* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Document
Bra–ket notation wikipedia , lookup
System of polynomial equations wikipedia , lookup
History of algebra wikipedia , lookup
Capelli's identity wikipedia , lookup
Cartesian tensor wikipedia , lookup
Quadratic form wikipedia , lookup
Jordan normal form wikipedia , lookup
Eigenvalues and eigenvectors wikipedia , lookup
Linear algebra wikipedia , lookup
Symmetry in quantum mechanics wikipedia , lookup
Singular-value decomposition wikipedia , lookup
Four-vector wikipedia , lookup
Determinant wikipedia , lookup
Non-negative matrix factorization wikipedia , lookup
Matrix (mathematics) wikipedia , lookup
Perron–Frobenius theorem wikipedia , lookup
System of linear equations wikipedia , lookup
Matrix calculus wikipedia , lookup
Needs Work • Need to add – HW Quizzes Chapter 13 Matrices and Determinants 13.1 Matrices and Systems of Equations A matrix is a rectangular array of numbers. We subscript entries to tell their location in the array rows a a a 11 12 13 row a a a 21 22 23 A a31 a32 a33 am1 am 2 am3 m n a1n a2 n a3n amn Matrices are identified by their size. 1 5 3 1 5 0 2 4 1 2 6 1 3 4 4 2 1 2 4 1 3 5 7 2 5 8 9 7 9 0 4 A matrix that has the same number of rows as columns is called a square matrix. a11 a12 a a 21 22 A a31 a32 a41 a42 a13 a23 a33 a43 a14 a24 a34 a44 3x 2 y 5 z 3 2 x y 4 z 2 x 4 y 7z 1 If you have a system of equations and just pick off the coefficients and put them in a matrix it is called a coefficient matrix. 3 2 5 1 4 Coefficient matrix A 2 1 4 7 3x 2 y 5 z 3 2 x y 4 z 2 x 4 y 7z 1 If you take the coefficient matrix and then add a last column with the constants, it is called the augmented matrix. Often the constants are separated with a line. 3 3 2 5 # 1 4 2 Augmented matrix A 2 1 4 7 1 Operations that can be performed without altering the solution set of a linear system 1. Interchange any two rows 2. Multiply every element in a row by a nonzero constant 3. Add elements of one row to corresponding elements of another row We are going to work with our augmented matrix to get it in a form that will tell us the solutions to the system of equations. The three things above are the only things we can do to the matrix but we can do them together (i.e. we can multiply a row by something and add it to another row). We use elementary row operations to make the matrix look like the one below. The # signs just mean there can be any number here---we don’t care what. 1 0 0 # 1 0 # # 1 # # # After we get the matrix to look like our goal, we put the variables back in and use back substitution to get the solutions. Suse row operations to obtain row echelon form: We already have the 1 where we need it. 1 2 1 1 3 5 1 3 2 6 7 1 The augmented matrix We’ll take row 1 and multiply it by -3 and add to row 2 to get a 0. The notation for this step is -3r1 + r2 we write it by the row we replace in the matrix (see next screen). x 2y z 1 3x 5 y z 3 2x 6 y 7z 1 Work on this column first. Get the 1 and then use it as a “tool” to get zeros below it with row operations. 1 0 0 # 1 0 # # 1 # # # 11 -3r1 + r2 03 22 2 11 11 51 1 2 30 6 77 11 -2r1 + r3 11 00 02 22 11 26 11 22 57 1 0 11 Now our first column is like our goal. -3r1 + r2 -3 -6 -3 -3 3 5 1 3 0 -1 -2 0 -2r1 -2 -4 -2 -2 + r3 2 6 7 0 2 5 -1 1 Now we’ll use -2 times row 1 added to row 3 to get a 0 there. - 1r2 1 1 1 2 0 1 2 0 0 2 5 1 We need a 1 in the second row second column so we’ll multiply row 2 by a -1 -2r2 + r3 0 -2 -4 0 0 2 5 -1 0 0 1 -1 -2r2 + r3 11 22 11 11 00 11 22 00 00 20 51 1 1 We’ll use row 2 with the 1 as a tool to get a 0 below it by multiplying it by -2 and adding to row 3 the second column is like we need it now Now we’ll move to the second column and do row operations to get it to look like our goal. 1 0 0 # 1 0 # # 1 # # # z column y column x column xx222y z11 1 y y 22z100 y2 z 1 equal signs 1 2 1 1 x 2 0 1 2 0 0 0 1 1 Substitute -1 in for z in second equation to find y Substitute -1 in for z and 2 for y in first equation to find x. Now we’ll move to the third column and we see for our goal we just need a 1 in the third row third column and we have it so we’ve achieved the goal and it’s time for back substitution. We put the variables and = signs back in. Solution is: (-2, 2, -1) 1 0 0 # 1 0 # # 1 # # # x 2y z 1 3x 5 y z 3 2x 6 y 7z 1 Solution is: (-2, 2, -1) This is the only (x, y, z) that make ALL THREE equations true. Let’s check it. 2 22 1 1 3 2 52 1 3 2 2 62 7 1 1 These are all true. Geometrically this means we have three planes that intersect at a point, a unique solution. To obtain reduced row echelon form, you continue to do more row operations to obtain the goal below. 1 0 0 0 1 0 0 0 1 # # # This method requires no back substitution. When you put the variables back in, you have the solutions. Let’s try this method on the problem we just did. We take the matrix we ended up with when doing row echelon form: -2r 3r32+r11 -2r3+r2 11 00 000 002 11 000 0 13 02 11 x 2y z 1 3x 5 y z 3 2x 6 y 7z 1 112 02 00 x 2, y 2, z 1 1 11 Let’s get the 0 we need in Notice when we put the variables the second column by and = signs back in we have the using the second row as a tool. solution Now we’ll use row 3 as a tool to work on the third column to get zeros above the 1. 1 0 0 0 1 0 0 0 1 # # # Let’s try another one: The augmented matrix: 3 2 2 6 2 3 4 0 7 3 2 1 11 0 -2r1+r2 2 -7r1+r3 7 07 r1-r2 11 553 10 33 222 66 848 12 12 0 216 2 143 If we subtract the second row from the first we’ll get the 1 we need for the first column. 3x 2 y 2 z 6 2x 3y 4z 0 7 x 3 y 2 z 1 We’ll now use row 1 as our tool to get 0’s below it. We have the first column like our goal. On the next screen we’ll work on the next column. 1 0 0 # 1 0 # # 1 # # # 3x 2 y 2 z 6 2x 3y 4z 0 6 1 1 2 0 5 8 12 0 10 16 43 -1/5r2 10r2+r3 11 11 22 66 8 12 12 8 00 11 5 5 0 10 16 5 5 43 0 0 0 7 x 3 y 2 z 1 We’ll now use row 2 as our tool to get 0’s below it. 19 INCONSISTENT - NO SOLUTION If we multiply the second row by a -1/5 we’ll get the one we need in the second column. Wait! If you put variables and = signs back in the bottom equation is 0 = -19 a false statement! 1 0 0 # 1 0 # # 1 # # # 5 6 1 2 3 1 4 3 1 r1-r3 1 3 2 3 4 3 4 1 5 One more: 2 1 1 1 1 5 5x 6 y z 4 2x 3y z 1 4x 3y z 5 1 3 2 11 1/3r2 0 1 1 1 1 -9r2+r3 0 9 0 09 09 Oops---last row ended up all zeros. Put variables and = signs back in and get 0 = 0 which is true. This is the dependent case. We’ll figure out solutions on next slide. -2r1+r2 -4r1+r3 1 3 2 11 02 33 13 13 04 93 19 59 1 0 0 # 1 0 # # 1 # # # 1 3 2 1 0 1 1 1 0 0 0 0 solve for x & y Let’s go one step further and get a 0 above the 1 in the second column x 3r2+r1 y x z2 y z 1 put variables back in z 1 0 1 2 0 1 1 1 0 0 0 0 zz No restriction on z x z2 y z 1 zz Infinitely many solutions where z is any real number 5x 6 y z 4 2x 3y z 1 4x 3y z 5 532 612 01 4 232 312 01 1 432 312 01 5 works in all 3 What this means is that you can The solution can be choose any real number for z and written: (z + 2, z + 1, z) put it in to get the x and y that go with it and these will solve the equation. You will get as many solutions as there are values of z to put in (infinitely many). x z2 y z 1 Let’s try z = 1. Then y = 2 and x = 3 Let’s try z = 0. Then y = 1 and x = 2 zz Infinitely many solutions where z is any real number HW #13.1 Pg 572 1-4, 6-7, 9-11, 15 HW Quiz 13.1 Wednesday, May 24, 2017 Row 1, 3, 5 1. 4 2. 6 3. 10 4. 15 Row 2, 4 1. 2 2. 4 3. 6 4. 10 Chapter 13 Matrices and Determinants Section 13.2 Addition and Subtraction of Matrices To Add and Subtract matrices To find the additive inverse of a matrix To compare matrices they must have the same dimensions and have the same entries in the same positions You can only add or subtract matrices when they have exactly the same dimensions 4. The operation is not possible The additive inverse of a matrix can be obtained by replacing each element by its additive inverse. Finding the Additive Inverse of a Matrix Find the additive inverse of the matrix 12 2 15 16 0 9 3 9 13 13 3 2 10 Find the Additive Inverse of each Matrix 5 0 2 1 4 3 2 5 1 3 4 3 2 1 5 4 2 7 6 3 0 10 8 Subtracting by finding the Additive Inverse Subtract by finding the Additive Inverse 2 3 2 1 1 3 4 2 Exercises for Example 4 Subtract by finding the Additive Inverse HW # 13.2 Pg 575 1-32 Chapter 13 Matrices and Determinants Section 13.3 Cramer’s Rule Objective: Evaluate a 2 x 2 Determinant A B Objective: Solve a system of 2 equations and 2 variables using Cramer’s Rule Objective: Evaluate a 3 x 3 Determinant C Objective: Solve a system of 3 equations and 3 variables using Cramer’s Rule D E HW #13.3 Pg 580 1-33 odd, 34-42 Chapter 13 Matrices and Determinants Section 13.4 Multiplying Matrices MULTIPLYING TWO MATRICES A B AB 4 X 3 3 X 5 4X5 4 rows 3 rows 3 columns 5 columns MULTIPLYING TWO MATRICES A B AB 4 X 3 3 X 5 4X5 4 rows 4 rows 5 columns 5 columns MULTIPLYING TWO MATRICES If A is a 4 X 3 matrix and B is a 3 X 5 matrix, then the product AB is a 4 X 5 matrix. MULTIPLYING TWO MATRICES A B m X n n X p m rows n rows n columns p columns AB mXp MULTIPLYING TWO MATRICES A B AB m X n n X p mXp m rows m rows p columns p columns MULTIPLYING TWO MATRICES If A is an m X n matrix and B is an n X p matrix, then the product AB is an m X p matrix. Finding the Product of Two Matrices Find AB if –2 3 A = 1 –4 6 0 and B= –1 –2 SOLUTION Because A is a 3 X 2 matrix and B is a 2 X 2 matrix, the product AB is defined and is a 3 X 2 matrix. To write the entry in the first row and first column of AB, multiply corresponding entries in the first row of A and the first column of B. Then add. Use a similar procedure to write the other entries of the product. 3 4 Finding the Product of Two Matrices A B AB 3X2 2X2 3X2 –2 3 1 –4 6 0 –1 3 –2 4 (– 2)(– 1) + (3)(– 2) (– 2)(3) + (3)(4) (1)(–1) + (– 4)(–2) (1)(3) + (– 4)(4) (6)(– 1) + (0)(– 2) (6)(3) + (0)(4) Finding the Product of Two Matrices A B AB 3X2 2X2 3X2 –2 3 1 –4 6 0 –1 3 –2 4 (– 2)(– 1) + (3)(– 2) (– 2)(3) + (3)(4) (1)(–1) + (– 4)(–2) (1)(3) + (– 4)(4) (6)(– 1) + (0)(– 2) (6)(3) + (0)(4) Finding the Product of Two Matrices A B AB 3X2 2X2 3X2 –2 3 1 –4 6 0 –1 3 –2 4 (– 2)(– 1) + (3)(– 2) (– 2)(3) + (3)(4) (1)(–1) + (– 4)(–2) (1)(3) + (– 4)(4) (6)(– 1) + (0)(– 2) (6)(3) + (0)(4) Finding the Product of Two Matrices A B AB 3X2 2X2 3X2 (– 2)(– 1) + (3)(– 2) (– 2)(3) + (3)(4) (1)(– 1) + (– 4)(– 2) (1)(3) + (– 4)(4) (6)(– 1) + (0)(– 2) (6)(3) + (0)(4) –4 6 7 – 13 –6 18 3x2 2x2 AB will be 3 x 2 2x2 2x2 AB will be 2 x 2 2x2 2x2 BA will be 2 x 2 Properties of Matrix Arithmetic • For any matrices A, B, C of dimensions appropriate for them to be added or multiplied. – Commutative Property of Addition • A+B=B+A – Associative Property • A + (B + C) = (A + B) + C • A(BC) = (AB)C Properties of Matrix Arithmetic • For any matrices A, B, C of dimensions appropriate for them to be added or multiplied. – Additive Identity • There exists a unique matrix O such that A+0 = 0 +A=A – Additive Inverse • There Exists a unique matrix –A such that A + (-A) = -A + A = 0 Properties of Matrix Arithmetic • For any matrices A, B, C of dimensions appropriate for them to be added or multiplied. – Distributive Property • A(B + C) = AB + AC Properties of Matrix Arithmetic • For any real numbers k and m k(A + B) = kA + kB (k + m)A = kA + mA (km)A = k(mA) HW #13.4 Pg 587-588 1-39 Odd, 40-46 HW Quiz 13.4 Wednesday, May 24, 2017 Row 1, 3, 5 Write the answer to 1. 9 2. 15 3. 17 4. 39 Row 2, 4, 6 Write the answer to 1. 9 2. 15 3. 17 4. 39 1 3 1 0 4 6 14 7 1. 0 1 3 1 3 2 14 7 1 2 3 8 8 8 3 1 0 1 3 6 3. 1 1 2 8 8 8 1 1 1 2 2 4 8 8 8 4 6 1 0 4 6 2. 3 1 3 1 0 1 1 0 0 0 1 0 0 0 1 Chapter 13 Matrices and Determinants Section 13.5 Inverses of Matrices To write a matrix equation equivalent to a system of matrices To determine when two matrices are multiplicative inverses and find the multiplicative inverse of a 2 x 2 matrix Two n x n matrices are inverses of each other if their product (in both orders) is the n x n identity matrix. AA-1 = A-1A = 1 To determine if two matrices A and B are inverses of each other you need to make sure AB = BA = I Determine if A and B are multiplicative inverses of each other 3 2 5 3 4 5 B A 2 6 1 3 5 10 No Determine if C and D are multiplicative inverses of each other 2 3 C 3 6 2 1 D 2 1 3 Yes Tell whether the matrices are multiplicative inverses of each other. Computing an Inverse Matrix 2 x 2 Let’s Prove it! Use the shortcut Find the inverse of the given matrix 2 1 1 1 2 3 4 1 2 7 15 2 3 3 5 1 1 5 15 1 0 3 2 1 5 5 HW #13.5 Pg 591-592 1-19, 21-25 Odd 13.6 Inverses and Systems Writing Linear Systems as Matrix Equations Consider the system Let A = Let X = Let B = Write the equation AX = B using the above matrices Coefficient Matrix Matrix of Constants Matrix of Variables For a linear system of equations written as a matrix equation the matrix A is the coefficient matrix of the system, X is the matrix of variables, and B is the matrix of constants. Write the system of linear equations as a matrix equation 2 1 x 1 3 2 y 0 3 4 x 4 4 5 y 7 4. x 2 y 3 z 4 2 x y z 1 6 5 x 3 3 2 y 3 4x y z 1 1 2 3 x 4 2 1 1 y 1 4 1 1 z 1 SOLUTION OF A LINEAR SYSTEM Let AX = B represent a system of linear equations. If the determinant of A is nonzero, then the linear system has exactly one solution, which is X = A-1B. SOLUTION OF A LINEAR SYSTEM Let AX = B represent a system of linear equations. If the determinant of A is nonzero, then the linear system has exactly one solution, which is X = A-1B. SOLUTION OF A LINEAR SYSTEM Let AX = B represent a system of linear equations. If the determinant of A is nonzero, then the linear system has exactly one solution, which is X = A-1B. Solve system of linear equations. 2 1 x 1 3 2 y 0 3 4 x 4 4 5 y 7 4. x 2 y 3 z 4 2 x y z 1 6 5 x 3 3 2 y 3 4x y z 1 1 2 3 x 4 2 1 1 y 1 4 1 1 z 1 HW 13.6 Pg 596-597 1-19 Odd 13-7 Using Matrices Two stores sell the exact same brand and style of a dresser, a night stand, and a bookcase. Matrix A gives the retail prices (in dollars) for the items. Matrix B gives the number of each item sold at each store in one month. Calculate AB and interpret the entries of AB Your Turn HW #13.7 Pg 600-601 1-7 Chapter 13 Test Review Part 1 Helpful Hints • What does it mean when the determinant of a matrix is 0 – In terms of a system of Linear Equations – In Terms of a Matrix • Rules of Multiplication/Addition/Equality • Scalar Multiplication • Shortcut for finding the inverse of a 2 x 2 matrix Solve Evaluate the determinant of the matrix. Use Cramer’s rule to solve the system of equations. Use Augmented Matrices to solve the system of equations. Use the matrix equation AX = B to solve the system. When does a matrix fail to have an inverse? Study all the challenge problems in the book Know how to multiply and add matrices Chapter 13 Test Review Part 2 This test will have only 1 part HW #R-13 Pg 609 1-14