|A = [aij] =||
The matrix A has m rows and n columns.
The order of A is m × n.
For any m × n matrices A and B,
A + B = [aij + bij]
A - B = [aij - bij]
For any m × n matrix A and scalar k,
kA = [kaij]
For any m × n matrices A, B and C and scalars k and l,
A + B = B + A
A + (B + C) = (A + B) + C
(kl)A = k(lA)
For any m × n matrix A, there exists a unique matrix -A such that
A + (-A) = 0,
where 0 is the m × n matrix [aij], where aij=0 for every i and j.
For any m × n matrices A and B and scalars k and l,
k(A + B) = kA + kB
(k + l)A = kA + lA
For any m × n matrix A, and n × p matrix B,
C = AB is the m × p matrix [cik], where cik = ai1b1k + ai2b2k + ... + ainbnk
In other words, to find the value of cell cik of the product matrix, C, you sum the n individual products of cells in the row i of matrix A and column k of matrix B.
Note: Matrix multiplication is not commutative. In other words, in general, AB ≠ BA
For matrices A, B, and C, assuming the indicated operations are possible:
A(BC) = (AB)C
A(B + C) = AB + AC
(B + C)A = BA + CA
Reminder: The commutative property does not hold for matrix multiplication.
Consider the set of m simultaneous linear equations in m variables v1, v2, ... vm,
a11v1 + a12v2
+ ... + a1mvm = c1
a211 + a222 + ... + a2mm = c2
am11 + am22 + ... + ammm = cm
The coefficients and constants can be put into an m × n augmented matrix A by setting n = m+1, and letting ain=ci.
The matrix rank is the maximum number of linearly independent rows or columns.
Matrix rank is invariant under row or column interchange.
For an m × n matrix A, rank [A] ≤ min(m,n).
For an n × n matrix A, rank [A] = n iff |A| ¹ 0.
To calculate matrix rank by hand, convert a matrix to its RREF -- Reduced Row Echelon Form -- and count the number of non-zero rows.
The determinant of a square matrix (usually denoted by writing the matrix between vertical bars) is the algebraic sum of all possible products where the number of factors in each product is the same as the number of rows or columns, each factor in a given product is taken from a different row and column, and the sign of a product is positive or negative depending upon whether the number of permutations necessary to place the indices representing each factor's position in its row or column in the order of the natural numbers is odd or even.
Example: consider this 4x4 matrix:
|A = [aij] =||
4P4=24 (there are 24 permutations of four things taken four at a time) so there are 24 products where each factor in a given product is taken from a different row and column. If the four products are taken from successive columns and rows 1, 2, 3, 4, the product is added to find the determinant. If two rows are interchanged (e.g. rows 1, 2, 4, 3) then the product is subtracted to find the determinant. Another way to look at it is this: If an even number of operations is needed to restore the rows to 1, 2, 3, 4 order, then the product is added to find the determinant. If an odd number of operations is needed to restore the rows to 1, 2, 3, 4 order, then the product is subtracted to find the determinant.
In each product, below, the four factors are taken from successive columns. If the order of the rows was achieved by an even number of pairwise interchanges, then a "plus" sign is attached to it. If the order of the rows was achieved by an odd number of pairwise interchanges, then a "minus" sign is attached to it:
To find the determinant of larger and larger matrices, you would need to contend with faster-than-exponential growth of the number of permutations, if you want to carry out the calculations as I have described thus far. So it is important to realize that the determinant of a matrix [aij] is equal to
a11 multiplied by the determinant of the matrix containing all but the row and column containing a11,
minus a12 multiplied by the determinant of the matrix containing all but the row and column containing a12,
plus a13 multiplied by the determinant of the matrix containing all but the row and column containing a13,
This suggests a recursive algorithm for finding the determinant of matrix A of order n:
If n==1 then return a11
set c = 0
for i = 1 to n,
matrix B = matrix A with row i and column 1 deleted.
c = c + (-1)(i-1) ai1 |B|
RREF -- Reduced Row Echelon Form
Vectors -- the "dot" product and the "cross" product, explained.
The webmaster and author of this Math Help site is Graeme McRae.