Matrix

REVIEW LAST LECTURE

Keyword Parametric form Augmented Matrix Elementary Operation Gaussian Elimination Row Echelon form Reduced Row Echelon form Leading 1’s Rank Homogeneous System

Goal of Elementary Operation To arrive at an easy system

Theorem 3 The number of leading 1’s

Homogeneous Equation When b = 0 What is the solution?

MATRIX REVIEW

Matrix Review column matrix Or column vector A has 2 rows 3 columns A is a 2 x 3 matrix a 22 a 13 c 21 Square matrix (number of row equals number of column

Matrix Review Scalar multiplication kA = [ka ij ]

Matrix Addition Rules

Transpose Swap the index of rows and columns A = [a ij ] A T = [a ji ]

Transpose Rule If A is an m x n matrix, then A T is n x m matrix (A T ) T = A (kA) T = kA T (A + B)T = A T + B T

Main Diagonal & Symmetric

Example A = 2A T Solve for A A = 2AT = 2[2AT]T = 2[a(AT)T] = 4A 0 = 3A Hence A = 0

Dot Product Step in multiplication We need to compute 3* * * 5 The multiplication of (3 -1 2) and (6 3 5) is called a dot product of row 1 and column 3

Identify Matrix A matrix whose main diagonal are 1’s and 0’s are elsewhere In most case, we assume that the identity matrix is a square matrix

Multiplication Rules In most case AB != BA (no commutative!!) In most case AB != BA (no commutative!!)

Example When AB = BA? (when will they commutes?) (A – B)(A + B) = A 2 – B 2

MATRIX AND LINEAR EQUATION

Matrix and Linear Equation factoring Matrix equation Linear equation 2 x 1 matrix 2 x 3 and 3 x 1 matrix

Matrix Equation A X B AX = B

Matrix Equation AX = B Coefficient matrix Constance matrix Solution

Associated homogeneous system Given a particular system AX = B There is a related system AX = 0 Called associated homogeneous system

Solution of a linear system Let X 1 be a solution to AX = B X 0 be a solution to AX = 0 X 1 + X 0 is also a solution of AX = B Why? A(X 1 + X 0 ) = AX 1 + AX 0 = B + 0 = B

Theorem 2 Suppose X 1 is a particular solution to the system AX = B of linear equations. Then every solution X 2 to AX = B has the form X 2 = X 1 + X 0 For some solution X 0 of the associated homogeneous system AX = 0

Proof X 1 is our particular solution to AX = B

Implication of Theorem 2 Given a particular system AX = B We can find all solutions by Find a particular solution to AX = B Reduce the problem into finding all solution to AX = 0

Example Find all solution to Gaussian Elimination gives parametric form x = 4 + 2t y = 2 + t z = t

Basic Solution Solve the homogeneous system AX = 0 Do the elimination

Basic Solution x1 = 2s + (1/5), x2 = s, x3 = (3/5)t, x4 = t

Basic Solution A basic Solution is a solution to the homogeneous system

Linear Combination The solution to the previous system is sX 1 + tX 2 Solutions in this form are called a linear combination of X 1 and X 2

Linear Combination Consider the previous solution

Linear Combination Consider the previous solution We can let r =t / 5… Hence, it is also another parametric form but [ ] T is a solution as well!! Hence, a scalar multiple of a basic solution is a basic solution as well

Relation to Rank A system AX = 0 Having n variable and m equation (A is m x n matrix) Suppose the rank of A is r Then there are n – r parameter (from theorem 3 of the last slide) We will have exactly n – r basic solutions Every solution is a linear combination of these basic solutions

BLOCK MULTIPLICATION

Multiplication by Block

Block Multiplication

Compatibility Block multiplication is possible when partition is compatible. i.e., size of partitioning allows multiplication of the block Can we divide here?

MATRIX INVERSE

Solving equation How to solve a scalar equation ax = b Multiply both side by 1/a ax/a = b/a x = b/a We need multiplicative inverse

Matrix Inverse

Example Find the inverse of Let If B is the inverse, we have AB = I Cannot be I

Existence of an Inverse From the previous example There is a matrix having no inverse!!! Zero matrix cannot have an inverse

Non-Square matrix What should be an inverse of non-square? Let A is m x n matrix What should be A -1 ? We can have B = n x m such that AB = I m and BA = I n But this gives m = n If m < n, there exists a basic solution X (a 1 x n matrix) for AX = 0 So X = I n X = (BA)X = B(AX) = B(0) = 0  contradict Non square matrix has no inverse

Theorem If B and C are both inverse of A, then B = C If we have inverse, it must be unique.

Proof Since B and C are inverses CA = I = AB Hence B = IB = (CA)B = C(AB) = CI = C

Inverse For A A -1 is unique A -1 is square

First introduction to Det of 2 x 2 matrix Det  determinant Det of is (ad – bc)

Adjugate of 2 x 2 Adjugate of B

Det and Inverse Let det adj e B So, if e != 0, we multiply it by 1/e gives A(1/e)B = I =(1/e)BA So, the inverse of a is (1/e)B AB = eI = BA

Determinant Det exists before matrix Det is used to determine whether a linear system has a solution

Inverse and Linear System We have AX = B We can solve by

Inversion Method A method to determine the inverse of A based on solving linear equation system We have A = 2 x 2 matrix We need to find A -1 We write the inverse as

Inversion Method We have AA -1 = I Gives Each are a system A is the coefficiency matrix

Solving A Find the equivalent systems in a reduced row echelon form Gives This can be done by elementary operation In fact, we do this at the same time for both equation

Inversion Method A short hand form [A I]  [I A -1 ] Double matrix

Matrix Inversion Algorithm If A is a square matrix There exists a sequence of elementary row operation that carry A to the identity matrix of the same size. This same series carries I to A -1

Conclusion Matrix Det Inverse