We could express the left side as a sum of vectors and obtain the Vector Form of a Linear System: a 12 a x n. a m2
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1 Week 22 Equations, Matrices and Transformations Coefficient Matrix and Vector Forms of a Linear System Suppose we have a system of m linear equations in n unknowns a 11 x 1 + a 12 x a 1n x n b 1 a 21 x 1 + a 22 x a 2n x n b 2 a m1 x 1 + a m2 x a mn x n b m We could express the left side as a sum of vectors and obtain the Vector Form of a Linear System: b 1 b 2 b m x 1 a 11 a 21 a m1 + x 2 a 12 a 22 a m2 + + x n That is, B is a linear combination of A 1 a 11 a 21 a m1 a 1n a 2n a mn b 1 b 2, A 2 b m a 12 a 22 a m2,, A n Notes 1 The original system of equations above is consistent if and only if B can be written as a linear combination of A 1, A 2,, A n 2 (x 1, x 2,, x n ) is a solution to the above system if and only if x 1 A 1 + x 2 A x n A n B Matrix-vector product: Given an m n matrix A [A 1 A 2, A n ] and a vector X, we define the matrix-vector product as AX x 1 A 1 + x 2 A x n A n We can then express the above system of equations as the single matrix equation a 11 a 12 a 1n x 1 b 1 a 21 a 22 a 2n x 2 b 2 a m1 a m2 a mn x n b m x 1 x 2 x m a 1n a 2n a mn or AX B 1
2 Example: Express the following system of equations as a vector equation and a matrix equation Is B [ ] 5 a linear combination of A 7 1 x 1 + 2x 2 5 x 1 + x 2 7 [ ] 1 and A 1 2 [ 2 ]? Transformations A transformation T : R n R m is a function involving vectors instead of scalars Recall: A function f from a set A to a set B is a rule that assigns to each element from A a unique element in a set B We write f : A B The set A is called the domain of f and B is called the range If f(a) b, we say b is the image of a under f, and that a is a pre-image of b A single equation ax b corresponds to a function f : R R with f(x) ax A system of equations AX B, where A is m n-matrix, corresponds to a transformation T : R n R m with T (X) AX A solution to a single equation is the pre-image of b under the corresponding function f A solution to a system of equations is the pre-image of B under the corresponding transformation T Single Equation 2x System x 1 + 2x 2 + x 7 x 1 + 5x 2 Transformation y f(x) 2x [ w1 w 2 ] x 1 [ ] T x 2 x1 + 2x 2 + x x x 1 + 5x 2 2
3 a 11 a 1n In general, if A is an m n-matrix, then T (X) AX with a m1 a mn x 1 x n x 1 T A x n a 11 x a 1n x n a 21 x a 2n x n a m1 x a mn x n is a transformation T : R n R m We call A the standard matrix of the transformation T Main Task: Given a transformation, find the standard matrix Example 4 is a transformation from to x 1 x 1 [ ] T x 2 A x 2 x1 + 2x 2 + x x x x 1 + 5x 2 (a) Identify the standard matrix A 1 (c) Find the image of the vector X 2 under T
4 Geometric Transformations in R 2 Geometrically, a transformation maps a vector into a new vector We will find the standard matrix for several geometric transformations in R 2 by expressing output components w i as linear combinations of input components x i Reflection about y-axis: [ ] [ ] x x T y y w 1 1x + 0y w 2 0x 1y [ ] [ ] [ ] w1 1 0 x w y [ ] 1 0 Standard matrix is 0 1 Reflection about x-axis: [ ] 1 0 Standard matrix is 0 1 Reflection about line y x: [ ] 0 1 Standard matrix is 1 0 Projection onto x-axis: [ ] [ ] x x T y 0 w 1 1x + 0y w 2 0x + 0y [ ] [ ] [ ] w1 1 0 x w y [ ] 1 0 Standard matrix is 0 0 Projection onto y-axis: [ ] 0 0 Standard matrix is 0 1 Rotations about the origin through an angle of θ (counterclockwise): [ ] cos θ sin θ Standard matrix is sin θ cos θ 4
5 2 Matrix Multiplication Let A be an m n matrix and let B be an n p matrix a 11 a 12 a 1n a 21 a 22 a 2n b 11 b 12 b 1j b 1p b 21 b 22 b 2j b 2p A and B a i1 a i2 a in b n1 b n2 b nj b np a m1 a m2 a mn If we think of the i th row of A as a row vector a i [a i1 a i2 a in ] and the j th column of B as the column vector b j [b 1j b 2j b nj ] T, then we define the (i, j)-entry of the matrix product AB as (AB) ij a i b j a i1 b 1j + a i2 b 2j + + a in b nj In order for this operation to be defined, we require the number of columns of A to be equal to the number of rows of B The size of the resulting matrix AB will be m p Example: We say that two square matrices A and B commute if AB BA Note: In general, matrix multiplication is NOT commutative Properties of Matrix Multiplication Suppose A, B and C are matrices of the appropriate sizes and k is a scalar 1 A(BC) (AB)C 2 A(B + C) AB + AC 5
6 (B + C)A BA + CA 4 k(ab) (ka)b A(kB) 5 (AB) T B T A T Exercise: Prove 2-5 using (i, j) notation The identity matrix, which we defined earlier, behaves like the number 1: If A is an m n matrix and I n denotes the n n identity matrix, then 24 Matrix Inverses AI n A and I m A A Another important property of the number 1 is that a number x divided by itself (if x 0) is equal to 1 That is, x x 1 There is no operation called division in matrix arithmetic Instead we multiply a matrix by its inverse Definition: x 1 x 1, if x 0 AA 1 I A 1 A, if A is invertible If A is a square matrix and AB I BA we say B is the inverse of A and we write B A 1 If we can find an inverse, we say A is invertible 6
7 Note: If B and C are both inverses of A, then B C Example: [ ] 2 5 B is the inverse of A 1 [ ] 0 0 A has no inverse 1 2 [ ] 5 since BA AB I 1 2 Good news: In general, AB may not be equal to BA BUT: if AB I then BA I as well Theorem: If A and B are invertible n n matrices, then AB is invertible and (AB) 1 B 1 A 1 Proof: (AB)(B 1 A 1 ) A(BB 1 )A 1 AIA 1 AA 1 I Theorem: If A is an invertible matrix and k is a non-zero scalar 1 (A 1 ) 1 A 2 (A T ) 1 (A 1 ) T (ka) 1 1 k A 1 Proof: 1 AA 1 I (ka)( 1 k A 1 ) k 1 k AA 1 1I I The following theorem gives us a quick way to test if a 2 2 matrix is invertible or not 7
8 Theorem [ ] a b If A, then A is invertible if and only if ad bc 0 In this case, c d A 1 1 ad bc [ d ] b c a Proof: [ a ] b 1 [ d ] b c d ad bc c a [ ] [ ] 1 a b d b ad bc c d c a [ ] 1 ad bc ab + ba ad bc cd dc cb + da ad bc 0 ad bc ad bc 0 [ ] ad bc Note that it is important that ad bc 0 in order for the inverse of a 2 2 matrix to exist We call this number ad bc the determinant of a 2 2 matrix We will see in the next chapter that this determinant can be defined for any square matrix and it will determine whether a matrix is invertible or not The inverse will exist if and only if the determinant is non-zero Example: Find the inverse of Solution: [ ] Linear systems of n equations in n unknowns: Consider the linear system AX B, where A is an invertible n [ n matrix ] We could solve this system using an augmented matrix OR x1 since we are solving for X, we can rearrange the equation AX B by multiplying both sides on the left by A 1 to obtain x 2 X A 1 B 8
9 Example: x + 2y 5 x + y 7 Inverse Transformations Recall: If f(x) 2x +, we can find f 1 (y) by solving for x y 2x + y 2 x f 1 (y) y 2 We can also find inverses of some transformations T Suppose W T (X) is defined by [ w1 [ ] 1 [ ] 2 1 Then, solving for X T 1 w1 (W ) we have 4 w 2 w 2 ] [ ] [ ] 2 1 x1 4 x 2 ] [ x1 x 2 Example: Find the clockwise rotation about the origin through an angle of θ Theorem: Let T : R n R n be a transformation Then T is invertible if and only if its standard matrix is invertible Challenge: What can we say for a transformation T : R n R m, where m n? 9
10 25 Elementary Matrices An elementary matrix is an n n matrix that can be obtained from the n n identity matrix I by performing a single elementary row operation Example 1 [ ] and are elementary matrices Fact: Performing an elementary row operation on an m n matrix A is equivalent to multiplying A on the left by the corresponding elementary matrix Example 2 Multiply A R + R on the left by the elementary matrix corresponding to the ERO Note: Every elementary matrix E is invertible and the inverse of E is simply the elementary matrix that transforms E back to I Inverse Matrix Theorem Let A be an n n matrix Then the following are equivalent: (a) A is invertible (b) A x 0 has just the trivial solution (c) The Reduced Row Echelon Form of A is I n Proof: (a b) If A is invertible, we can multiply both sides of A x 0 on the left by A 1 to obtain A 1 A x A 1 0 or I x 0 so x 0 and we have just the trivial solution (b c) If A x 0 has just the trivial solution, then the system of equations in reduced form is x 1 0 x 2 0 x n 0 10
11 or 1x 1 + 0x x n 0 0x 1 + 1x x n 0 0x 1 + 0x x n 0 or x x x n and the RREF of A is I (c a) If the RREF of A is I, then there is a sequence of say k EROs that reduce A to I Each ERO is equivalent to multiplying A on the left by an elementary matrix Thus, there are k elementary matrices E 1, E 2, E k such that E k E 2 E 1 A I This tells us A 1 E k E 2 E 1 and so A is invertible A Method for Computing A 1 Looking at the equation E k E 2 E 1 A I, we multiply both sides on the right by A 1 to obtain E k E 2 E 1 I A 1 This tells us that the same sequence of EROs that reduce A to I will reduce I to A 1 We must keep track of the EROs that reduced A to I and apply these same EROs in the same order to I to give us A 1 We can apply these EROs simultaneously to a double matrix [A I] to produce [I A 1 ] If A does not reduce to I, we must have a row of 0s and so A is not invertible Example Find the inverse of (a) A (b) B
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