2018 Fall 2210Q Section 013 Midterm Exam II Solution

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1 08 Fall 0Q Section 0 Midterm Exam II Solution True or False questions points 0 0 points) ) Let A be an n n matrix. If the equation Ax b has at least one solution for each b R n, then the solution is unique for each b. True. If the equation Ax b has at least one solution for each b R n, then A is invertible, and thus the solution is unique for each b. ) Let A be an n n matrix. If the rows of A are linearly independent, then the columns of A span R n. True. If the rows of A are linearly independent, then A is invertible, and thus the columns of A span R n. ) If A and B are n n matrix, then A B is the inverse of AB. False. The inverse of AB is B A. 4) If A and B are n n matrix, then deta T ) detb T ) detab) T ). True. deta T ) detb T ) deta) detb) detab) detab) T ). ) A subset H of a vector space V is a subspace of V if the zero vector is in H. False. We need H to be stable under addition and taking scalar multiples. 6) If a linear transformation T : R m R n is onto, then the column space of the standard matrix A for T is the entire R m. True. 7) The null space of A is the solution set of the equation Ax 0. True. 8) If dim V n, then any subset of n vectors in V spans V if and only if the subset is a linearly independent subset. True. Any spanning set of n vectors is a basis, and any linearly independent set of n vectors is a basis. 9) If dim V p, then there exists a spanning set of p + vectors in V. True. This is because, we can always take a basis which consists of p vectors) and then add another arbitrary vector, and still get a spanning set. 0) If two matrices A and B are row equivalent, then their row spaces are the same. True.

2 Short Problems points points) ) Let A be a 4 7 matrix. What are the minimal possible dimension of NulA)? Please give an example for the minimal possible dimension of NulA). Solution. The rank of A is at most min{4, 7} 4. So dim NulA) n ranka). so dim NulA) is at least. For example A There are three variables; so dim NulA). ) Inside P {Polynomials of degree }, find a nontrivial) linear relation among p + t, p t, p 4 + t. Solution. We write c p + c p + c p 0. So c + t) + c t) + c 4 + t) 0 Comparing the coefficients, we get { c + c + 4c 0 c c + c 0 ) ) ) ) x So we have solution x. Taking x, we get linear relations x p p + p 0. ) If A is a -matrix with det A. What is deta) )? Solution. We have deta) ) deta) ) deta) ) 8 ) 40.

3 0 Problem I points) Consider the matrix A. ) points) Compute the determinant of A. Remark: your determinant should be nonzero.) ) 7 points) Why does the computation in ) shows that A is invertible? Find the inverse of A. The numbers for A are a little larger than I had expected.) ) points) What are the column space of A and the null space of A? Hint: this is a simple question given your answers to ) and ).) Solution. ) We compute ) A is invertible because deta) 0. We compute its inverse as follows So we have A 0 ) Since A is invertible, the column space of A is the entire R. The null space of A is {0}.

4 Problem II 0 points) ) 7 points) Compute the area of the parallelogram P with vertices at ) ) ) ) 0,,,. 0 6 ) points) If T : R R is a linear transformation given by ) 4 7 x x What is the area of T P)? ) Solution. ) We pick the vertex. After drawing the pictures, we see that the ) ) 0 0 adjacent vertices are and. So we compute the side vectors of the parallelogram 6 P. ) ) ) ) ) ) 0, So AreaP) 6 6. ) The linear transformation will scale any area by a factor of So the area of T P) is. 4

5 Problem III points) Consider the matrix A 4 0. ) points) If ColA) is a subspace of R k, what is k? ) points) If NulA) is a subspace of R l, what is l? ) points) Find a basis for each of NulA), RowA), and ColA). 4) 4 points) Determine if the vector u 0 is in NulA). ) 6 points) Determine if the vector v Solution. ) ColA) is a subspace of R 4 ; so k 4. ) NulA) is a subspace of R ; so l. ) We perform a row reduction process on A: 4 0 is in ColA) The columns,, 4 are pivot columns, and columns and correspond to free variables. A basis of RowA) is given by ), ), ). A basis of ColA) is given by, 0,.

6 For ColA), we solve Ax 0 to get x x 0 x x x 4 x 0 x x x 0 0 x x 0 So a basis of ColA) is given by 0 0 0, ) For this, we compute Au So u belongs to NulA). ) For this, we solve 4 0 So the system is consistent, and therefore, v ColA)

7 Problem IV points) Consider the bases B {b, b } and C {c, c } of R given by ) ) ) ) 4 7 b, b, c, c. 7 ) points) Give a reason why B {b, b } is a basis of R ). ) 4 points) What is the B-coordinate of the vector v? ) 8 ) points) Which vector in R has B-coordinate? 4) points) What is the change-of-coordinates matrix from C to B? Solution. ) Note that b and b are not multiple of each other. So they form a linearly independent set. In R, any linearly independent set of vectors forms a basis. So B is a basis of R. ) For this, we solve ) 8 ) ) So [v] B. ) The vector is ) ) 4) For this, we solve ) So we get P B C ). ) ) 0 ) ) ) ) 0 0 ) 0 7

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