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QUIZ # 2 *This quiz’s worth 40 points. Please organize your paper. * Turn this quiz by 3/13 Show all the steps 1) (8) Find the determinant of A by reducing the matrix to a reduced row echelon form, using EROs ( elementary row operation) 2 2 4 A = 3 4 5 2 − 2 2 2) (8) If |A|= 4 and A, B are 3 3 matrices, det (3B )2 A −3 = 3 . What is |B|? 8 3) (8) Find the distance of the point (3,4) to the line 2 y − x = 5 4) (10) Follow the result of above question, find the distance of the point (3,−2,1) to the line 2 x − y + 3z = 2 Note: You only use the cross product for this question 5) (6) Given a b c d e f = 6. g h i Find 5a 5b 5c d /2 e/2 f /2 g − 2d h − 2e i − 2 f Math 410, Linear Algebra Chapter 1.7 Chapter 1.7 Diagonal, Triangular and Symmetric Matrices Fall 2020 Example 1: The product of a matrix A and with its transpose is symmetric 2 1 T A = −1 3 2 1 −1 3 A = 1 4 0 1 3 − 2 1 4 2 2 1 − 1 3 1 AAT == 1 4 0 1 3 − 2 1 4 − 1 3 1 3 4 − 2 0 1 1 4 1 3 15 9 15 4 − 2 = 9 18 − 1 0 1 15 − 1 40 1 4 Example 2: 3 0 0 0 1 / 2 0 1 / 8 A = 0 1 / 3 0 = 0 1 / 27 0 0 0 0 1 / 4 0 1 / 64 Example 3: Determine whether the matrix is symmetric A = aij a) aij = 2i 2 + 2 j 3 . Since a ji = 2 j 2 + 2i 3 # 2i 2 + 2 j 3 not symmetric b) a ij = 2i + 2 j Since a ji = 2 j + 2i = 2i + 2 j they are symmetric Math 410, Linear Algebra I) Chapter 1.8 Fall 2020 Chapter 1.8 Linear Transformation 1) Recall a vector in R n is called order n-tuples. It is represented by a column. But for convenient we can write in a comma-delimited form (or sometimes I physics: Contravariant vector) x1 : x= : : x n or x = ( x1 , x 2 ,……, x n ) And the standard basis vector associated with order n-tuples is In R 3 e = (e1 , e2 , e3 ) Example 1: We write a vector in R 3 in a standard basis vector as 1 0 0 x = (3,−2,5) = 30 − 21 + 50 0 0 1 Recall that in Cal 1 a function f(x) is a transformation from a value x in R to a value f(x) in R We say f : x → f ( x) In linear algebra we deal the same transformation with vectors in R n T : Rn → Rm 2) If the transformation is a matrix, we denote it as T A The matrix transformation, transforms a vector x in R n to a vector w in R m . The most easy way to find a standard matrix transformation is to observe the matrix operations Properties: For a matrix transformation T with u.v be vectors and k be a scalar The proof is specially easy and straight forward. Students try c) and d) Example 2: Consider a system of linear equation: A matrix transformation TA : R 4 → R 3 That is we can write 1 2 − 3 1 − 5 1 − 3 T A ( X ) = AX = 4 1 − 2 1 = 3 0 5 −1 4 0 8 2 Example 3: Some linear special transformations Zero matrix transformation T0 ( X ) = 0 X = 0 Example 4: Identity matrix transformation ( linear ) TI ( X ) = IX = I transforms a matrix to it-self 3) It is also obvious that the properties of matrices multiplications are compatible. Thus if TA ( X ) = TB ( X ) → AX = BX then A = B 4) Find a standard matrix: Example 5: If TA : R 2 → R 4 and 2x − 3y 2x − 3y x 3x x 3x T = Note, I usually write this instead of T y = x + 2 y It is the same y x + 2y − 2x + y − 2 x + y Then the matrix representation of T is 2 − 3 0 3 , where A = 1 2 − 2 1 2 1 3 Or, slowly, we see T (e1 ) = T = 0 1 − 1 − 3 0 0 and T (e2 ) = T = 1 2 1 − 4 Example 6: Find the image of X = under the transformation in example 5: 5 2 − 3 7 0 − 4 − 12 − 4 3 = T = 2 5 6 5 1 − 2 1 13 Example 7: Find a transformation that satisfies” 1 3 − 2 3 T is 2×2 and T = and T = 2 2 3 4 Think this as c TA = 1 c2 c3 . We have c4 c1 c3 c2 1 2 c = and 1 c4 2 3 c3 c2 − 2 3 = c4 3 4 So that c1 c3 c And 1 c3 c2 1 2 = c1 + 2c2 = 2; c3 + 2c3 = 3 c4 2 3 c2 − 2 3 = − 2c1 + 3c2 = 3 and − 2c3 + 3c4 = 4 c4 3 4 Now you can solve 2 2×2 systems with 4 unknowns: or − 2c1 + 3c 2 = 3 and − 2c3 + 3c 4 = 4 1 − 2 0 0 0 c1 2 3 0 0 c 2 3 = 0 1 2 c3 3 0 − 2 3 c 4 4 2 0 You can solve this 4×4 system. Or simply we can solve it as the example 1.6 A = Te1 | Te2 as in the example 7(1.8) in the textbook Here we have an augmented matrix for an 2 (2×2) system with the same coefficients 3 1 0 0 1/ 7 1 2 2 3 1 2 2 3 1 2 2 − 2 3 3 4 0 7 7 10 0 1 1 10 / 7 0 1 1 10 / 7 We have c1 = 0 c2 = 1; c4 = 10 / 7 and c3 = 1 / 7 Therefore 1 0 A = 1 / 7 10 / 7 Some Standard matrix simple transformations 5) x − x 1 0 Reflection about x-axis T = = y y 0 − 1 6) Reflection about y-axis x − x 1 0 T = = y y 0 − 1 7) Reflection about the line y = x x y 0 1 T = = y x 1 0 8) Reflection about the line xy plane x x 1 0 0 T y = y = 0 1 0 z − z 0 0 − 1 9) Reflection about the line xz plane x x 1 0 0 T y = − y = 0 − 1 0 Graph is similar to the above case z z 0 0 1 10) Reflection about the line yz plane x − x − 1 0 0 T y = y = 0 1 0 Graph is similar to the above case z z 0 0 1 11) Orthogonal Projection on x – axis 11) x x 1 0 T = = y 0 0 0 Orthogonal Projection on y – axis x 0 0 0 T = = y x 0 1 12) Orthogonal Projection on xy plane or ( xz plane , yz plane ) x x 1 0 0 T y = y = 0 1 0 z 0 0 0 0 13) Rotation operators: a) by counter-clockwise ( positive rotation) by angle θ cos R = sin b) − sin cos by clockwise ( negative rotation) by angle θ cos R− = − sin sin cos Example 8: What is the domain and co-domain of A has size 6 × 3. The domain is in R 3 and the co-domain is in R 6 Example 9: What is the domain and co-domain of A if a 2 1 c 0 0 −1 1 b 2 1 −1 c 2 e e 1 d The domain is in R 4 and the co-domain is in R 5 4 3 Example 10: Use matrix multiplication to find the image of the vector (3, −4) when it is rotated about the origin through an angle of θ =30º counter-clockwise 3 3 + 2 3 cos 30 − sin 30 3 2 T = = − 4 sin 30 cos 30 − 4 3 − 2 3 2 II) 1.9 Composition of matrices transformations If TA : R m → R n and TB : R k → R m then the composition of TAB : (k m) (m n) = k n We write TBA = TB (TA X ) = TB TA (X ) This is the composition transformation: TB TA (X ) 4) Find a standard matrix: Example 11: Given transformations T1 : R 3 → R 2 and T2 : R 2 → R 3 T1 ( x, y, z) = (3x − 2 y − z,−2x + y + 2z) T2 ( x, y) = (2x + 5 y,−2x + y, x + 3 y) Find a standard matrices T2 T1 ( X ) and T1 T2 ( X ) x x 3 − 2 − 1 3 − 2 − 1 y → A1 = We see that T1 y = − 2 1 2 − 2 1 2 z z 2 5 2 5 x x and T2 = − 2 1 → A2 = − 2 1 y 1 3 y 1 3 2 5 9 10 3 − 2 − 1 − 2 1 = T2 T1 ( X ) = A2 A1 = 2 − 4 − 3 − 2 1 1 3 2 5 − 4 1 8 3 − 2 − 1 = − 8 5 4 and T1 T2 ( X ) = A1 A2 = − 2 1 2 1 3 − 2 1 − 3 1 5 Note: * In general the multiplication is not commute, thus we have T1 T2 T2 T1 * The rotation transformations are commute Example 12: Given transformations cos20 − sin 20 cos30 − sin 30 and R2 = R1 = sin 20 cos20 sin 30 cos30 We will see by inspection that if we rotate a vector by 20º then rotate it 30º more is the same as rotating it by 30º then rotate 20º more. The total is 50º rotation. cos20 − sin 20 cos30 − sin 30 cos30 − sin 30 cos20 − sin 20 = sin 20 cos20 sin 30 cos30 sin 30 cos30 sin 20 cos20 cos20 − sin 20 cos30 − sin 30 cos50 − sin 50 = sin 20 cos20 sin 30 cos30 sin 50 cos50 III) The inverse of the transformation matrices The inverse of TA T2 I s defined as TA−1 = TA−1 The inverse of the transformation T A is related to T A −1 −1 −1 TATA = TA

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