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UNFORMATTED ATTACHMENT PREVIEW

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) = 30 − 21 + 50 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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