IMPORTANT NOTE: These are the answers. In the Exercise 1.2-30 exam you need to add justifications Find the polynomial of degree 3 whose graph Example (3.3-27) passes through the points (0,1),(1,0),(-1,0), (2,-15) Determine whether the following vectors form a basis of R4 (1,1,1,1), (1,-1,1,-1),(1,2,4,8),(1,-2,4,-8) Answer: Yes, (the matrix with these vectors as columns is invertible) Find the inverse of the rotation Let T be a clockwise rotation in R2 matrix. by π/2 followed by an orthogonal projection onto the y axis. 1. Find the matrix of T. cos(a) -sin(a) Answer= cos(a) sin(a) 2. Determine whether T is invertible 3. Find im(T) and ker(T) sin(a) cos(a) -sin(a) cos(a) Answer: It was given in class. For the matrix A below, find Find the inverse of the matrix and 1 2 a b all the 2x2 matrices X that check your answer. Interpret your satisfy the equation A.X=I2. 3 5 result geometrically. b -a Answer: The matrix is a reflection about a line L followed by a scaling by (a2 + b2 )1/2. The inverse is the same reflexion, followed -5 2 Answer: by a scaling (a2 + b2 )-1/2 3 -1 a b (a2 + b2 )-1/2 b -a 1 3.2-46 Find a basis of the kernel and image of the matrix. (2.4-31)For which values of the Determine the dimensions of the constants a, b and c is the kernel and image. following matrix invertible? Determine the rank. Justify your answers. There are 0 a b no values of a, b, c 1 2 0 3 5 -a 0 c that make the matrix invertible 0 0 1 4 6 -b -c 0 Done in class Note: The problem was first Example (3.3-31) stated that T was from R to Give an example of a 5 x 4 matrix R4 such. In this case, the Let V be the subspace of R4 answer is there are no A with dim(ker A)=3. defined by the equation x1- transformations (why? Prove Compute dim(im A). x2 + 2x3 + 4 x4 = 0 this!). Find a linear transformation T One answer for a transf. 3 4 1 0 0 0 from R to R such that from R3 to R4 such is The image has ker(T)={0}, im(T)=V. 0 1 0 0 dimension 2. 0 0 0 0 Describe T by its matrix. 1 0 0 0 0 0 0 1 2 0 0 0 0 0 0 -1 2 0 0 -1 Example (3.3-29) Find a basis of the subspace of R3 defined by the equation 2x1+3x2 + x3 =0 Answer: (3,-2,0), (0,1,-3) 2.