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MA122 Midterm

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  1. MA122 Midterm Rosanna Lo - Coordinator Shawn Lucas – Tutor Greg Krikorian - Tutor

  2. About Us • SOS started in 2004 • Raise money for children in Latin America • Many volunteer opportunities • Check out our website: http://www.studentsofferingsupport.ca/ • Video!

  3. Shawn Lucas Program: BBA w/Math & Comp Sci Minor Year: Second Hometown: Waterdown Favourite Team: Blue Bots (Obviously..) Qualifications: Taken all first year math courses; A+ in all Love: Corny Math Jokes

  4. Greg Krikorian • 2nd year BBA/Financial Math • From Kingston/Pittsburgh/Vancouver/Ottawa • I.A. for MA129 • Favourite Hockey Team is the Penguins

  5. About this Session • Welcome to Linear AlgeBRO • Listen! • Participate! • Ask questions! • Yell at Us System • What does little mermaid wear? • Algae-bra

  6. Agenda • Chapter 1 – Systems of Linear Equations and Matrices • Linear Systems • Gaussian Elimination • Matrices & Matrix Operations • Algebraic Properties of Matrices • Elementary Matrices & Inverse • Matrix Inverses & Linear System • Chapter 2 – Determinants • Cofactor Expansion of Determinants • Evaluating Determinants • Properties of Determinants 6

  7. Chapter 3 – Euclidean Vector Spaces • Vectors in 2-, 3-, n-space • Norm, Dot Product, & Distance in • Study Tips

  8. Chapter 1:Systems of Linear Equations and Matrices

  9. Linear Systems of Equations • What is a Linear Equation? • of the form • What is a Linear System? • two or more linear equations using the same variables • solution must satisfy each equation in the system

  10. How do we Solve? • How do we solve a Linear System? • Can use substitution & elimination • Can draw a graph • Or.. • Augmented Matrices!

  11. Matrix Forms • Row Echelon Form (REF) • Leading 1s • Leading 1 in an upper row must be to the left of the leading 1 in the row below it • Once we get into this form, can use back substitution 11

  12. Matrix Forms (ctd.) • Reduced Row Echelon Form (RREF) • In row echelon form and • All entries except leading 1s are zero • Once we get into this form, can use Gauss-Jordan Elimination 12

  13. Using Matrices to Solve • Elementary Row Operations • Multiply a row by any nonzero constant • Interchange any two rows • Replace a row by itself plus a multiple (+ or -) of another • Caution: DO NOT combine the first and last; it’s a shit storm • Gaussian Elimination • Step 1: Get leading 1 in first row • Step 2: Get zeroes for all entries below the first leading 1 • Step 3: Repeat Process with second row, and so on

  14. Using Matrices to Solve (ctd.) • Gaussian Elimination • Step 1: Get leading 1 in first row • Step 2: Get zeroes for all entries below the first leading 1 • Step 3: Repeat Process with second row, and so on • Now we have REF, but we need RREF: • Use EROs to introduce zeroes above leading 1s • ERO = Elementary Row Operations

  15. Example • Solve the system of equations

  16. Solution • Step 1: Get leading 1 in first row • Step 2: Get zeroes for all entries below the first leading 1 • Step 3: Repeat Process with second row, and so on Now we have REF, but we need RREF!!

  17. Example • Solve the system of equations • Solution:

  18. Matrix Operations • Addition/Subtraction • Two matrices A and B can only be added if: • They are the same size • How does this work? • Add numbers in same position in each matrix 18

  19. Example • Find the matrix C, where C = A + B, given the following matrices: A = B = 19

  20. Solution C = A + B C = + C = 20

  21. Matrix Operations • Scalar Multiplication • The matrix A can also be multiplied by a constant, say c • Multiply each entry in the matrix by c • Quick Example: • Find the matrix cA if A = and c = 2 • Solution: 21

  22. Matrix Operations • Matrix Multiplication • Suppose we have two matrices, A and B, and we wish to multiply them to form a new matrix AB.. • Important: The number of columns in A must equal the number of rows in B • e.g is 2 x 3, and so it can only be multiplied by a matrix that is 3 x n: “The entry of the jth row and kth column of the new matrix AB is the dot product of the jth row of A with the kth column of B.” = 22

  23. Example • Find the matrix C = AB if A = B = C = 23

  24. Matrix Operations • Transpose • For a matrix A, the transpose is denoted AT • How? Turn rows into columns and columns into rows • E.g. Find the transpose of • Solution: Rotate 45° clockwise and then swap columns OR; Your R1 is now C1, R2 is now C2, etc. - Solution: 24

  25. Matrix Operations • Trace • tr(A) • sum of the entries on the main diagonal of A 25

  26. Algebraic Properties • Inverse • The matrix A has an inverse matrix B if: • BA = AB = I • Recall, I is the identity matrix: • How do we find the inverse? • Row reduce the matrix [ A | I ] until the left part is in RREF • You will end up with a matrix of the form [ I | B ] where B = A-1 • Note that if the RREF of A is not I, then A is not invertible 26

  27. Example • Determine whether A = is invertible, and if so, determine its inverse.

  28. Solution R1-R2 R2-R1 R3-2R2 R3-2R1 R1+2R3 -R3

  29. Example • Determine whether A = is invertible, and if so, determine its inverse. • Solution: • Check: =

  30. Algebraic Properties • Using the Inverse to Solve a System • Suppose we have a system of equations Ax=b • E.g. • If we multiply both sides of the equation Ax=b by A-1, we get: A-1 Ax= A-1 b Ix= A-1 b x= A-1 b • So now we have another way to solve for x besides row reducing!

  31. Example • Solve the system of equations x1+x2+2x3 = 2 x1+2x2+2x3 = 7 2x1+4x2+3x3 = 5

  32. Solution • Step 1: Put into a Matrix • Step 2: Find the Inverse A-1(we found this already in the last example) • Step 3: Multiply the Inverse by b A= b= =

  33. Example • Solve the system of equations x1+x2+2x3 = 2 x1+2x2+2x3 = 7 2x1+4x2+3x3 = 5 • Solution: x1 = - 21, x2 = 5, x3 = 9

  34. Algebraic Properties • Matrix Polynomials • For a given polynomial , we define the matrix polynomial in Ato be • In other words, we substitute A for x and replace a0 with a0I • In better words; Change x to A (input matrix A), and any constants in your equation change to the identity matrix

  35. Example • Given and A = find p(A). • Solution:

  36. Chapter 2:Determinants

  37. The Determinant • What is the determinant? • For a 2x2 matrix: • det(A) = a11a22-a12a21 • For a 3x3 matrix: • det(A) = ai1Ci1+ai2Ci2+ai3ci3 “Cofactor Expansion” • Where C is the cofactor

  38. Cofactor • What is the cofactor? • First, some notation: • Let A(j,k) be the 2x2 matrix obtained from A by deleting row j and column k • Then the cofactor Cjk = (-1)(j+k)·det[A(j,k)] • Checkerboard method on board

  39. Example • Find the determinant of .

  40. Solution • Recall two important formulae: • det(A) = ai1Ci1+ai2Ci2+ai3ci3 • Cjk = (-1)(j+k)·det[A(j,k)] • Choose to expand along the last row: • det(A)= 1·C31+6·C33 • det(A)= 1·(-1)4·det[A(3,1)] +6·(-1)6·det[A(3,3)] • det(A)= • det(A)= (-1)(5)-(3)(1) + 6[(4)(3)-(2)(-1)] • det(A)= 76

  41. Properties of Determinants • Adjoint • Create a matrix of the cofactors for each number in the original matrix • Find the transpose of this matrix Original Matrix A Matrix of Cofactors Transpose

  42. Example • Find the adj(A) for A = • Solution: • The cofactors are C11 = -6 C12 = -3 C13 = 20C21 = 8 C22 = 4 C23 = -10 C31 = -8 C32 = 21 C33 = -40

  43. Solution • Solution: • The cofactors are Transpose C11 = -6 C12 = -3 C13 = 20C21 = 8 C22 = 4 C23 = -10 C31 = -8 C32 = 21 C33 = -40

  44. Properties of Determinants • We can use the adjoint to find the inverse of a matrix:

  45. Example • Find the inverse of A = • We know adj(A) from the last example…what’s det(A)? • Recall, we found C21 = 8, C22 = 4, C23 = -10 • Using this we know (Cofactor expansion middle)det(A) = (6)(8) + (8)(4) + (3)(-10)det(A) = 50 • And so…

  46. Solution

  47. Cramer’s Rule • Suppose we have a system of linear equations Ax = b… • Also suppose det(A) is not zero… • Then, we know the system has a unique solution, and that solution is: • What is An? It’s the matrix A, except that we substitute the entries in the nth column with the entries in b

  48. Example • Use Cramer’s rule to solve the system of equations:x – 4y + z = 64x – y + 2z = -12x + 2y – 3z = -20 • Step 1: Put into a matrix A = B =

  49. Solution A1 = A2 = A3 =

  50. Solution