BAI CM20144 Applications I: Mathematics for Applications Mark Wood

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BAI CM20144 Applications I: Mathematics for Applications Mark Wood cspmaw@cs.bath.ac.uk

BAI Quick refresher Worked examples Short test Book questions Linear Algebra with Applications – G. Williams Tutorial Structure

BAI Quick refresher Worked examples Short test Book questions Linear Algebra with Applications – G. Williams Revision requests E-mail me the week before if possible Tutorial Structure

BAI Gauss-Jordan Elimination

BAI Start with a system of linear equations Gauss-Jordan Elimination

BAI Start with a system of linear equations Write down the augmented matrix Gauss-Jordan Elimination

BAI Start with a system of linear equations Write down the augmented matrix Use elementary row operations Multiply a row by a nonzero constant Add a multiple of one row to another Swap two rows Gauss-Jordan Elimination

BAI Start with a system of linear equations Write down the augmented matrix Use elementary row operations Multiply a row by a nonzero constant Add a multiple of one row to another Swap two rows Aiming for reduced echelon form Rows of zeros are at the bottom All the other rows have leading 1s Each leading 1 is to the right of the one above Every column which contains a leading 1 has zeros elsewhere Gauss-Jordan Elimination

BAI Three possibilities Unique solution – read from right-hand column Many solutions – need to use parameters No solutions – will get 0 = 1 Gauss-Jordan Elimination

BAI Three possibilities Unique solution – read from right-hand column Many solutions – need to use parameters No solutions – will get 0 = 1 Can solve for many systems simultaneously by appending columns to the right-hand side of the augmented matrix. Gauss-Jordan Elimination

BAI Three possibilities Unique solution – read from right-hand column Many solutions – need to use parameters No solutions – will get 0 = 1 Can solve for many systems simultaneously by appending columns to the right-hand side of the augmented matrix. Algorithm Examples… Gauss-Jordan Elimination

BAI 4x 1 + 8x 2 – 12x 3 = 44 3x 1 + 6x 2 – 8x 3 = 32 -2x 1 - x 2 = -7 Gauss-Jordan Elimination

BAI 2x 1 - 4x 2 + 12x 3 - 10x 4 = 58 -x 1 + 2x 2 - 3x 3 + 2x 4 = -14 2x 1 - 4x 2 + 9x 3 - 6x 4 = 44 Gauss-Jordan Elimination

BAI x 1 - x 2 + 2x 3 = 3 2x 1 - 2x 2 + 5x 3 = 4 x 1 + 2x 2 - x 3 = -3 2x 2 + 2x 3 = 1 Gauss-Jordan Elimination

BAI x 1 - x 2 + 3x 3 = b 1 2x 1 - x 2 + 4x 3 = b 2 -x 1 + 2x 2 - 4x 3 = b 3 For b 1 = 8, 0, 3 in turn b 2 11 1 3 b 3 -11 2 -4 Gauss-Jordan Elimination

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