Finding the Reduced Row Echelon Form of a Matrix

We will perform Gauss-Jordan Elimination to obtain the Reduced Row Echelon Form of the given matrix.

Recall that we can perform any of the following three row operations on the matrix:

1. Switch two rows.
2. Multiply a row by any non-zero constant.
3. Add a scalar multiple of one row to any other row.

The top left element is a pivot, so the rest of the elements in the first column must be zero. We can subtract multiples of first row from second and third row so that the first column becomes $\left[ \begin{array}{c} 1 \\ 0 \\ 0 \\ \end{array}\right]$ .

$\left[ \begin{array}{c c c} 1 & 4 & -5\\ 6 & 4 & 0 \\ 2 & 2 & -1 \\ \end{array}\right] \ce{->[\large R_{2} - 6R_{1}]} \left[ \begin{array}{c c c} 1 & 4 & -5\\ 0 & -20 & 30 \\ 2 & 2 & -1 \\ \end{array}\right] \ce{->[\large R_{3} - 2R_{1}]} \left[ \begin{array}{c c c} 1 & 4 & -5\\ 0 & -20 & 30 \\ 0 & -6 & 9 \\ \end{array}\right]$

Now that the leftmost column is $\left[ \begin{array}{c} 1 \\ 0 \\ 0 \\ \end{array}\right]$ , the middle element can be made 1 by dividing the second row by -20 .

$\left[ \begin{array}{c c c} 1 & 4 & -5\\ 0 & -20 & 30 \\ 0 & -6 & 9 \\ \end{array}\right] \ce{->[\large \text{Divide the second row by } -20]} \left[ \begin{array}{c c c} 1 & 4 & -5\\ 0 & 1 & -1.5 \\ 0 & -6 & 9 \\ \end{array}\right]$

We will now make the second column of the matrix $\left[ \begin{array}{c} 0 \\ 1 \\ 0 \\ \end{array}\right]$ by adding scalar multiples of the second row to the first and third rows.

$\left[ \begin{array}{c c c} 1 & 4 & -5\\ 0 & 1 & -1.5 \\ 0 & -6 & 9 \\ \end{array}\right] \ce{->[\large R_{1} - 4R_{2}]} \left[ \begin{array}{c c c} 1 & 0 & 1\\ 0 & 1 & -1.5 \\ 0 & -6 & 9 \\ \end{array}\right] \ce{->[\large R_{3} + 6R_{2}]} \left[ \begin{array}{c c c} 1 & 0 & 1\\ 0 & 1 & -1.5 \\ 0 & 0 & 0 \\ \end{array}\right]$

We have now completed Gauss-Jordan elimination and have obtained the Reduced Row Echelon Form of the given matrix. $\text{rref}(A) = \left[ \begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & -1.5 \\ 0 & 0 & 0 \\ \end{array} \right]$

The sum of the entries is $1 + 1 + 1 - 1.5 = \boxed{1.5}$ $_\square$