Chapter 4: Q47E (page 196) URL copied to clipboard! Now share some education! a.Find the change of basis matrix S from the basis B considered in Exercise 28 to the standard basisU=(1,t,t2)ofP2considered in Exercise 27.b.Verify the formula SB=AS for the matrices B and A you found in Exercises 28 and 27,respectively.c.Find the change of basis matrix from U to B Short Answer Expert verified a.The solution isS=(1−1101−2001)b. Verifiedc.The solution isS−1=(111012001) Step by step solution 01 Step 1:Definition for the B matrix of transformation T Consider a linear transformation T from V to V, where V is an n-dimensional linear space .Let B be a basis of V.Consider the linear transformationLB∘T∘LB−1 fromRntoRn with standard matrix BWhich impliesBx→=LB[T(LB−1(x→))]∶Äx→ â¶Ä‰in â¶Ä‰RnThis matrix B is called the B matrix of transformation T. 02 Step 2:Definition for the change of basis matrix of transformation T Consider two bases U and B of an n-dimensional linear space V.Consider the linear transformationLU∘LB−1 from Rnto Rnwith standard matrix S,meaning thatSx→=LU[LB−1(x→)] â¶Ä‰âˆ¶Ä â¶Ä‰â€‰x→ â¶Ä‰in â¶Ä‰RnThis invertible matrix S is called the change of basis matrix from B to U,sometimes denoted bySB→U 03 (a) Solution for the change of matrix S Consider the bases as B and U as followsB=[1 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰âˆ’1 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰10 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰2 â¶Ä‰â€‰â¶Ä‰â€‰âˆ’40 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰0 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰4]U=[1 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰âˆ’1 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰10 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰2 â¶Ä‰â€‰â¶Ä‰â€‰âˆ’40 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰0 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰4]Now by inspection we find out the change of matrix from B to USB→U=[a−b+cb−2cc]SB→U=(1−1101−2001)Hence the solution. 04 (b) Solution for the SB=AS Consider the matrix as followsB=[1 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰âˆ’1 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰10 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰2 â¶Ä‰â€‰â¶Ä‰â€‰âˆ’40 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰0 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰4]A=[1 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰âˆ’1 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰10 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰2 â¶Ä‰â€‰â¶Ä‰â€‰âˆ’40 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰0 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰4]Here both the A and B matrix is same.S=(1−1101−2001)SB=(1−1102−4004)[1 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰âˆ’1 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰10 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰1 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰âˆ’20 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰0 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰1]SB=(1−2502−8004)Similarly,AS becomes as followsAS=[1 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰âˆ’1 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰10 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰1 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰âˆ’20 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰0 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰1](1−1102−4004)AS=(1−2502−8004)Hence the proof 05 (c) Solution for the change of matrix from U to B Consider the bases as B and U as followsB=[1 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰âˆ’1 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰10 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰2 â¶Ä‰â€‰â¶Ä‰â€‰âˆ’40 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰0 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰4]U=[1 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰âˆ’1 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰10 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰2 â¶Ä‰â€‰â¶Ä‰â€‰âˆ’40 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰0 â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰4]Now by inspection we find out the change of matrix from U to BS−1U→B=[a+b+cb+2cc]S−1U→B=(111012001)Hence the solution. Unlock Step-by-Step Solutions & Ace Your Exams! Full Textbook Solutions Get detailed explanations and key concepts Unlimited Al creation Al flashcards, explanations, exams and more... Ads-free access To over 500 millions flashcards Money-back guarantee We refund you if you fail your exam. Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!