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Unique Solution
Many Solutions
No Solution
No Free Variable
Free Variable
0 = #
General Solution
{ x1 = 3x2 + 5
{ x2 is free
Vector Equation
x1[3] + x2[8] = [1]
[5] + [3] = [2]
Vector in span {v1, v2}
v = c1v1 + c2v2
A times x
x1*A1 + x2*A2 + x3*A3
A1 = column 1
Matrix Equation
[A] [x] = [b]
Linearly Independent
rows <= columns
only trivial solution
no multiplies
no 0 vector
T(x) is linear if…
T(0) = 0
Standard Matrix
A = [ T(e1), T(e2), T(e3) ]
One to One
One solution
Linearly Independent
Inverse of 2x2
A^-1 = 1/detA [d -b, -c a]
Solve x with Inverse
x = (A^-1) B
Find Inverse
[ A | I ] => [ I | A^-1 ]
Onto
Is Anxn inversable
Do the columns of A span R^n
Needs a pivot in every row
Is triangle inversable
If diagonal multiple is non zero
Subspace
Any set with a zero vector
H = span{ v1, v2, v3 }
u+v and cu in H
Column Space
ColA = span{ a1, a2, a3 }
Rm, m = rows
Null Space
All solutions for Ax=0
Rn, n = columns
Basis for ColA
- RREF
- Take pivot columns, discard free columns
- Use original A, not RREF
Basis for NullA
- Ax = 0
- RREF
- Parametic form
- Vectors go into { v1, v2, v3 }
Basis for Subspace
The linearly independent set,
no matching
Rank
RankA + dim(NullA) = # of columns
Cofactor Expansion
detA = a11C11 + a12C12 + a13C13
Cij = (-1)^i+j detAij
Row Operations
Addition, detB = detA, detE = 1
Multiplication, detB = k detA, detE = k
Swapping, detB = -detA, detE = -1
Cramer’s Rule
xi = detAi (b) / detA
detA but replace column Ai with b
Adjugate
adjA =
C11 … Cn1
C1m … Cnm
Adjugate Invertable Matrix Rule
For square invert
A^-1 = (1 / detA) adjA
Cofactor
Cij = (-1)^i+j * detAij
Coordinate Vector of x relative to B
xB = [c1, c2, c3]
x = c1v1 + c2v2 + c3v3
Find xB
Parametric Form
x = p + x3v
x = [#] + x3[# in x3]
_
x = 3 + x2[5]
x1 = 3 + 5x2
