Question

(10) 5. (a) Find Ker(T) and Rng(T), dimensions, bases, and give a geometrical description of each if T:R3 R3, T(x) = Ax, and 24-1 A = -1 -2 1/2 4 8-2

Answer 1

R? Solution: Given, T:13 T(x) = AX by, 2 4 also given A= - 2 - 1 V2 - 2 8 kert X EIR² : AX O 2 Now, - A= 4 R2 = R2+1 / Ri -1 -2 42 R3 = R3 - 2 R1 4 8 - 2 Let, X = E IR X2 23 Now, AX=0 imply 2 x4 + 482 - x3 = 0 Let, az=c and x2 = d (where Cad some scalars) From we get, 2x = c - 4d > 24 = 2 - 2d V2 Hence, - 2d + d X = d *) C un td 2 ¿' ( ) + d (3) Luts Hence, basis of Ke (t) = { (6): (?)}

Hence, Kert = Basis X = (X1) X21 23) EIR² : 2x + 4x2-x3=0} of kerT= {(9).(3) dimensions of Key T = 2 Geometrical description of kertois, kert represents a plane passing through origin Through origin in IR3. Now) Range T = {T(X) : XEIR3} = {AX : X EIR3} if x = 21 X2 AX = 2 4 al - 2 12 -2 ) x3 4 8 2 - 24 + xz V2 + - 2 1 / 2 span ( 2 ) 2 because, span { ( ) } 8 V2 -2 - Ž 6)

Hence, Range(r) = span {G)] dimension of Range (1) basis of Range 117 - ? ? geometrical de sesi ption of Range (T) is :- it sepoesents a line passing Through origin in 1R².