MATH 2210Q MIDTERM EXAM I PRACTICE PROBLEMS

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1 MATH Q MIDTERM EXAM I PRACTICE PROBLEMS Date and place: Thursday, November, 8, in-class exam Section : : :5pm at MONT Section : 9: :5pm at MONT 5 Material: Sections,, 7 Lecture 9 8, Quiz, Worksheet 9 8, and the practice exam and additional practice problems below Policy: No calculators will be allowed at the exam Format: The actual exam will have the same format as the practice problems (except possibly the points division within subproblems) The content will be reasonably similar to the practice exam and the additional practice problems (For the midterm II, at least five of the ten T/F questions are the original questions from the worksheets The T/F problems I gave in the practice exam are new problems) Practice Exam: True/False Questions ( points = points) You do not need to provide reasonings Correct = points, blank = point, incorrect = point () If A and B are n n matrices, then (A B)(A + B) = A B () If A is matrix and the equation Ax = invertible () If two rows of a -matrix A are the same, then det A = has a unique solution, then A is () If B is formed by replacing Row of A by the sum of Row and Row of A, then det B = det A (5) If B is formed by replacing one row of A by a linear combination of the other rows of A, then det B = det A (6) det( A) = det(a) (7) If V is a p-dimensional vector space and S = {v,, v p } be a set spanning V Then S cannot be linearly dependent (8) If H is a subspace of R, then there is a matrix A such that H = Col(A) (9) If A is a m n matrix and the linear transformation x Ax is onto, then rank(a) = m () A change-of-coordinates matrix is always invertible Short Problems (5 points = 5 points) () Give the inverse of the following matrix 7

2 () Find vectors that span the following subspace { a + b + c H = b + c a + c } ; for a, b, c R a + b () If A = a b c d e f is a matrix with det A = 5 What is the determinant of g h i Problem I (5 points) For the matrix A =, 9a 6b c 6d e f? g h i () (6 points) Compute its determinant () ( points) Determine det(( A) ) () (6 points) Consider the parallelepiped P with one vertex at the origin, and adjacent vertices, 8, 9, and consider the linear transformation T : R R given by x Ax What is the volume of T (P)? Problem II ( points) Consider P = {Polynomials of degree } with basis B = { + t, t, + t + t } () ( points) If a polynomial has B-coordinate What is this polynomial? () (7 points) Find the B-coordinate of the polynomial + t + t Problem III (5 points) Consider the matrix A = () ( points) If Col(A) is a subspace of R k, what is k? () ( points) If Nul(A) is a subspace of R l, what is l? () ( points) Find a basis of Nul(A), Row(A), and Col(A) () ( points) Determine if the vector u = is in Nul(A) (5) (6 points) Determine if the vector v = 5 is in Col(A)

3 Problem IV (5 points) Consider the following vectors in R 5 b =, b =, c =, c = 5 6 Let B = {b, b } and C = {c, c } be two bases of R 7 () ( points) What is the B-coordinate of the vector v =? 7 () ( points) Which vector in R has B-coordinate? () ( points) What is the change-of-coordinate matrix P B C from C to B? () ( points) What is the change-of-coordinate matrix P C B from B to C? Additional practice problems Problem Compute the determinant and the inverse of the following matrix A = Problem Compute the following determinant 7 Problem Determine the rank of the matrix A = and dim Nul(A)

4 True/False Questions Solution to the Practice Exam: () If A and B are n n matrices, then (A B)(A + B) = A B False Note that (A B)(A + B) = A(A + B) B(A + B) = A + AB BA B In general, AB BA, so (A + B)(A B) need not be equal to A B () If A is matrix and the equation Ax = has a unique solution, then A is invertible True By the property of homogeneous/non-homogeneous equations, if Ax = has a unique solution, then Ax = has a unique solution Therefore A is invertible () If two rows of a -matrix A are the same, then det A = True () If B is formed by replacing Row of A by the sum of Row and Row of A, then det B = det A True (5) If B is formed by replacing one row of A by a linear combination of other rows of A, then det B = det A False Such formed matrix B must have linearly dependent rows; and therefore, we always have det B = Maybe it is good to give an example to illustrate the difference between () and (5): For (), we are aiming to explain: a b c d e f g h i = a + d + g b + e + h c + f + i d e f g h i For (5), we are aiming to explain: d + g e + h f + i d e f g h i = (6) det( A) = det(a) False If A is n n, then det( A) = ( ) n det(a) (7) If V is a p-dimensional vector space and S = {v,, v p } be a set spanning V Then S cannot be linearly dependent True If the number of vectors of a spanning set is equal to the dimension, the set is linearly independent (8) If H is a subspace of R, then there is a matrix A such that H = Col(A) True Since H is a subspace of R, then H has a spanning set with vectors Putting these spanning vectors as columns of A and maybe adding zero columns at the end, gives a matrix Clearly H = Col(A)

5 (9) If A is a m n matrix and the linear transformation x Ax is onto, then rank(a) = m True If x Ax is onto, the column space of A is exactly R m ; and so rank(a) = dim Col(A) = m () A change-of-coordinates matrix is always invertible True Short Problems (5 points = 5 points) () Give the inverse of the following matrix 7 Solution We have A = d b = det A c a 7 7 = () Find vectors that span the following subspace { a + b + c H = b + c a + c } ; for a, b, c R a + b Solution We have { a + b + c } H = b + c a + c = a +b +c ; for a, b, c R = Span a + b 7 7/ = / {,, Remark: If I ask for a basis of H, we need to make a computation to see these three vectors are linearly dependent () If A = a b c 9a 6b c d e f is a matrix with det A = 5 What is the determinant of 6d e f? g h i g h i Solution We solve this by the following list of equalities 9a 6b c 6d e f g h i = a b c d e f g h i = a b c d e f g h i = 5 = 8 Here, for the first equality, we take out a common factor of from the first row (making 9a to a, 6b to b, and c to c), and take out a common factor of from the second row For the second equality, we take out a common factor of from the first column and a common factor of from the second column Problem I (5 points) For the matrix A =, () (6 points) Compute its determinant () ( points) Determine det(( A) ) 5 }

6 () (6 points) Consider the parallelepiped P with one vertex at the origin, and adjacent vertices, 8, 9, and consider the linear transformation T : R R given by x Ax What is the volume of T (P)? the system consistent? When the system is consistent, give the solution of the system Solution () We compute the determinant as follows = = 9 = () We have 9 = = 8 det(( A) ) = ( det( A)) ( = ( ) 8 ) = 6 Here the first equality follows from the fact that det(b ) = det(b) for a matrix B For the second equality, we note that A means to multiply every entry of A by This is equivalent to multiply the first row by, and then multiply the second row by, and finally multiply the last row by So going from det A to det( A), we need to multiply the determinant by, one factor for each row The cubic power corresponds to the matrix having size () The volume of the parallelepiped P is 8 9 = ( ) = 8 The linear transformation T will scale volume by det(a) So the volume of T (P) is 8 8 = 86 (The actual exam will have smaller numbers in the computation) Problem II ( points) Consider P = {Polynomails of degree } with basis B = { + t, t, + t + t } () ( points) If a polynomial has B-coordinate What is this polynomial? () (7 points) Find the B-coordinate of the polynomial + t + t Solution () The polynomial is () We write Comparing coefficients, we get ( + t) + t + ( + t + t ) = 7 + t + 5t + t + t = c ( + t) + c t + c ( + t + t ) = c + c = c + c = c + c 6

7 So we have [ + t + t ] B = Problem III (5 points) Consider the matrix A = () ( points) If Col(A) is a subspace of R k, what is k? () ( points) If Nul(A) is a subspace of R l, what is l? () ( points) Find a basis of Nul(A), Row(A), and Col(A) () ( points) Determine if the vector u = is in Nul(A) (5) (6 points) Determine if the vector v = 5 is in Col(A) Solution () Col(A) is a subspace of R ; so k = () Nul(A) is a subspace of R 5 ; so l = 5 () We perform a row reduction process on A: The columns, are pivot columns, and columns, and 5 correspond to free variables A basis of Row(A) is given by ( ), ( ) A basis of Col(A) is given by, 5 For Nul(A), we solve Ax = to get x x + x x x x = x x x 5 = x + x + x 5 x 5 x 5 7

8 So a basis of Nul(A) is given by,, () For this, we compute = = + + So u Nul(A) (5) For this, we solve So the system is inconsistent, and therefore, v does not belong to Col(A) Problem IV (5 points) Consider the following vectors in R 5 b =, b =, c =, c = 5 6 Let B = {b, b } and C = {c, c } be two bases of R 7 () ( points) What is the B-coordinate of the vector v =? 7 () ( points) Which vector in R has B-coordinate? () ( points) What is the change-of-coordinate matrix P B C from C to B? () ( points) What is the change-of-coordinate matrix P C B from B to C? Solution () For this, we solve ( ) So [v] B = () The vector is () For this, we solve So we get P B C = = + 9 = ( ) ( )

9 () The change-of-coordinates matrix from B to C can be computed by taking the inverse: P C B = (P B C ) = = = = 9

10 Solutions to Additional practice problems Problem Compute the inverse and the determinant of the following matrix A = Solution For determinant, we compute = 8 = = ( ) = We compute A as follows So we have 6 A = Problem Compute the following determinant 7 Solution We could compute by expansion by rows and columns: 7 = 7? +?? Now this becomes a lower triangular matrix; so we get = Problem Determine the rank of the matrix A = and dim Nul(A)

11 Solution We first compute the column space of A: Col(A) = Span {,,, } = Span {, } Here the second equality, we removed the zero vector and the duplicated vector Now the two vectors are not multiple of each other; so they are linearly independent and form a basis of Col(A) It them follows that rank(a) = dim Col(A) =, and therefore dim Nul(A) = dim rank(a) =