bai cm20144 applications i: mathematics for applications mark wood [email protected]
TRANSCRIPT
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
BAIGauss-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
4x1 + 8x2 – 12x3 = 44
3x1 + 6x2 – 8x3 = 32
-2x1 - x2 = -7
Gauss-Jordan Elimination
BAI
2x1 - 4x2 + 12x3 - 10x4 = 58
-x1 + 2x2 - 3x3 + 2x4 = -14
2x1 - 4x2 + 9x3 - 6x4 = 44
Gauss-Jordan Elimination
BAI
x1 - x2 + 2x3 = 3
2x1 - 2x2 + 5x3 = 4
x1 + 2x2 - x3 = -3
2x2 + 2x3 = 1
Gauss-Jordan Elimination
BAI
x1 - x2 + 3x3 = b1
2x1 - x2 + 4x3 = b2
-x1 + 2x2 - 4x3 = b3
For b1 = 8 , 0 , 3 in turn
b2 11 1 3
b3 -11 2 -4
Gauss-Jordan Elimination