Question

inteb CORE 17 20. The matrices in the last two Exercises were the standard matrices of the operators [proji] and refli], respectively, where L is a line through the origin in R2 with unit direction vector (a, b) See Exercise 25 in Section 2.2. Give a geometric argument as to why one of these matrices is invertible and the other matrix is not invertible. Explain also the geometric significance of the inverse of the invertible matrix. For Exercises (21) to (24): Forthe given linear transformation T: R2 → R2 : (a) find [T]; (b) find I7 1J, if 7 is invertible, and if so: (e) explicitly give a formula for T(x,)) and finally, (d) check that T (TKx, y)(x,y). 21. 7((x,y))=(3x-7),-4x + 9y) 23. Tx,y) (3x + 5y, 5x +9y) and only if either T is one-to-one or T is onto. 26. Let's play: [3][2-7]lstheresult ngma us invertible? a. Compute: 1 u1 nnd 1 x2 matrices and multiplying th
21 please

Answer 1

T: R^2 rightarrow R^2 be a linear transformation T(x , y) = (3x - 7y , -4x + 9y) T(1, 0) = (3, -4) = 3(1, 0) - 4 (0, 1) T(0, 1) = (-7, 9) = -7 (1, 0) + 9 (0, 1) Therefore [T] = [3 -4 -7 9] Let xi = (x , y) elementof ker T Therefore T(x , y) = 0 Rightarrow (3x - 7y , - 4x + 9y) = 0 Rightarrow 3x - 7y = 0 - 4x + 9y = 0 Rightarrow x = y = 0 Therefore ker T = {(0, 0)} Therefore T is one to one \r\n By Rank-nullity theorem dim (ker T) + dim (I_m T) = 2 Rightarrow 0 + dim (I_m T) = 2 Rightarrow dim (I_m T) = 2 Therefore I_m T = R^2 Therefore T is onto Therefore T is Invertible