consider in r 3 the basis b v1 0 3 1 v2 1 2 2 v3 4 1 0 and b w1 1 0 0 w2 2 1 1 w3 3 2 1 a find t

To find the matrix of the change of basis from B to B', we need to express the vectors of B in t

Consider in r 3 the basis b v1 0 3 1 v2 1 2 2 v3 4 1 0 and b...

Consider in ℝ³ the basis B = {v₁ = (0, 3, 1), v₂ = (1, -2, 2), v₃ = (4, 1, 0)} and B' = {w₁ = (1, 0, 0), w₂ = (2, 1, 1), w₃ = (3, 2, 1)}. a) Find the matrix of the change of basis from B to B'. b) If vB = (3, 1, 0), calculate vB'. c) If vB = (3, 1, 0), calculate vE, where E is the canonical basis of ℝ³.

Transcript

00:01 Hi here for the given question.

00:03 We are given that there is a basis p which is equal to b1, which is 0 3 1 similarly b2 is equal to 1 minus 2 2 again.

00:16 B3 is equal to 4 1 0 now here we have b dash which is equal to w 1 which is equal to 1 0 0 similarly w 2 is equal to 2 1 1 similarly w 3 is equal to 3 2 and 1 now here we need to find a matrix that changes pieces from b to b dash.

00:42 So here matrix which changes pieces from b to b dash and here now in order to do that we need to write the matrix in a canonical form as w 1 is equal to a 1 1 multiplied with v 1 plus a 2 1 multiplied with b 2 plus a 3 1 multiplied with b 3 similarly w 2 is equal to a 2 1 multiplied with b 1 plus a 2 2 multiplied with b 2 plus a 2 3 multiplied with b 3.

01:21 Similarly w 3 is equal to a 1 3 multiplied with p 1 plus a 2 3 multiplied with b 2 plus a 3 3 multiplied with b 3.

01:33 So now here using the values we can say that the matrix form will be w 1 w 2 w 3 equals to v 1 v 2 v 3 multiplied with p.

01:46 Now here we need to find the matrix p in order to find the matrix p here.

01:50 We have value of v 1 is equal to 0 3 1 v 2 is equal to 1 minus 2 2 whereas the value of v 3 is equal to 4 1 0 similarly w 1 is equal to 1 0 0 w 2 is equal to 2 1 1 and w 3 is equal to 3 2 1.

02:13 So using this values our above equation can be written as 1 is equal to a 2 1 plus 4 a 3 1 similarly 0 is equal to a 1 1 plus a 2 2 plus 2 a 3 2 similarly 0 is equal to a 1 1 plus 2 a 2 2 plus a 3 2 again here we can write 0 is equal to a 1 1 plus 2 a 2 2 again 3 is equal to 3 a 1 1 plus minus 2 a 2 1 plus a 3 1 similarly 1 is equal to a 1 1 minus 2 a 2 1 plus 2 a 3 1 0 is equal to 4 a 1 1 plus 2 a 2 1 again 0 is equal to a 1 1 minus 2 a 2 1 1 0 is equal to a 1 1 plus a 2 2 plus 4 a 3 2 other 0 is equal to 3 a 1 1 minus 2 a 2 1 plus 2 a 3 2 again 1 is equal to a 1 1 minus 2 a 2 1 plus a 3 2 again 0 is equal to 4 a 1 1 plus 2 a 2 1 so solving this we have value of a 1 1 is equal to 1 by 5 similarly a 2 1 is equal to 2 by 5 value of a 3 1 is equal to minus 3 by 5 the value of a 2 2 is equal to 3 by 5 a 3 2 is equal to minus 4 by 5.

03:59 So here further rest of the values are 0 so here using this the matrix p can be written as p is equal to 1 by 5 0 0 2 by 5 3 by 5 0 minus 3 by 5 minus 4 by 5 0 so here this is the solution now here for the next part...

Question

Please give Ace some feedback