Matrices

Matrices
A rectangular array of numbers of the form (?(a_11&?&a_1n@?&?&?@a_m1&?&a_mn ))
is called m × n matrix, with m rows and n columns. We count rows from the top and columns from the left.
Matrices addition: Suppose that the two matrices
A=(?(a_11&?&a_1n@?&?&?@a_m1&?&a_mn )) and B=(?(b_11&?&b_1n@?&?&?@b_m1&?&b_mn ))
both have m rows and n columns. Then we write
A+B=(?(a_11+b_11&?&a_1n+b_1n@?&?&?@a_m1+b_m1&?&a_mn+b_mn ))
and call this the sum of the two matrices A and B.
Example 1: Suppose that A=(?(1&-2@3&0@-5&7)) and B=(?(2&3@-1&6@5&-4))
Then A+B=(?(1+2&-2+3@3+(-1)&0+6@-5+5&7+(-4) ))=(?(3&1@2&6@0&3))
Remark: Two matrices of different sizes cannot be added.
Example 2: We do not have a definition for adding the matrices
A=(?(4&2&-1@0&1&3@6&-2&3)) and B=(?(-2&3&0@1&-1&5))
Multiplication by scalar : Suppose that the matrix A=(?(a_11&?&a_1n@?&?&?@a_m1&?&a_mn ))
has m rows and n columns, and that c ? R. Then we write
c A=(?(?ca?_11&?&?ca?_1n@?&?&?@?ca?_m1&?&?ca?_mn ))
Example 3: Suppose that
A=(?(3&0&-1@1&2&4)) then 3A=(?(9&0&-3@3&6&12))

Matrix multiplication: Suppose that
A=(?(a_11&?&a_1n@?&?&?@a_m1&?&a_mn )) and B=(?(b_11&?&b_1p@?&?&?@b_n1&?&b_np ))
are respectively m × n matrix and n × p matrix. Then the product AB will make sense and it will be m × p matrix.
AB=(?(q_11&?&q_1p@?&?&?@q_m1&?&q_mp )) ,
where for every i=1,2,?,m and j=1,2,?,p, we defined
q_ij=?_(k=1)^n??a_ik b_kj ?=a_i1 b_1j+a_i2 b_2j+?+a_in b_nj
Example 4: Find AB and BA if
A=(?(1&-1&2@3&0&-2)) and B=(?(2&1@4&-3@-1&0))
Solution: AB is going to make sense. It is the product of 2 × 3 by 3 × 2 and the result is going to be 2× 2.
AB=(?(q_11&q_12@q_21&q_22 ))
q_11=1×2+(-1)×4+2×(-1)=-4
q_12=1×1+(-1)×(-3)+2×0=4
q_21=3×2+0×4+(-2)×(-1)=8
q_22=3×1+0×(-3)+(-2)×0=3
AB=(?(1&-1&2@3&0&-2))(?(2&1@4&-3@-1&0))=(?(-4&4@8&3))
BA=(?(2&1@4&-3@-1&0))(?(1&-1&2@3&0&-2))
=(?(2×1+1×3&2×(-1)+1×0&2×2+1×(-2)@4×1+(-3)×3&4×(-1)+(-3)×0&4×2+(-3)×(-2)@(-1)×1+0×3&(-1)×(-1)+0×0&(-1)×2+0×(-2) ))
=(?(5&-2&2@-5&-4&12@-1&1&-2))
Transpse of a Matrix:
Let A be m × n matrix. Then A^T,the transpse of A is the matrix obtained by interchanging the rows and columns of A.
For example:
If A=(?(1&2&-1@0&3&5)) then A^T=(?(1&0@2&3@-1&5))
Determinants:
If A is n × n square matrix is called a matrix of order n.With such a matrix we associate a number called the determinant of A and written det?A.
For n=2 we have this definition:
If A=(?(a_11&a_12@a_21&a_22 )) then det?A=|?(a_11&a_12@a_21&a_22 )|=a_11×a_22-a_21×a_12
For n=3 we have this definition:(we will choose first row)
If A=(?(a_11&a_12&a_13@a_21&a_22&a_23@a_31&a_32&a_33 )) then det?A=|?(a_11&a_12&a_13@a_21&a_22&a_23@a_31&a_32&a_33 )|
det?A=a_11 |?(a_22&a_23@a_32&a_33 )|-a_12 |?(a_21&a_23@a_31&a_33 )|+a_13 |?(a_21&a_22@a_31&a_32 )|
Remark : we can choose any row or any column of A.
Example 6: Find det?A for
A=(?(2&1&3@3&-1&-2@2&3&1))
Solution:
det?A=a_11 |?(a_22&a_23@a_32&a_33 )|-a_12 |?(a_21&a_23@a_31&a_33 )|+a_13 |?(a_21&a_22@a_31&a_32 )|
det?A=2|?(-1&-2@3&1)|-1|?(3&-2@2&1)|+3|?(3&-1@2&3)|
det?A=2{(-1)×1-3×(-2)}-1{3×1-2×(-2)}+3{3×3-2×(-1)}
det?A=2×5-7+3×11=36



Minors and Cofactors
Suppose we are given a matrix A of order n , and a_ij is one of its entries. The minor M_ij is defined to be the determinant of the matrix obtained by deleting the row i and the column j from the matrix.
The cofactor A_ij=(-1)^(i+j) M_ij
Example 7: Find the cofactor of a_(23 ) if A=(?(2&1&3@3&-1&-2@2&3&1))
A_23=(-1)^(2+3) M_23=-|?(2&1@2&3)|=-(6-2)=-4
Adjoint of Matrix
The adjoint of a matrix A of order n is obtained by taking the transpose of the cofactor matrix A and we write adj(A).
To find the adjoint of a matrix, first find the cofactor matrix of the given matrix. Then find the transpose of the cofactor matrix.
Example 8: Find adj(A) for A=(?(2&1&3@3&-1&-2@2&3&1))
Solution:
A_11=(-1)^(1+1) |?(-1&-2@3&1)|=5 A_12=(-1)^(1+2) |?(3&-2@2&1)|=-7
A_13=(-1)^(1+3) |?(3&-1@2&3)|=11 A_21=(-1)^(2+1) |?(1&3@3&1)|=8
A_22=(-1)^(2+2) |?(2&3@2&1)|=-4 A_23=(-1)^(2+3) |?(2&1@2&3)|=-4
A_31=(-1)^(3+1) |?(1&3@-1&-2)|=1 A_32=(-1)^(3+2) |?(2&3@3&-2)|=13
A_33=(-1)^(3+3) |?(2&1@3&-1)|=-5
A_ij=(?(5&-7&11@8&-4&-4@1&13&-5))
adj(A)=A_ij^T=(?(5&8&1@-7&-4&13@11&-4&-5))
Identity matrix
The identity matrix of order n denoted I_n is a diagonal matrix with all ones on the diagonal
I_2=(?(1&0@0&1)) ,I_3=(?(1&0&0@0&1&0@0&0&1)) ,I_4= (?(?(1&0@0&1)&?(0&0@0&0)@?(0&0@0&0)&?(1&0@0&1)))
If A is an m x n matrix, then
I_m A= A and A=AI_n
If A is a square matrix of order n , then
I_n A=AI_n=A
Inversion of matrices
If A is a square matrix of order n , then its inverse is given by
A^(-1)=1/det(A) adj(A) ; det(A)?0.
The inverse has the special property that
AA^(-1)=A^(-1) A=I
Example 9: Find A^(-1) for A=(?(5&2&1@0&6&3@8&4&7))
Solution:
det(A)=5(42-12)-2(0-24)+1(0-48)=150
A_11=(-1)^(1+1) |?(6&3@4&7)|=30 ,A_12=(-1)^(1+2) |?(0&3@8&7)|=24
A_13=(-1)^(1+3) |?(0&6@8&4)|=-48 , A_21=(-1)^(2+1) |?(2&1@4&7)|=-10
A_22=(-1)^(2+2) |?(5&1@8&7)|=27 , A_23=(-1)^(2+3) |?(5&2@8&4)|=-4
A_31=(-1)^(3+1) |?(2&1@6&3)|=0 , A_32=(-1)^(3+2) |?(5&1@0&3)|=-15
A_33=(-1)^(3+3) |?(5&2@0&6)|=30
A_ij=(?(30&24&-48@-10&27&-4@0&-15&30))
adj(A)=(?(30&-10&0@24&27&-15@-48&-4&30))
A^(-1)=1/det(A) adj(A)
A^(-1)=1/150 (?(30&-10&0@24&27&-15@-48&-4&30))


If
A=(?(2&0&-1@1&-1&3@0&2&1)) and B=(?(1&-1&2@3&0&-1@2&1&-2))
1. Find AB and BA
2. Find A^(-1) and B^(-1)