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### q1S04soln

Course: MAT 202, Spring 2008
School: Princeton
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Word Count: 1045

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202 MAT Spring 2004, Quiz 1 on Chapters 1 and 2 1. (8 pts) Find all solutions to the given system by using elementary row operations to bring the matrix of the system into reduced row echelon form. Be sure to indicate which variables are free variables and which are leading variables. x + 2y + u + v + 3w = 0 2x + 4y + 5u + 2v + 3w = 0 3x + 6y + 3u - 4v + 2w = 0 Since the system is homogeneous (only zero appears on...

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202 MAT Spring 2004, Quiz 1 on Chapters 1 and 2 1. (8 pts) Find all solutions to the given system by using elementary row operations to bring the matrix of the system into reduced row echelon form. Be sure to indicate which variables are free variables and which are leading variables. x + 2y + u + v + 3w = 0 2x + 4y + 5u + 2v + 3w = 0 3x + 6y + 3u - 4v + 2w = 0 Since the system is homogeneous (only zero appears on the right hand side) we can take the coefficient matrix and reduce it completely to solve. We conclude that x = -2y - 3w, u = w and v = -w. The variables x, u and v are leading variables since they correspond to lead 1's in the reduced matrix. The other variables y and w are free. In vector form the solution is all vectors in R5 of the form 1 2 0 0 3 1 2 1 1 3 2 4 5 2 3 - 0 0 1 0 -1 0 0 0 1 1 3 6 3 3 9 x y u v w = s -2 1 0 0 0 +t -3 0 1 -1 1 2. (10 pts) Find all vectors b so that the system below is solvable. Interpret your answer geometrically. 2x1 - x2 + 4x3 = b1 x1 + x 2 - x 3 = b 2 7x1 + x2 + 5x3 = b3 Form the augmented matrix and reduce: Once more step gives 2 -1 4 | b1 0 -3 6 | b1 - 2b2 1 1 -1 | b2 1 -1 | b2 - 1 1 -1 | b2 1 - 0 1 -2 | 2b2 /3 - b1 /3 7 1 5 | b3 0 -6 12 | b3 - 7b2 0 1 -2 | 7b2 /6 - b3 /6 1 1 -1 | b2 2b2 /3 - b1 /3 0 1 -2 | 0 0 0 | b2 /2 - b3 /6 + b1 /3 So the original system will be solvable if and only if b2 /2 - b3 /6 + b1 /3 = 0 or in other words, if and only if b lies in the plane 2x + 3y = z. Geometric interpretation. The vector b can be written as a linear combination of the vectors (2, 1, 7), (-1, 1, 1) and (4, -1, 5) that form the columns of the coefficient matrix if and only if b is of the form 2b1 + 3b2 = b3 . This is because each of the column vectors already lies in this plane. We can get any vector in that plane by taking linear combinations of these vectors, but we can never move out of the plane this way. 3. (10 pts) Let T (x) = Ax be a linear transformation from R2 to R2 and suppose that T 3 2 = 3 2 and T 7 5 = 7 5 + 3 2 = 10 7 Find the matrix A of this transformation. Geometrically, what kind of transformation is this? Since A[v1 |v2 ] = [Av1 |Av2 ] it must be that A 3 7 2 5 = 3 10 2 7 We can solve this equation if we multiply through by A-1 . Using the formula a b c d we compute that A-1 = 5 -7 -2 3 A= -1 = 1 ad - bc d -b -c a and so 5 -7 -2 3 -5 9 -4 7 3 10 2 7 = This transformation is a shear along the line L consisting of all multiples of the vector (3, 2). Since T (3, 2) = (3, 2) we know that any vector of the form (3t, 2t) will be sent to itself. We are also told that the transformation behaves like a shear on the vector (7, 5) since it gets pushed parallel to the line L. This is enough to define the transformation everywhere, and so it must be a shear along the line L. 4. (10 pts) Find the matrix for reflection across the plane x+y-2z = 0. Is this matrix invertible? There lots are of ways to do this problem. The most common successful approach was the following: Let P denote orthogonal projection onto this plane and let R denote reflection across the plane. Then it is easy to see that R(x) + x = 2P (x). So first compute the projection matrix and then use it to compute the reflection matrix. To project onto the plane we use the fact that (1, 1, -2) is a normal vector to the plane. We compute the component of x that is perpendicular to the plane by projecting x onto the normal line. Then we subtract the normal component from x to find the component in the plane. This is the projection of x onto the plane. Let n denote any normal vector to the plane. Then xn P (x) = x - n nn This leads to the matrix that projects onto the plane 5/6 -1/6 2/6 5/6 2/6 Projection Matrix = -1/6 2/6 2/6 2/6 We can use this to compute the reflection matrix 2/3 -1/3 2/3 2/3 2/3 Reflection Matrix = -1/3 2/3 2/3 -1/3 This matrix is invertible. Its inverse is itself since geometrically it is clear that R(R(x)) = x. 5. (12 pts) Determine whether the following statements are true or false. If the statement is true, explain your reasoning. If the statement is false then find a counterexample. (a) Suppose that A is an n n matrix and v is a vector in Rn such that the system Ax = v has no solutions. Then the system Ax = 0 will have infinitely many solutions. TRUE. Ax = v has no solutions tells us that there must be a zero row in the reduced row echelon matrix of A. Therefore the rank of A is less than n so at least one of the n variables must be free. The system Ax = 0 is always solvable, and because of the free variable we will get infinitely many solutions. (b) If A, B and C are NONZERO 2 2 matrices and AB = AC then B = C. FALSE. If A is invertible then AB = AC implies that B = C. So for a counterexample we must search among noninvertible matrices. The simplest example is 1 0 0 0 2 3 0 0 = 2 3 0 0 = 1 0 0 0 2 3 4 5 (c) If A is a 2 2 matrix and A2 = I2 then A = I2 . FALSE. We need a matrix A that is its own inverse. We could figure out what all 2 2 matrices with this property are or we can just try to think up a simple example. We 0 1 would work. Another already know that any reflection would work, so A = 1 0 -1 0 easy to think up counter-example would be 0 -1 (d) If two matrices A and B have the same shape and the same rank, then each can be obtained from the other by a sequence of elementary row operations. FALSE. We need for example two rref matrices of the same shape and rank that are not equal. The easiest example is A= 1 0 0 0 and B = 1 1 0 0 Both are 2 2 matrices of rank 1 and clearly row operations can't transform them to each other. If we don't like to use words like clearly, we could argue as follows: the solutions to Ax = 0 and to Bx = 0 are not the same. Since row operations won't change the solution to the homogeneous system, we conclude that it is impossible to transform Ax = 0 to the system Bx = 0.
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