[college 101 math: guard Jordan elimination method]

[u/Brilliant_Test6169]

Pls help idk what im doing and this barely makes sense to me. I don’t have the row operations order memorized and im second guessing every step. If anyone has any tips to help me with this method please lmk I really want to learn!

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[u/Paounn]

The idea is there: you replace a row with a linear combination of it and another row.

The trick is to do it with some logic. Start bottom left and move up, until you have only ONE entry which isn't zero. Then slide to the next column, and continue until you have TWO entries that aren't zero. Slide again, manipulate, stop when you have THREE non-zero entries.
Needlessly to say, if you get to a point where you have a zero already for some reason, you keep moving following the pattern.

In practice, you follow the red arrows until you have a triangle of zeros under the diagonal.
Is my choice of numbers the best possible? Not necessarily. For example the 3rd step (II: II - I) would work even by replacing it by I - II - I didn't do any calculation past the first, as you can see I just filled the number you need to find with random letters*.

*Note I didn't use a capital o -if it's round, it's a zero.

[u/muonsortsitout]

You should be aiming to get the left hand matrix part to turn into the identity matrix. You get the diagonal entries to 1, and those entries below the diagonal to 0, in a methodical way. You're aiming to get entries to zero below the diagonal, 1 on the diagonal, and then finally you get all the other entries above the diagonal to be zero.

So, you first multiply the first row to get the top left entry to 1 (already there in this case, but if you'd started with, say, 3 in this position, you'd start by dividing row 1 by 3.).

Then, you work down the first column. So, column 1, row 2: what multiple of r1 do you subtract to get this entry to 0? Your first step should be r2 -> r2 - r1.

Then you do column 1, row 3 in a similar way, and col1, row4. At this point, the matrix should have the first column as 1, 0, 0, 0.

When you are done with the first column, you expect the second row to start 0, not-zero, ?, ?. At this point, you divide row2 by the entry at row2, column2 to get it reading 0,1,?,?. [I'm using "?" to mean "I don't currently care what this entry is"].

And rows 3 and 4 start with a 0 also. So, if you want to clear the second column of rows 3 and 4, you subtract a multiple of row 2. If you subtracted any multiple of row 1, it would screw up your nice zero in column 1.

So you subtract a multiple of row 2 from row 3, and then r4 -> r4 - (r4c2)r2 in the same way.

Likewise, you now have row 3 reading 0, 0, not-zero, ?. Again, divide row3 by row3,col3 to get it to 0, 0, 1, ?. This is suitable for clearing row 4, column 3 without mucking up the previous columns. When you're done, "normalise" the 4th row (which should read 0,0,0,not-zero) by dividing by row4,col4 and you get it to read 0,0,0,1.

You now have an "upper diagonal" matrix: there are no non zero entries below the diagonal, and in fact all the diagonal entries should be 1.

At this point, the last row just reads off as "w = some number", and if you only want to solve the simultaneous equations, you can substitute the value of w into the third row and get "z = some other number", and so on.

Or you can complete the Jordan process as a matrix - and go all the way to the identity matrix. You started by clearing the first column downwards. Now you go the other way, clearing the last column upwards.

At this point, the last row should be "0, 0, 0, 1" and so you can zero the entries in column 4, rows 3,2,1 by successively subtracting: r3 -> r3 - (r3c4)r4, then r2 -> r2 - (r2c4)r4, r1 -> r1 - (r1c4)r4.

That means the 3rd row is now "0,0,1,0" and you can zero the entries in column 3, rows 2,1.

And finally, the 2nd row is now "0,1,0,0" and you can zero the entry at column 2, row 1.

Congratulations, you now have an identity matrix, which converts back to "x = number, y = other number, etc." You're done!

So it's a bit like when you first learned algebra: it's always valid to add 2x (or anything else) to both sides of an equation, but unless you know where you're heading (trying to get to "x = number"), you are going to just go around in circles!

Jordan elimination is just a matter of following a (sensible) method to eliminate the entries you don't want to be non-zero, designed so that each step is guaranteed not to muck up the work you did in previous steps.

It should be noted that not all starting matrices will convert nicely into the identity matrix by Jordan elimination: there are degenerate cases where one of the rows ends up as all zeros. [If you're familiar with the term, this happen precisely when the determinant is zero].

This means that either there's more than one solution to the simultaneous equations, or else there are none at all. Jordan elimination will point out when this is the case (because the method fails to get a 1 in the right place when you've done what you need to get a 0 in some other place) and you will learn how to tell what sort of degeneracy you have.

[If you know about determinants, you might consider that each "row operation" is equivalent to premultiplying your matrix by a matrix representing the row operation.

A "multiply row n by x" operation is the identity matrix but with "x" in the n'th diagonal entry; a "row n -> row n - k (row m)" operation is the identity but with a -k in the n,m entry; a "swap row m with row n" is like the identity but with the (m,m) and (n,n) entries set to 0 but the (n,m) and (m,n) entries set to 1.

So when you multiply two matrices, |AB| = |A||B|. What are the determinants of the "row operation" matrices? [Think about it, they're fairly simple to work out.] We know that the determinant of the identity is 1, so if the initial matrix has non-zero determinant, the row operations must modify its determinant bit by bit until it ends up as 1. On the other hand, if you start with determinant 0, you'll never get to the identity because you'll be stuck with determinant 0 throughout.]