POLI270 - Linear Algebra

Transcription

1 POLI7 - Linear Algebra Septemer 8th Basics a x + a x a n x n b () is the linear form where a, b are parameters and x n are variables. For a given equation such as x +x you only need a variable and only the first power. Through substitution, elimination of variables, and matrix methods you get x + x 7 x + x () For n equations and n unknowns you get a x + a x... a n x n b a x + a x... a n x n b () Matrix arrays of numbers are treated as one mathematical object, and the elements of the matrix has it s own row and column number. For example, given the matrix 7 A (4) 7 8 we get a (order n, m is given in row, column format) A vector is an ordered set arranged in row or column. A symmetric matrix is defined by a ij a ji Diagonal matrix Scalar matrix (same value in all diagonals) Identity matrix Equality of matrices. For A B a ij b ij Transposition of matrix A is indicated by A and defined by making the nth row into the nth column (a ij a ji ) Addition and subtraction of matrices require the involved matrices to have the

2 same dimensions/comfortable for addition A + B B + A (A + B) + A A + (B + C) (A + B) A + B () Example: [ ] [ ] a a A b b, B a a b b [ ] (6) a + b A + B a + b a + b a + b Scalar multiplication KA ((Ka ij )) Vector multiplication ( ) a b c ( a b c ) (7) Matrix Multiplication: n by m matrix A and m by p matrix B gives n by p matrix C where each element is the dot product. [ ] [ ] [ ] [ ] 6 ()() + ()(7) ()(6) + ()(8) 9 (8) ()() + (4)(7) ()(6) + (7)(8) 4 The number of columns in the first matrix must be equal to the number of rows in the second matrix in order for multiplication to make sense. Multiplication is not commutative. This means that for matrices A and B, AB does not equal BA Example: A [ ], B 4 [ ] 4 AB [ ] BA 6 4 Multiplication of matrices also follow these laws: (AB)C A(BC) 4 6 [ ] (9) A(B + C) AC + AC (AB) B A ()

3 [ ] The Identity Matrix is if post multiplied will be self Row Operators by premultiplying by modified identity matrices To interchange rows and : () To multiply a row by a scalar 8 8 () To add twice second to the first 4 ()!! Column operations are performed by postmultiplying. Echelon matrix - each row begins with more zeroes than previous row (upper triangle. Diagonals are pivots. (i.e. 4 reduced echelon form all ) 6 No division, but yes inverse (aa ( ) a ( ), a ) BA AB I so must be n by n If B exists, A is called invertable/nonsingular, otherwise B is called singular. If the determinant does not equal, then it is nonsingular (x multiply by cofactor) [ ] [ ] 6 4 inverse 4 4 (4) 4 Sequence of row operators that transform A into I is premultiplying matrix B To get a matrix into echelon form ( upper triangle approach of text ) [ ] () 4 Subtract x first row from second row [ ] [ ] 4 4 [ ] (6) Subtract second row from first [ ] [ 4 ] [ ] (7)

4 [ ] [ ] [ ] 4 Product of is inverse If row or column of A is all zero A ( ) does not exist If row or column is multiplie of another, then the inverse does not exist (A ) A (A ) (A ) (AB) B A (8) Gaussian Elimination Algorithm to Inversion and Systems of Equations Augmented matrix: A A I (9) Start at top row. Interchange rows to get non zero entry. If not possible, there is no inverse.. Divide each element by diagonal element/pivot to get I in pivot position. 4. Add multiples of row to lower rows to sweep on non zeros below pivot.. Do for all rows. 6. Begin at bottom and work to sweep out non zeros above pivot. (a) Subtract twice the first row from second and for times first row from third. () (b) Switch and (c) 8 second row, 8 4 () third row 8 () 4

5 (d) Subtract third row from first 8 () (d) Add second to first 8 8 (4) If initially x x + x x x x 4 x 4x + 4x + 4x x () Solving Systems of Linear Equations For we use x + x x x (6) giving us For example: A AX A B IX A B X A B [ ] as the coefficient matrix and [ ] as the column vector. (7) x a + x b c [ ] [ ] [ ] x + x x + x 4x + 9x [ ] ( ) [ x 4 9 x ] (8) (9)

6 To solve, we multiply both sides by the inverse giving us ( x ) x [ [ ] 4 9] () To get the inverse of the coefficient matrix, we first set up the augmented matrix A I: [ ] () 4 9 Subtract twice the first row from the second and divide. Then subtract. times the second row from the first: [ ] [ ] [ ] () Multiplying that through the column vector, we get ] is equal to [ ] [ () ( ) x + x (4) Example : x + x + x x + x + x x x + 9x 9 Finding the inverse of the coefficient matrix, we get: 9 x x x Finally, solving for the variables, we get x x x () (6) (7) 6

7 In order to have a unique solution, there must be no linear dependence. For example, the vectors, 6 are linearly dependent. These three matrices (call them a, b and c respectively) can be rewritten [ ] [ ] [ ] (8) 4 as a + b c all three are not linearly independent Form into a matrix, derive the echelon form and if one rows are all zeroes they are linearly dependent R + R 7R + R (9) R is 6 (4) 6 The rank of the matrix is the number of non zero rows in echelon matrix There is a unique solution if the rank is the same as the number of unknowns. If there is a greater rank to the echelon form, then there are more equations than unkowns and the matrix is overdetermined (no solution). If there is less rank than the variables the the matrix has an infinite number of solutions; underdetermined as there are more unknowns than equations. Example : x x x + x 6x + x (4) Augment 8 (4) Giving a unique solution of x and x (4) 7

8 Example : x + x 4 x + x x + x 4 x + x (44) Solution: (4) (46) Giving us x + x 4, x x 4, and x + x 4. This is undetermined and there are an infinite number of solutions. Example : a + b + c a + b + c a + b + c 6 (47) Solution: (48) 6 This gives us a+b+c. This is overdetermined, and thus has no solution. 8