lecture_slides_1.4-1.5.pdf - Math 220 Robert Creese Robert Creese 1 32 1.4 The Matrix Equation Ax = b Robert Creese 2 32 1.4 Product of a matrix and

# lecture_slides_1.4-1.5.pdf - Math 220 Robert Creese Robert...

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Math 220 Robert Creese September 12, 2019 Robert Creese September 12, 2019 1 / 32 1.4 The Matrix Equation: A x = b Robert Creese September 12, 2019 2 / 32 1.4 Product of a matrix and vector A x Let A be a m × n matrix, with each column denoted by a 1 , a 2 , · · · , a n . A = a 1 a 2 · · · a n If x is in R n , then we can define the product of A times x written A x , as the linear combination of the columns of A using the corresponding entires of x as weights. A x = a 1 a 2 · · · a n x 1 x 2 . . . x n = x 1 a 1 + x 2 a 2 + · · · + x n a n This is only defined when A has the same number of columns as x has entries (rows). Robert Creese September 12, 2019 3 / 32 1.4 Example: Multiplying a matrix and vector To evaluate a matrix times a vector we can write the equivalent expression using vectors and evaluate the expression as in Sec. 1.3. 4 2 - 1 3 2 5 1 - 2 = 1 4 - 1 2 + ( - 2) 2 3 5 = 4 - 1 2 + - 4 - 6 - 10 = 0 - 7 - 8 2 4 1 2 - 1 3 4 - 3 5 = 4 2 2 + ( - 3) 4 - 1 + 5 1 3 = 1 26 Robert Creese September 12, 2019 4 / 32 1.4 Vector Operations as a matrix times a vector Given v 1 , v 2 , v 3 , v 4 write 2 v 1 + 4 v 2 - 9 v 3 + v 4 as a matrix times a vector. Just use the equation definition of a matrix times a vector a 1 a 2 · · · a n x 1 x 2 . . . x n = x 1 a 1 + x 2 a 2 + · · · + x n a n . This leads to 2 v 1 + 4 v 2 - 9 v 3 + v 4 = v 1 v 2 v 3 v 4 2 4 - 9 1 Robert Creese September 12, 2019 5 / 32 1.4 Matrix Equation Given any linear system an equivalent vector equation can be found, for example, x 1 - 3 x 2 + 5 x 3 = 1 2 x 1 + x 2 - 4 x 3 = - 3 is equivalent to x 1 1 2 + x 2 - 3 1 + x 3 5 - 4 = 1 - 3 This is also a matrix equation , an equation written as A x = b 1 - 3 5 2 1 - 4 x 1 x 2 x 3 = 1 - 3 Robert Creese September 12, 2019 6 / 32 1.4 Matrix Equation: Finding a solution set If A is a m × n matrix with columns a 1 , . . . , a n and b in R m , the matrix equation A x = b has the same solution set as the linear system whose augmented matrix is a 1 . . . a n b The main method to solve any matrix equation A x = b is to find the reduced echelon form of the corresponding augmented matrix. Robert Creese September 12, 2019 7 / 32 1.4 Existence of solutions to A x = b A x = b has a solution if and only if b is a linear combination of the columns of A , { a 1 , . . . , a n } Section 1.3 covered the problem is a specific b in Span { a 1 , . . . , a n } ? This is equivalent problem to asking if A x = b has at least one solution. A new problem to consider: Is A x = b consistent for all possible b ? Robert Creese September 12, 2019 8 / 32 1.4 Example: Consistency Does A x = b have a solution for all possible b where A = 2 1 4 6 ? Begin by writing the corresponding augmented matrix ( b = ( b 1 , b 2 )) and finding echelon form 2 1 b 1 4 6 b 2 Replace row 2 by r 2 - 2 r 1 2 1 b 1 0 4 b 2 - 2 b 1 The echelon form is consistent. Therefore A x = b has a solution for all possible b in R 2 .  #### You've reached the end of your free preview.

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