Answers to practice questions for Midterm 1

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1 Answers to practice questions for Midterm Paul Hacking /5/9 (a The RREF (reduced row echelon form of the augmented matrix is So the system of linear equations has exactly one solution given by x =, y =, and z = Geometrically, each of the equations defines a plane in R, and the intersection of the three planes is one point (b The RREF of the augmented matrix is So the system of linear equations has a line of solutions given by x = 5 t, y = + 4t, z = 5 t (the variable t is free Geometrically, each equation defines a linear space of dimension in R 4, and the intersection of these is a line in R 4 The vector equation is equivalent to a system of linear equations in variables The RREF of the augmented matrix is 4 So there are no solutions (the last row corresponds to the equation = Geometrically, the vectors v, v, v lie in a plane P in R, and the vector

2 b does not lie in P So any combination x v + x v + x v also lies in P and in particular cannot be equal to b The RREF of the augmented matrix is So the solutions are given by x = x x 4 and x = x 4 (the variables x and x 4 are free The equation Ax = c does not have a solution for every vector c in R changing b to c will only change the last column of the RREF above, and for some choice of c we will get last row of the form ( d for some d solutions Then for this choice of c the equation Ax = c has no 4 (a S is reflection in the y-axis, T is rotation about the origin through an angle of π/4 radians clockwise, U is reflection in the line y = x, V is a vertical shear (see p 6 64 of the textbook (b The matrix of S U is ( ( = The matrix of T T is ( ( = The matrix of T U T is ( ( ( ( ( = ( Geometrically, S U and T T are both rotation about the origin through an angle of π/ radians clockwise, T U T is reflection in the x-axis

3 5 (a The matrix of a rotation T : R R about the origin through an angle of θ radians anticlockwise is ( cos θ sin θ sin θ cos θ In our case θ = π/ so the matrix is ( (b The reflection U : R R in the line through the origin with direction v is given by the formula ( x v U(x = v x v v ( ( x In our case v = so, writing x = we have y ( ( ( x + y x U(x = = 5 y 5 ( So the matrix of U is ( x + 4y 4x + 8y ( x = y 5 ( x + 4y = 4x + y 5 Alternatively, we can ( first write down a vector perpendicular to the line, for example n =, and use the formula U(x = x ( x n n n n for U : R R the reflection in the line through the origin perpendicular to the vector n (c The equation x + y + z = of the plane can be written as x n = x, where x = y and n = So we see that the vector n z given by the coefficients of the equation is a normal vector to the plane (a vector perpendicular to every vector lying in the plane The ( 4 x 4

4 projection V : R R onto the plane through the origin with normal vector n is given by the formula ( x n V (x = x n n n In our case this gives matrix (d Recall that the matrix of a linear map T : R n R m has columns T e,, T e n, where e i denotes the vector with ith entry and all other entries Now using this recipe for W (draw a picture! gives matrix cos(π/ sin(π/ = sin(π/ cos(π/ 6 (a The line is the line through the origin in some direction (Note: if we project onto a line that does not go through the origin then we don t get a linear map, because for a linear map T we always have T ( = To find the direction, just compute the image T (v of any vector under the map T that will be a vector in the direction of the line For example if we take v = ( So the line has direction ( then T (v = ( 4 6 = (b To find the plane we observe that for any vector v in R the difference T (v v will be a normal vector to the plane Taking v = we find T (v v = = = ( 4

5 7 So the plane of reflection is the plane through the origin with normal vector, in other words, the plane with equation x + y z = (c A vector x in the direction of the axis of rotation satisfies U(x = x This is a system of linear equations for x Solving for x we find that x is a multiple of So the axis of rotation is the line through the origin in the direction (a To find the inverse of A we apply the row reduction algorithm to the matrix (A I formed by A and the identity matrix This gives a matrix (I B, and then B = A is the inverse of A In this way we compute 5 7 (b The system of linear equations can be written as Ax = b where b = This has solution x = A b = 7 8 The matrix of T is the matrix with columns given T, T, T, that is, the matrix ( Now we can compute the images of the vertices,,,,,,, 5

6 of the unit cube under the map T : they are ( ( ( ( ( (,,,,,, ( ( 4, 4 The image of an edge of the cube is the line segment joining the images of the two vertices on the edge 6

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