Question

linear algebra

do all parts A,B,C and D please

1. Let B = {bi, b2)- and C-(C1 , С2)- 111,12 be two ordered bases for R2 and VE then perform each of the following tasks. (a) Write v as then set up the augmented matrix for this linear combination and put your matrix in reduced row echelon form (not row echelon form) using pencil and paper calculations. Use your answer to state the coordinate vector VB (b) Write v as Use precisely the same technique in part (a) to find Vc (c) Using pencil and paper calculations to find the transition matrix PcB (or using your author's notation, [/g) (d) Using pencil and paper caleulations, show thait That is, take you answer for (vB from part (a) and multiply it by the transition matrix found in part (c) and show that the answer is the the coordinate vector [vlc found in part (b). Rather than just saying that vlPc-BVB by listing the matrices and vectors, show some steps to convince your reader that you understand the multiplication

Answer 1

B = {b_1 b_2} - { [1 1] [- 2 2] and c = implies {c_1 c_2} = { [- 3 1] [1 2]} v = [- 1 5] it v = a_1 b_1 + a_2 b_2 [- 1/5] = a_1 [1 1] + a_2 [- 2 2] [- 1 5] = [a_1 - 2a_2 a_1 + 2a_2] therefore a_1 - 2a_2 = -1 a_1 + 2a_2 = 5 implies matrix 1 1 - 2 2] [a_1 a_2] = [- 1 5] augmented matrix is [1 1 - 2 2 - 1 5] now we have to write is as reduced row echelon form R_2 rightarrow R_2 - R_1 [1 0 - 2 4 - 1 6] R_2 - 1/4 R_2 [1 0 - 2 1 - 1 6/4] \r\n R_1 rightarrow 2R_2 [1 0 0 1 2 3/2] therefore a_1 = 2 a_2 = 3/2 therefore v = 2 b_1 + (3/2) b_2 therefore [v]_B =