Matrices MM1 module 3, lecture 3

1. MatricesMM1 module 3, lecture 3 David Godfrey

2. Matrices lecture 3 objectives • After this lecture you should have a clear understanding of: • Solving linear equations in 2 variables; • Solving linear equations in 3 variables; • Gaussian elimination for solving linear equations in n variables; • Ill-conditioning; • The nature of solutions for linear equations.

3. Linear equations… • A set of n linear equations in n variables, xican be written in matrix form, Ax = b:

4. …Linear equations • e.g. the 2  2 system of linear equations • can be written in matrix form, Ax = b: • To solve, we premultiply the equation by A–1, A–1Ax = A–1b, giving x = A–1b:

5. Matrix equations… • We can use this method for matrix equations of the form AX = B: e.g. find the matrix X if • We can get rid of the matrix to the left of X by premultiplying each side of the equation by the inverse • Thus

6. …Matrix equations • Similarly for XA = B: e.g. find the matrix X if • We can remove matrix A by postmultiplying each side of the equation by its inverse, A–1 • Thus

7. Solving n nlinear equations • We can solve a general system of n nlinear equations by writing it in matrix form, AX = B, and premultiplying the matrix equation by the inverse of the matrix of coefficients, A–1. • i.e. if AX = Bthen A–1AX = A–1B, but A–1A = Iso X = A–1B • However, this method is almost never used because it is inefficient and prone to rounding errors. • We shall investigate Gaussian elimination, a more efficient and accurate method for solving linear equations.

8. Visualising n nlinear equations • 3 variables • Solution is at the intersection of 3 planes • 2 variables • Solution is at the intersection of 2 lines

9. Gaussian elimination… • Consider a system of 3  3 linear equations in matrix form, Ax = b: • To make book-keeping simpler, we represent the system by an augmented matrix:

10. …Gaussian elimination… • We can zero the first column by subtracting a21/a11 times the first row from the second row, and subtracting a31/a11 times the first row from the third row (primes indicate changed values) • Similarly, we can now zero the second column by subtracting a’32/a’22 times the first row from the third row (double primes indicate changed values), forming an upper triangular matrix:

11. …Gaussian elimination… • The last row represents an equation in a single variablea”33x3 = b”3which can be solved as x3 = b”3 / a”33 • The second row represents an equation in two variablesa’22x2 + a’23x3 = b’2 • Since the variable x3 has already been found in the previous step, x2 can be solved asx2 = (b’2 – a’23x3) / a’22

12. …Gaussian elimination • The first row represents an equation in three variablesa11x1 + a12x2 + a13x3 = b1 • Since the variables x2 and x3 have already been found in the previous steps, x1 can be solved asx1 = (b1 –  a12x2 – a13x3) / a11 • This process of solving an upper triangular matrix equation is called back substitution.

13. Gaussian elimination: example 1… • Solve the system of equations • Represent the system as an augmented matrix: • We introduce a lower triangular matrix, L, to record row multipliers in the • positions

14. …Gaussian elimination: example 1… • Calculate the row multipliers and record them in the L matrix, l21 = a21/a11 and l31 = a31/a11 • Zero the first column by subtracting l21 = a21/a11 times the first row from the second row, and subtracting l31 = a31/a11 times the first row from the third row

15. …Gaussian elimination: example 1… • Calculate the row multiplier and record it in the L matrix, l32 = a’32/a’22 • Zero the second column by subtracting l32 = a’32/a’22 times the second row from the third row, forming an upper triangular matrix

16. …Gaussian elimination: example 1 • Back substitute • The last row represents an equation in a single variable, a”33z = b”3, which can be solved as z = b”3 / a”33 = –7 / –7 = 1 • The second row represents an equation in two variables, a’22y + a’23z = b’2, which can be solved asy = (b’2 – a’23z) / a’22 = (–6 + 2) / (–2) = 2 • The first row represents an equation in three variablesa11x + a12y + a13z = b1, which can be solved asx1 = (b1 –  a12y – a13z) / a11 = (4 – 2 – 1) / 1 = 1 • The solution is thus x = 1, y = 2, z = 1.

17. Gaussian elimination: example 2… • Solve the system of equations • Represent the system as an augmented matrix:

18. …Gaussian elimination: example 2… • Calculate the row multipliers and record them in the L matrix, l21 = a21/a11 and l31 = a31/a11 • Zero the first column by subtracting l21 = a21/a11 times the first row from the second row, and subtracting l31 = a31/a11 times the first row from the third row

19. …Gaussian elimination: example 2… • Calculate the row multiplier and record it in the L matrix, l32 = a’32/a’22 • Zero the second column by subtracting l32 = a’32/a’22 times the second row from the third row, forming an upper triangular matrix

20. …Gaussian elimination: example 2 • Back substitute • The last row can be solved as z = b”3 / a”33 = 0.6 / 0.3 = 2 • The second row can be solved asy = (b’2 – a’23z) / a’22 = (–11 + 6) / (–5) = 1 • The first row can be solved asx = (b1 – a12y – a13z) / a11 = (7 – 5 – 6) / 2 = –2 • The solution is thus x = –2, y = 1, z = 2.

21. Gaussian elimination: example 3… • Solve the system of equations • Work to 4 significant figures and give the answer to 3. Check the answer by substituting into the original equation. • Represent the system as an augmented matrix:

22. …Gaussian elimination: example 3… • Calculate the row multipliers and record them in the L matrix, l21 = a21/a11 and l31 = a31/a11 • Zero the first column by subtracting l21 times the first row from the second row, and subtracting l31 times the first row from the third

23. …Gaussian elimination: example 3… • Calculate the row multiplier and record it in the L matrix, l32 = a’32/a’22 • Zero the second column by subtracting l32 = a’32/a’22 times the second row from the third row, forming an upper triangular matrix

24. …Gaussian elimination: example 3… • Back substitute • The last row can be solved as z = –39,880,000 / –200,400,000 = 5.025 • The second row can be solved asy = (–200,000 + 199,995) / (0.001) = –5,000 • The first row can be solved asx = (b1 – a12y – a13z) / a11 = (10 + 0.5 – 9.99975) / 0.0001 = 5,003 • The computed solution is thus x = 5,003, y = –5,000, z = 5.03

25. …Gaussian elimination: example 3 • Check the computed solution against the original equations • Numerical rounding has caused these errors. • Working to more significant figures givesx = –7.939 c.f. 5,000y = 5.922 c.f. –5,000z = 5.025 c.f. 5.03

26. Geometric interpretation • Ill-conditioning, as seen in example 3, can be interpreted geometrically: • The shaded red area represents uncertainty in the lines’ positions due to rounding errors. • The solid red area represents where the intersection may be found due to these errors. • The closer to parallel the lines are, the larger the possible error.

27. Geometric interpretation: 1 solution • Gaussian elimination to the formwhere none of the diagonal elements are zero • We get the usual form of the augmented matrix and can fine a unique solution by back substitution

28. Geometric interpretation: 0 solution • Gaussian elimination to the formwhere b3 is not zero • We end up with 0 z = non-zeroi.e. no solution can exist because at least 2 planes, or 3 lines of intersection, are parallel

29. Geometric interpretation: ∞ solution • Gaussian elimination to the formi.e. b3 is zero • We end up with 0 z = zeroi.e. an infinite number of solutions exist because this is always true for any value of z

30. Examples • Performing Gaussian elimination brings us to the augmented matrices shown. Discuss the nature of the solutions • 1 z = 1  z = 1Unique solution • 0 z = 1 ???No solution • 0 z = 0 !!!Infinite solutions

31. Matrices lecture 3 objectives • After this lecture you should have a clear understanding of: • Solving linear equations in 2 variables; • Solving linear equations in 3 variables; • Gaussian elimination for solving linear equations in n variables; • Ill-conditioning; • The nature of solutions for linear equations.