# Having trouble understand Row Echelon Form

I'm having trouble understanding Row Echelon Form. I'm trying to solve the system

$-2x - 10y - 29z = 5$

$-4x - 19y -56z = -3$

$x + 5y + 15z = 3$

it has the solution $x = ? , y = ? , z = ?$

I put the system into a augmented matrix; I then performed the row operations:

1) $R_2 = 2R_1 - R_2$

2) $R_3 = R_1 + 2R_3$

3) $R_1 = R_1/-2$

4) $R_2 = R_2/-1$

I obtained: $\begin{bmatrix}1&5&29/2&5\\0&1&-27&-8\\0&0&1&8\end{bmatrix}$

REF says :

1) All nonzero rows are above any rows of all zeros

2) Each leading entry of a row is in a column to the right of leading entry above it

3) ALL entries in a column below a leading entry is zero

I'm satisfying all of these requirements for REF form am I not? Why then is $z = 8$ not true??

You made an arithmetic error in your second step. You apparently forgot to multiply the final column entry by two when calculating $R_3=R_1+2R_3$.

$\begin{array}{ccccc}&-2&-10&-29&5\\+2(&1&5&15&3)\\\hline &-2&-10&-29&5\\+&2&10&30&\color{red}{6}\\\hline &0&0&1&11\end{array}$

You appear to have done $5+3$ instead of $5+2\cdot 3$ for the final entry.

The full row reduction process using the same proposed steps that you suggest

$\begin{bmatrix}-2&-10&-29&5\\-4&-19&-56&-3\\1&5&15&3\end{bmatrix}$

$2R_1-R_2\mapsto R_2$

$\begin{bmatrix}-2&-10&-29&5\\0&-1&-2&13\\1&5&15&3\end{bmatrix}$

$2R_3+R_1\mapsto R_3$

$\begin{bmatrix}-2&-10&-29&5\\0&-1&-2&13\\0&0&1&11\end{bmatrix}$

$-\frac{1}{2}R_1\mapsto R_1$ and $-1R_2\mapsto R_2$

$\begin{bmatrix} 1&5&\frac{29}{2}&-\frac{5}{2}\\0&1&2&-13\\0&0&1&11\end{bmatrix}$

This should have been the matrix you arrived at by the end of your work shown so far. Notice second row third column entry and all fourth column entries are incorrect due to arithmetic mistakes.

Continuing: $R_2-2R_3\mapsto R_2$ $\begin{bmatrix}1&5&\frac{29}{2}&-\frac{5}{2}\\0&1&0&-35\\0&0&1&11\end{bmatrix}$ $R_1-5R_2-\frac{29}{2}R_3\mapsto R_1$ $\begin{bmatrix}1&0&0&13\\0&1&0&-35\\0&0&1&11\end{bmatrix}$

• Wow, #facepalmself – Shammy Jun 23 '16 at 16:42
• @Shammy depending on the level of accuracy required, a mistake like this can usually be mostly forgiven in a testing scenario in math. Out of 15 marks, I would mark off at most one for this mistake. Unfortunately, in some fields, accuracy is necessary. I understand arithmetic mistakes suffer large penalties in engineering exams. – JMoravitz Jun 23 '16 at 16:46
• I think I've made another mistake because now RREF solutions aren't coming out right for $x$ and $y$ I did the operations $1)R_2 = R_2 + 27R_3$ ... $2)R_1 = R_1 + -29/2R_3$... $3)R_1 = R_1 + -5R_2$.. To get the RREF of: $\begin{bmatrix}1&0&0&-3209/2\\0&1&0&289\\0&0&1&11\end{bmatrix}$ so that the solutions were $x = -3209/2, y = 289, z = 11$ for some reason, only $z$ is correct. Is my algebra wrong again? – Shammy Jun 23 '16 at 16:55
• @shammy it would appear so. wolfram – JMoravitz Jun 23 '16 at 16:57
• @shammy any time you have multiple equations and unknowns. Finding best fit curves to fit data. Wikipedia may provide more historical uses. Honestly, calculators and software can do most things better than humans, but understanding how to do it without aid still has some merit – JMoravitz Jun 23 '16 at 17:36