Linear Algebra Section 1.5: Solutions of linear systems

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Linear system is homogenous if

it is of the form Ax = 0 vector in matrix equation form.

Parametric vector equations (through a line and through a plane)

line: x = t(v) plane: x = su + tv

N(A)

solution set of Ax=0

When Ax=b has no solution,

the solution set is empty

Trivial solution

x = 0; of Ax = 0

homogenous linear system of size m by n has infinitely many solutions when...

...when n > m either infinitely many or exactly one solution when n = m

Why should A(cv + dw) = 0 for each pair of scalars "C" and "D"?

A(cv + dw) = = A(cv) + A(dw): Distribity property Rn = c(Av) + d(aw): Compatitbility property of Rn = c(0) + d(0) = 0 + 0 = 0

Is a homogenous solution consistent or inconsistent and why?

ALWAYS consistent because you can always write down the solution; talking all variable to be a zero vector.

Solutions of non-homogenous linear system

Ax = b, b ≠ 0

A is nonsingular if

N(A) = 0

A is singular if

N(A) = infinite

If b cannot equal zero, can Ax=b be a plane through the origin?

No. The equation of a plane through the origin has Ax=b and MUST = 0.

Equation of a plane through the origin:

Po = v1x + v2y + v3z = 0

THEOREM: Suppose Ax=b is consistent and p is a particular solution Hints: p, v, N(A)

The the solution set of Ax=b is the set of all vectors of the form: p + v where v ⋲ N(A) => p + N(A)

Does a homogenous linear system have a non-trivial solution? Why?

Yes because it has a free varibale

Let "A" be an m by n matrix and let "v" and "w" be vectors in Rn with Av = 0 and Aw = 0. Why must A(v+w) must be the zero vector?

a(v+w) = = av + aw = 0 + 0 = 0 This is the distributive property of Rn


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