Thread: Solving linear system using matrices a, ex from book-- I'm missing something!

Hello all!
Thanks in advance for looking at this for me.

I seem to missing a step somewhere as I am working through the example provided by this book.
Mathematics for 3D Game Programming and Computer Graphics - Eric Lengyel - Google Books

The link shows the exact problem and solution. I am having a hard time understanding how the matrix in figure 2.18 transitions into the matrix shown in figure 2.19. I have been stumped on this for literally hours working ahead and back again trying to get my work to match the examples. Hopefully, Im not just having a brain fart, but I was hoping someone would be able to provide a better explanation, specifically what we are doing the the first row of the matrix when the text saw we are effecting the second row.

Thanks
Glock

?? They tell you exactly how to go from 2.18 to 2.19: "Moving to the second row, we multiply by to obtain a leading entry of 1".
The matrix in 2.18 is .

Since the "reduced echelon" form that we are shooting for has "1"s on the diagonal, we need to change that " " to "1". We can do that by multiplying by its reciprocal , just as they said. Of course, we are doing "row operations" here so mutliplying that number by means multiplying every number in the second row by : , , , and giving 2.19:

Thanks for the quick response!

The problem is that figure 2.19 shows a difference in the first row without any explanation, or that I can see manipulation. UNLESS I am missing something.

Look at the first row of the matrix shown on 2.19. How/why?

Thanks,
Glock

ok, the idea is we want to solve n equations in n unknowns "simultaneously". so say n = 3, so we have 3 such equations:

ax + by + cz = d
ex + fy + gz = h
kx + my + nz = p

we can write this a bit more "compactly" as the matrix equation:

A(x,y,z) = (d,h,p), where A is the matrix:

[a b c]
[e f g]
[k m n].

the idea behind "row-reduction" is that doing "row operations" doesn't change "the answers we get for x,y,z". some people don't trust this, at first. let's use a simpler example:

3x + 2y + z = 13
2x + y + 2z = 13
x - y + z = 2

ok, i will say ahead of time that (x,y,z) = (1,3,4) works.

now normally, what we would try to do is: "eliminate one variable". so we might multiply equation 1 by 2, and equation 2 by 3, and then subtract the "new" equation 2, from the "new" equation 1:

(multplying both sides of an equation doesn't change anything).

6x + 4y + 2z = 26
-6x - 3y - 6z = -39
________________
0x + y - 4z = -13 <---this equation only has 2 variables, y and z.

matrix-wise, this amounts to multiplying the first row by 2, and the second row by 3, to get:

[ 6 ..4 ..2 | 26 ]
[-6 -3 -6 | -33]
[ 1 -1 ..1 | ...2]

now the next step (where we add our "new" equations) corresponds to adding row 1 to row 2. we don't change row 1, ONLY row 2. this gives us:

[6 4 .2 | ..26]
[0 1 -4 | -13]
[1 -1 1 | ...2] (the dots are just "spacers" here, not decimal points).

at this point, we can replace row 1, by 1/6 row 1:

[1 2/3 1/3 | 13/3]
[0 ..1. .-4 | ..-13]
[1 .-1 ...1 | .....2]

all we've done is replace our first two equations with 2 different equations. note that these still work for x = 1, y = 3, z = 4:

1(1) + (2/3)(3) + (1/3)(4) = 1 + 2 + 4/3 = 3/3 + 6/3 + 4/3 = 13/3

1(3) + (-4)(4) = 3 - 16 = -13

to eliminate the "x" in the third equation, we can just subtract row one from row three, and replace what was in row three, with this "new" equation:

[1 2/3 1/3 | 13/3]
[0 ..1. ..-4 | .-13]
[0 -5/3 2/3 |-7/3] <---notice how the "bottom right" is now 2 equations in two unknowns. so basically, we're going to "repeat the process":

we multiply the last row by -3/5, and get:

[1 2/3 .1/3 | 13/3]
[0 ..1. ..-4 | ..-13]
[0 ..1. -2/5 | .7/5]

and subtract row 2 from row 3 (we keep row 2, but we replace row 3):

[1 2/3 ..1/3 | 13/3]
[0 ..1. ...-4 | ..-13]
[0 ..0. 18/5 | 72/5] <--one equation, one unknown! we know z!

just to make it nice and neat, we replace row 3 by 5/18 times row 3:

[1 2/3 1/3 | 13/3]
[0 ..1. ..-4 | ..-13]
[0 ..0. ...1 | .....4] <---z is 4.

essentially, we're done, now. z = 4, and y - 4z = -13, we have that y - 16 = -13, so y = 3. and from:

x + (2/3)y + (1/3)z = 13/3, we get:

x + 2 + 4/3 = 13/3
x + 2 = 9/3 = 3
x = 1, as we knew going in. but we can "clean up the matrix" (putting it in a "standard form"), by putting in as many 0's as possible (0's in a matrix are good, they reduce calculation steps). so now we "work up", replacing row 2 by: 4 times row 3 + row 2:

[1 2/3 1/3 | 13/3]
[0 ..1 ...0 | .....3] <---this makes it obvious y = 3
[0 ..0 ...1 | .....4]

we can "clear out the 1/3" in the first row, by subtracting 1/3 row 3 from row 1, and replacing row 1 with this subtraction:

[1 2/3 0 | 3]
[0 ..1. 0 | 3]
[0 ..0. 1 | 4]

just one more thing to do: "clear out the 2/3" in the first row, by replacing row 1 with: row 1 - (2/3)*(row 2):

[1 0 0 | 1]
[0 1 0 | 3]
[0 0 1 | 4]

which translates to the 3 equations:

x = 1
y = 3
z = 4.