Getting matrices in to echlon form

[colerelm]

When using gaussian elimination to reduce a matrix to its echlon form, I clearly understand why the steps are taken when I see the answer, but I am very weak at coming up with my own steps when trying to reduce a matrix to echlon form. Here is an example of what a problem I am currently working on looks like:



Is there any tricks/strategies I should be using when trying to reduce a matrix? I always end up taking the wrong steps and having to start over. ANY help would be appreciated.

Thanks

 

[mmm4444bot]

I'm not following the first row operation, and I can see only the matrices; your descriptions of the steps are too small.

There's more than one echelon form: row echelon form (ref) and reduced row echelon form (rref). The result you posted is ref.

There may be tricks (foresight that comes with experience, really) for reducing the number of steps with some matrices, but I've never paid much attention to shortcuts unless its obvious.

Gaussian Elimination is very mechanical. I always start by getting a 1 in the upper-left (row1,col1), and then I use that 1 to zero-out all the entries below it. Next, I get a 1 in row2,col2 and use it to zero-out all the entries below it. And so on, with row3,col3.

Here are the row operations that I would have done on your matrix.

-2R1 + R2 --> R2

-R1 + R3 --> R3

(1/7)R2 --> R2

-6R2 + R3 --> R3

 

[HallsofIvy]

The basic tip you need is to be systematic:
1) work on one column at a time, from left to right.
2) work downward from the pivot column.
That way, you will always have "0" above and to the left so you will never change the numbers you already have correct.
And, of course,
3) always do row operations.

Your initial matrix is \(\displaystyle \begin{bmatrix}1 & -4 & -3 & -7 \\ 2 & -1 & 1 & 7 \\ 1 & 2 & 3 & 11\end{bmatrix}\)

Noting that the "pivot" (on the diagonal) for the first column is 1 and I have 2 and 1 below it, I know I need to subtract 2 and 1 to get "0"s there. Because I must use row operations, I will subtract 2 times the first row from the second row and subtract 1 times the first row from the third row. That gives:
\(\displaystyle \begin{bmatrix}1 & -4 & -3 & -7 \\ 0 & 7 & 7 & 21 \\ 0 & 6 & 6 & 10\end{bmatrix}\)

Now, the "pivot" for the second column is 7 and the number below it is 6. We can make that 0 by subtracting 6/7 times the second row from the third. That gives:
\(\displaystyle \begin{bmatrix}1 & -4 & -3 & -7 \\ 0 & 7 & 7 & 21 \\ 0 & 0 & 0 & -8\end{bmatrix}\)

Another thing we could have done is divide the second row by 7 so that we have "1" at the pivot, then just subtract that row times whatever happens to be below the pivot from the third row. That would give:
\(\displaystyle \begin{bmatrix}1 & -4 & -3 & -7 \\ 0 & 1 & 1 & 3 \\ 0 & 0 & 0 & -8\end{bmatrix}\)

 

[Deleted member 4993]

I use a spreadsheet program - like MS_Excel - to keep "book-keeping" straight.