row echelon form has $3$ leading $1$'s and thus the matrix has rank $3$, and thus the Rank-Nullity Theorem implies it has nullity $1$.