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Solution of a system is the set of points/combinations of variables that
satisfy an equation like 3x+2y=1.
[3 2][x
y] = 1 is equivalent
A := [3 2]
X := [x y]^T
b := 1
AX = b
Some systems like:
x - 2y = 0
x + y = 1
x - y = 2
have no solutions.
This can be written as a a matrix too.
Asking "does this system have a solution" is equivalent to asking:
Can [0 1 2]^T be written as a linear combination/in span of [1 1 1]^T
and [-2 1 -1]^T
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How to solve Ax = b where A is an mxn matrix, x is a nx1 matrix of free
variables, and b is a mx1 matrix of constants.
Ex.
A = [ 3 1 0 4
0 1 2 5
0 0 -1 1 ]
x = [ x1 x2 x3 x4 ]^T
b = [ 0 1 2 ]
Find the free variable, and then back-substitute
x4 = t
-x3 + x4 = 2 --> x3 = t - 2
x2 + 2*x3 + 5*x4 = 1 --> x2 + 2*(t-2) + 5*t = 1 --> x2 = 5 - 7t
3*x1 + x2 + x4 = 0 --> 3*x1 + 5 - 7t + t = 0 --> x1 = -5/3 + 2t
Pivot element := first non-zero element in a matrix row AND everything
below it is zero
A matrix is in *row echelon* form iff pivot elements are to the left of
all pivot elements in a lower row.
Therefore, all non-zero rows have a pivot but not every column.
Note: column echelon form also exists.
Generally, solving Ax = B can be done by solving
CAx = Cb where C is an invertible mxm matrix.
Single row operations used in Gaussian elimination are a subset of these
operations.
Ex.
A = 3x4 matrix.
Multiply 1/3 to first row: 1/3 R1 -> R1.
C = 3x3 matrix.
[ 1/3 0 0
0 1 0
0 0 1 ] * A
C^{-1} =
[ 3 0 0
0 1 0
0 0 1 ]
Switch rows R2 and R3:
C = [ 1 0 0
0 0 1
0 1 0 ]
C = C^{-1}
Add cR1 to R3:
C = [ 1 0 0
0 1 0
c 0 1 ]
C^{-1} = [ 1 0 0
0 1 0
-c 0 1 ]
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