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?
$\endgroup$ 21 Answer
$\begingroup$We perform the row operations: \begin{align*} \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & 2 & 4 & 2 \\ 2 & 0 & -4 & 1 \\ \end{bmatrix} & \sim_{R_2 \gets R_2-R_1} \begin{bmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 3 & 1 \\ 2 & 0 & -4 & 1 \\ \end{bmatrix} \\ & \sim_{R_3 \gets R_3-2R_1} \begin{bmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 3 & 1 \\ 0 & -2 & -6 & -1 \\ \end{bmatrix} \\ & \sim_{R_3 \gets R_3+2R_2} \begin{bmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 3 & 1 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \\ \end{align*} to obtain the row echelon form (in agreement with the OP's work).
It doesn't really make sense to talk about consistency here; it's just a matrix, not a system of equations.
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 $1$.
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