Sec1Sess14

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matrices can not always be inverted.

inverse = <math>A^{-1} = 1 / det(A) * adj(A)</math> where adj() is adjoint matrix

when determinant is 0, division by 0 can not occur and matrix inverse in not possible

Ax = b
Matrix is homogeneous if b = 0 and inhomogeneous if not.

inhomogeneous: <math> det(A) </math> is not 0, Ax = b has unique solution <math>x = A^{-1}*b</math>
homogeneous: <math> det(A) </math> is not 0, Ax = 0 has only trivial solution x = 0

HOMOGENEOUS CASE:
AX = 0
Trivial solution:
origin (0,0,0) is always the solution. The three planes pass through the origin
X = A-1 = 0

<math> det(A) = 0 </math> when the normals of all variables equal 0, <math> det(N_1, N_2, N_3) = 0 </math>
in that case <math> N_1, N_2, </math> and <math> N_3 </math> are in the same plane and therefore are coplanar.
(parallelpiped has volume 0)

line through origin (O) perpendicular to plane <math> N_1,\ N_2,\ N_3 </math> is parallel (//) to all 3 planes and in fact contained within them. Therefore there are infinitely many solutions

Non-trivial solution:
<math> N_1 </math> x (cross product) <math> N_2 </math> is parallel to <math> N_1 </math> and <math> N_2 </math> but also parallel to <math> N_3 </math>
The <math> N_1 </math> x <math> N_2 </math> vector goes along the line of intersections.

General Case: AX=B
if <math> det(A) <> 0 </math> then there is a unique solution <math> X=A^{-1}*B </math>.
if <math> det(A) = 0 </math> then either there is no solution or infinitely many solutions. One can tell if there are infinitely many if one can find any individual solutions.

consistent: for an inhomogeneous system if it has solutions it is consistent and it is inconsistent otherwise

In other phrasing:
Homogeneous systems: Ax = 0 has non-trivial solutions ⇔ (if and only if) <math> det(A) = 0</math>.
Inhomogeneous systems: Ax = b has the unique solution <math> x = A^{-1} b </math> , if <math> det(A) <> 0 </math>.
If |A| = 0, then Ax = b usually has no solutions, but does have solutions for some b.

Homogeneous system example:
x + 2y + 3z = 0
4x + 5y + 6z = 0
7x + 8y + cz = 0

Inhomogeneous system example:
x + 2y + 3z = 1
4x + 5y + 6z = 2
7x + 8y + cz = 3.

origin vectors lie in a plane if det(A) = 0

A^-1 exists only when det(A) <> 0

In a homogeneous system when det(A) = 0 a solution can be found by finding orthogonal vectors to the plane represented by each of the equations in
the problem. Since det(A) = 0 all rows in the A matrix in Ax = 0 are in a plane. The orthogonal vectors can be found by taking the cross product
of the first two (or any two) rows in the matrix A.

Orthogonal vectors finding example:
<1, 2, 3> × <4, 5, 6> = <<1 2 3><4 5 6>> = <−3, 6, −3>.

Ways to help describe |A|:
Singular: |A| = 0
Nonsingular or invertible if |A| <> 0

ill-conditioned: a problem that is highly sensitive to small changes in coefficients