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Title: Linear Algebra - Linear Systems
Description: These notes cover Linear Systems
Description: These notes cover Linear Systems
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Systems of Linear Equations
We’ve seen these before
...
The
general form of a linear system is
a1,1 x1
a2,1 x1
...
...
...
+ ···
+ ···
am,1 x1 + am,2 x2 + · · ·
+ a1,n−1 xn−1
+ a2,n−1 xn−1
...
...
...
= b1
= b2
...
...
As before, the objective is to find values
of x1 , x2 ,
...
There are three different numbers of solutions you can get
...
You can find an infinite number of solutions
...
Definition: A system of equations that has NO solutions is called inconsistent
...
Beyond the one type of result, inconsistency and NO answers, you can have either a unique
solution (ONE answer) or an infinite set of them
...
It has
to include ALL solutions
...
It’s usually best written in parametric form (it can be a line, a plane, or bigger,
but we’ve seen those)
...
So, NO constant terms
...
Example:
Here’s a system of equations:
3x + 2y − z = 2,
x − 4y + 4z = −1,
−2x + z = 0
...
Here’s another: it could be the result of a spanning question:
3
2
−1
2
1 x + −4 y + 4 z = −1
...
0
−1
−2
0
Example Here’s what a nearly fully solved system looks like:
x1 − x2 + x3 = 1
x3 = −1,
x2 + 2x3 = 2
each value written in terms of a constant or in terms of later variables
...
Done
...
You may have stared with more equations, but they got canceled out or something, so you
now have 3 variables, 2 equations
...
x2 = 2 − 2t
x1 = 1 + x2 − t = 3 − 3t
x1
3 − 3t
x2 = 2 − 2t
t
x3
so
3
−3
=⇒ = 2 + −2 t,
0
1
which is the parametric form of the solution
...
It HAS no solution, since it’s been broken down
into a contradiction 0 = 1
...
Here’s another:
x1 + 2x2 − x3 + 4x4 = 0
x3 − x4 = 1
which has 4 variables and 2 equations, and NO contradictions
...
This means x3 = 1 + s
...
= +
1 0
1
0
0
1
Parametric form of the solution
...
xn values
...
x1 − 2x2 + x3 − 4x4
x2 − x3 + x4
x3 + x4
x4
becomes
1 −2
1 −4 2
0
1 −1
1 1
0
0
1
1 0
0
0
0
1 1
=2
=1
=0
=1
or
1 −2
1 −4 2
0
1 −1
1 1
...
Notice the pattern, with the equations starting (as in, having non-zero values) further
right on the lower rows
...
Each row starts with a
different variable, x1 for the first, x2 for the second, etc
...
We start at the bottom, with x4 = 1
...
Next,
x2 = 1 + x3 − x4 = 1 − 1 − 1 = −1
...
Easy enough
...
As it happens: there’s even a name for this sort of arrangement:
Definition: An augmented matrix is in Row Echelon Form if the following three requirements are met
...
• All leading ones are to the right of those above them and to the left of those below
them
...
Definition: Leading ones in a REF augmented matrix are frequently called Pivots
...
A column with a pivot/leading one is a
pivot column and the unknown (x value or whatever) associated with that column is a pivot
3
variable
...
Pivot variables are written as dependent variables
...
, 0 1 0 2 , 0 0 , 0 1 , 0
0 0
0 1
0 0 1 1
0
0 0 0 0
0 0
0 0
Lets write out the general solution of that last one
...
Make it x4 = t, so
x3 + x4 = 2 so x3 = 2 − x4 = 2 − t
...
Our final step is
x1 = 1 + 3x2 − 2x3
=⇒
x1 = 1 + 3s − 2(2 − t)
=⇒
x1 = −3 + 3s + 2t
...
e
...
This would involve having
ONLY free variables accompanying the leading one in a row
...
• Each pivot (leading one) has only zeros above and below (it’s the only non-zero term
on its column)
...
We got those results
earlier, but with more difficulty
...
That happens
...
It’s
actually quite simple
...
Row Operations:
1
...
Have
them switch places
...
Row Multiplication: take a row and multiply it by a NON-ZERO real number
...
Row Addition: take a real multiple of a row and add that to ANOTHER ROW
...
Those three operations are all we need
...
• They are reversible
...
If you merely list the equations differently, it doesn’t change the solution
...
The other two are harder
...
Hint: there was a relevant question
in Assignment 3 to Row Multiplication and Addition not changing the solution
...
Theorem: two linear systems have the same general solution if and only if they are equivalent
...
Find the leftmost column that contains a non-zero term
...
2
...
3
...
4
...
5
...
Go back to step 1, using only
the rows below the top one
...
1
0
0 −2 −1
1 −2 −4
2 −3
Step 1: the first column has non-zero values
...
Step 3: divide by 2 to give us a leading one on the top left
...
Some
examples are given:
1
R
(1)/2
(1) = 1 (1)
R1 = 1 R1
1 −1 −2
0 −2
2 1
2
2
−1
1
2
1
3
...
Again,
It is generally recommended to call your row operations
...
1 −1 −2
0 −2
0
+R1
+(1)
0
0
1
1
...
We’ve written x1 in terms of the others, and now we apply Step 5 and
IGNORE that row, and concentrate on the x1 free system in the lower three rows
...
Here’s the system at this
point, with a horizontal line indicating what we can ignore:
1 −1 −2
0 −2
0
0
0
1
1
...
Step 2: the second row
(we’re looking at) has a one in that column, so switch them:
1 −1 −2
0 −2
0
R2 ⇔ R3
(2) ⇔ (3)
1
2 −2
1
...
Step 4: subtract
0, −1 times the top row (we’re paying attention to) from the last two
...
0
0
0
1
1
+R2
0
0
0
0
0
Step 5 has us ignore that row, which, again, has a leading one that has only zeros under it
...
0
0
0
1
1
0
0
0
0
0
Step 1: the leftmost non-zero term (that we’re paying attention to) is on the fourth column
...
Step 3: it’s one, no need to divide
...
Step 5: here’s the matrix:
1 −1 −2
0 −2
0
1
2 −2
1
...
7
Title: Linear Algebra - Linear Systems
Description: These notes cover Linear Systems
Description: These notes cover Linear Systems