I am struggling with reducing an augmented matrix to reduced row echelon form.

The given linear system is \begin{cases} 3x-2z=-3 \\ -2x+z=-2 \\ -z=2 \\ \end{cases}

I can write this as an augmented matrix: \begin{array}{ccc|c} 3&0&-2&-3\\ -2&0&1&-2\\ 0&0&-1&2 \end{array}

But I am unable to get past this to the reduce row echelon form. I am also supposed to find the solution to the linear system but that isn't making any sense to me either.

  • $\begingroup$ Should it have been\begin{array}{ccc|c} 3&0&-2&-3\\ -2&0&1&-2\\ 0&0&0&-2 \end{array} $\endgroup$ – aneorddot May 3 '14 at 3:03

$$\begin{array}{ccc|c} 3&0&-2&-3\\ -2&0&1&-2\\ 0&0&-1&2 \end{array}$$ Multiply the first row by 1/3 to put a pivot at $1,1$:

$$\begin{array}{ccc|c} 1&0&-2/3&-1\\ -2&0&1&-2\\ 0&0&-1&2 \end{array}$$

Add 2 times first row to the second row to clear out the pivot column:

$$\begin{array}{ccc|c} 1&0&-2/3&-1\\ 0&0&-1/3&0\\ 0&0&-1&2 \end{array}$$

Multiply the 3rd row by -1 to put a pivot at 3,3:

$$\begin{array}{ccc|c} 1&0&-2/3&-1\\ 0&0&-1/3&0\\ 0&0&1&-2 \end{array}$$

Clear out third pivot column, first add 2/3 of the 3rd row to the first:

$$\begin{array}{ccc|c} 1&0&0&-7/3\\ 0&0&-1/3&0\\ 0&0&1&-2 \end{array}$$

Second, add 1/3 of the third row to the second to finish clearing the pivot column:

$$\begin{array}{ccc|c} 1&0&0&-7/3\\ 0&0&0&-2/3\\ 0&0&1&-2 \end{array}$$

This is the RRE form of your augmented matrix. Note that your equation never had any solutions from the start, as the RRE indicates on the second row: $0 = -2/3$. Also note that most teachers will probably think that adding extra rows and columns of zeros to a matrix is a mistake (and it is if you don't know why it is ok).


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