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Matrices

December 8, 2016
Published By : Pratik Kataria
Categorised in:

Cofactor of an element $a_{ij}$ in |A|:

$A_{ij} = (-1)^{i+j} M_{ij}$

$|A| = |\table a_{11},a_{12},a_{13};a_{21}, a_{22},a_{23};a_{31},a_{32},a_{33}|$

e.g. $A_{11} =(-1)^{1+1}M_{11} = |\table a_{22},a_{23}; a_{32}, a_{33}|$


Adjoint of a matrix, adj. A = Transpose of cofactor matrix i.e. A’

|A| = 0 -> singular matrix
|A| ≠ 0 – > non-singular matrix

Inverse of A, if A is square matrix and |A| ≠ 0:


$A^{-1} = 1/{|A|} adj. A$

Multiplication of matrix by scalar:

$A = |\table 1,2;3,4|$
$(-1)A =|\table -1,-2;-3,-4| $

i.e. all elements get multiplied by that scalar

Multiplication of matrices:

If columns of A = rows of B, then product AB exists and is computed as:

$A = [\table 1,2,3;2,3,4;]_{2×3}$

$B = [\table 2,1,1]_{3×1}$

$AB = [\table 1(2)+2(-1)+3(3);2(2)+3(-1)+4(3)] = [\table 9;13]_{2×1}$

Note:

(A’)’ = A

(AB)’ = B’A’
(A+B)’ = A’ + B’

If A is square matrix, it can be written as:

$A = 1/2 (A + A’) + 1/2 (A-A’)$

$(AB)^{-1} = B^{-1}A^{-1}$

$ (A^{-1})’ = (A’)^{-1} $

|AB| = |A| . |B|

Note: Elementary transformations do not alter the rank of a matrix, hence we use the sign of equivalence

Rank of matrix:

(1)Echelon form e.g.:

$A = [\table 1,2,3,4,5;0,1,2,3,4;0,0,1,2,3;0,0,0,0,0]$

For rank, after reducing matrix to echelon form,

if m = total no. of rows in A and k = total no. of rows which contain all zero elements

ρ(A) = m – k

In above e.g. ρ(A) = 4 – 1 = 3

(2) Normal Form:

$[I_r]$ $[\table I_r,0]$
$[\table I_r;0]$ $[\table I_r,0;0,0]$

Here, I denotes identity matrix and r denotes rank

(3) Non-singular P & Q such that PAQ is in normal form

Consider 3×4 order matrix,

$ A_{3×4} = I_3 A I_4 $

Aim is to reduce A to normal form by using elementary transformations

Note 1: Perform row transformations only on left hand side matrix A and on $I_3$, keep $I_4$ unaltered.

Note 2: Perform column transformation only on left hand side matrix A and on $I_4$, keep $I_3$ unaltered.