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Matrix
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REVIEW LAST LECTURE
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Keyword Parametric form Augmented Matrix Elementary Operation Gaussian Elimination Row Echelon form Reduced Row Echelon form Leading 1’s Rank Homogeneous System
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Goal of Elementary Operation To arrive at an easy system
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Theorem 3 The number of leading 1’s
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Homogeneous Equation When b = 0 What is the solution?
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MATRIX REVIEW
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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
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Matrix Review Scalar multiplication kA = [ka ij ]
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Matrix Addition Rules
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Transpose Swap the index of rows and columns A = [a ij ] A T = [a ji ]
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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
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Main Diagonal & Symmetric
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Example A = 2A T Solve for A A = 2AT = 2[2AT]T = 2[a(AT)T] = 4A 0 = 3A Hence A = 0
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Dot Product Step in multiplication We need to compute 3*6 + -1 * 3 + 2 * 5 The multiplication of (3 -1 2) and (6 3 5) is called a dot product of row 1 and column 3
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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
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Multiplication Rules In most case AB != BA (no commutative!!) In most case AB != BA (no commutative!!)
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Example When AB = BA? (when will they commutes?) (A – B)(A + B) = A 2 – B 2
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MATRIX AND LINEAR EQUATION
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Matrix and Linear Equation factoring Matrix equation Linear equation 2 x 1 matrix 2 x 3 and 3 x 1 matrix
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Matrix Equation A X B AX = B
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Matrix Equation AX = B Coefficient matrix Constance matrix Solution
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Associated homogeneous system Given a particular system AX = B There is a related system AX = 0 Called associated homogeneous system
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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
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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
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Proof X 1 is our particular solution to AX = B
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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
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Example Find all solution to Gaussian Elimination gives parametric form x = 4 + 2t y = 2 + t z = t
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Basic Solution Solve the homogeneous system AX = 0 Do the elimination
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Basic Solution x1 = 2s + (1/5), x2 = s, x3 = (3/5)t, x4 = t
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Basic Solution A basic Solution is a solution to the homogeneous system
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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
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Linear Combination Consider the previous solution
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Linear Combination Consider the previous solution We can let r =t / 5… Hence, it is also another parametric form but [1 0 3 5] T is a solution as well!! Hence, a scalar multiple of a basic solution is a basic solution as well
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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
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BLOCK MULTIPLICATION
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Multiplication by Block
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Block Multiplication
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Compatibility Block multiplication is possible when partition is compatible. i.e., size of partitioning allows multiplication of the block Can we divide here?
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MATRIX INVERSE
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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
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Matrix Inverse
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Example Find the inverse of Let If B is the inverse, we have AB = I Cannot be I
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Existence of an Inverse From the previous example There is a matrix having no inverse!!! Zero matrix cannot have an inverse
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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
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Theorem 2.3.1 If B and C are both inverse of A, then B = C If we have inverse, it must be unique.
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Proof Since B and C are inverses CA = I = AB Hence B = IB = (CA)B = C(AB) = CI = C
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Inverse For A A -1 is unique A -1 is square
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First introduction to Det of 2 x 2 matrix Det determinant Det of is (ad – bc)
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Adjugate of 2 x 2 Adjugate of B
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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
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Determinant Det exists before matrix Det is used to determine whether a linear system has a solution
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Inverse and Linear System We have AX = B We can solve by
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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
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Inversion Method We have AA -1 = I Gives Each are a system A is the coefficiency matrix
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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
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Inversion Method A short hand form [A I] [I A -1 ] Double matrix
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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
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Conclusion Matrix Det Inverse
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