Here are some of the definitions and examples used in Linear Algebra and specifically the linear algebra calculators available.
Let A, B, and C represent n x n matrices.
Example of a 3 x 3 Matrix:
1 2 3 1 1 2 0 1 2
For our case, we are using n x n matrices, so this isn’t a problem.
The equation for multiplying two matrices is : (elementwise)
[AB]ij = SIGMA [A]ik[B]kj
Where the SIGMA summation goes from k=1…n
A example element from our 3×3 Case. To get the first element in our solution matrix c11
c11 = (a11 * b11) + (a12 * b21) + (a13 * b31)
Where aij and bij are from matrices A, B respectively.
For example:
A = | a b c d e f g h i |
AT = | a d g b e h c f i |
det(A) = SIGMA (±)a1j1 a2j2. . .anjn
Where SIGMA is our summation over all permutations j1 j2 … jn of the set S={1, 2, …, n }.
The sign is + or – according to whether the permutation is even or odd.
Example: In our 3×3 case it is a little easier, and boils down to :
det(A) = aei + cdh + bfg - ceg - bdi - afh
Where are matrix first row is a b c , 2nd row d e f, and 3rd row, g h i
Calculation Technique: For the n x n the calculation of the determinant, by definition, is based upon a factorial number of calculations with respect to the size of the matrix. ie. a 3×3 matrix would have 6 calculations (3!) to make, whereas a 20×20 matrix would have 2.43 x 10^18 calculations (20!).
So instead of brute forcing the calculations, I first do some operations on the matrix, which converts it to a upper triangular matrix, and then calculate the determinant by multiplying down the diagonal, since everything below is 0, this will give the determinant.
The cofactor of an element aij from matrix A is :
aij = (-1)i + j * det (A'), where A' is the matrix obtained from "omitting" the ith and jth rows, of matrix A.
Calculation Technique: The inverse was obtained using the Theorem:
Aadj(A) = det(A)In
Which when manipulated gives you:
A-1 = (1 / det(A)) * adj(A)