Fundamentals of Matrix Algebra, 2011a

Chapter 1
Systems of Linear Equaons
the coefficients and the constants; the names of the variables really didn’t maer. In
Example 1 we had the following three equaons:
r + b + g = 30
r = 2g
b = r + g
Let’s rewrite these equaons so that all variables are on the le of the equal sign
and all constants are on the right. Also, for a bit more consistency, let’s list the variables
in alphabecal order in each equaon. Therefore we can write the equaons as
b + g + r = 30
− 2g + r = 0
−b + g + r = 0
. (1.6)
As we menoned before, there isn’t just one “right” way of finding the soluon to
this system of equaons. Here is another way to do it, a way that is a bit different from
our method in Secon 1.1.
First, lets add the first and last equaons together, and write the result as a new
third equaon. This gives us:
b + g + r = 30
− 2g + r = 0
2g + 2r = 30
A nice feature of this is that the only equaon with a b in it is the first equaon.
Now let’s mulply the second equaon by − 1 2
. This gives
b + g + r = 30
g − 1/2r = 0
2g + 2r = 30
Let’s now do two steps in a row; our goal is to get rid of the g’s in the first and third
equaons. In order to remove the g in the first equaon, let’s mulply the second
equaon by −1 and add that to the first equaon, replacing the first equaon with
that sum. To remove the g in the third equaon, let’s mulply the second equaon by
−2 and add that to the third equaon, replacing the third equaon. Our new system
of equaons now becomes
b + 3/2r = 30
g − 1/2r = 0
3r = 30
Clearly we can mulply the third equaon by 1 3
and find that r = 10; let’s make
this our new third equaon, giving
b + 3/2r = 30
g − 1/2r = 0
r = 10
.
.
.
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