sample answer attached below

Let the basis vectors be given by a = [2, -2] and b = [-1, 2]. Give the image of a general coordinate vector [x, y] ER2. Give your answer as a row vector whose components are linear functions of x and y. T ([x, y]) = 1, ,Step 2/2 Let the basis vectors be given by 51' = [2, 2] and E = [3, 1] . Give the image of a general coordinate vector [55,31] 6 R2. Give your answer as a row vector whose components are linear functions of a: and y. Answer T([w,y]) = [Dimilj] x Wrong x 3-1; 17-:1: 31-31 T , = 5' 6-, ([5631]) 2 2 t (B y 4 4 The most important ingredient we need is the transformation matrix T5. Because we have the basis vectors of a and their images, we can calculate T5 from the matrices ETC! and (135: T5 = ETa ' ass Here, 035 is the transition matrix from the standard basis to the basis a. It is the inverse of the transition matrix ESQ from the basis a to the standard basis: use = 580:1 We form ESQt by placing the given basis vectors as columns in a matrix: asav H: i') The inverse is given by abe = Next, we form : Ta by placing the given images of the basis vectors as columns in a matrix: 2 0 E Ta = (T(a) T(b) ) = 2 9 5 Then we find 2 0 Te = =Ta . abe = 2 N / H IA / CO = OUT When T: operates on a column vector, it produces another column vector: N/8 3.y 2 TE 5 . x - 6 . y 31 17.a 31.2 4 4 4 Written as a row vector, as requested, the answer becomes: T ([x, y]) = 3 . y 2 5 . x - 6 . y, 17 . x 31 . y 2 4 4