Question

please answer the following question with detailed step

1 1. Consider vi = 2 V2 = a and v3 = -1 (a) Find the value(s) of a such that 01,02 and v3 are linearly dependent and write Vi as a linear combination of v2 and 03, if possible. (b) Suppose a = 0, write v = 2 as a linear combination of v1, V2 and 03. (c) Suppose a = 0, use the Gram-Schmidt process to transform {V1, V2, V3} into an orthonormal set of vectors {W1, W2, W3} and write v = 2 as a linear combination of w1,w2 and w3.

Answer 1

(1) Given v= 2 2 = a, vz = -1 1 3 (a) Consider the vectors as 44+ a 12 + av = 0 (1) 2 (1 0 = 2 +a, a +az -1 = 0 1-2) 1 3 10 a + 2a + a ) 2a +aa, - az = 0 (-2ą + a2 +3az( So, a + 2a, + a2 = 0,24 +ad; - az = 0, -2a + az +3a2 = 0 The matrix form of the equ's as AX = 0 (1 2 3) (1 2 3) SO, A = 2 a -12-285* 1 O 2 a-4 3 -7 - 1-2 i 3 R, +2R, O 5 9), 9 ) R )15 elo o 9-A) 9a-1 a-4 1 2 = AX=0= 0 a-4 3 -7 0 az a-4 On comparing -2=0=9a-1=0=a=1 a-4 Plug a value in AX = 0 we get 0 lo e olla olul 0 0 So, here a =ka;, = , here ag = 2 = ; +0 So, for a= the vectors are linearly dependent.

(b)let us consider v = 441 + a2 + a, v (3 (1) (1) (3) (4 + 2a + az = 2 = a 2 + a, 0 +a-1 = 2 = 20, - az 1) (-2) (1) (3) (1) (-20, +az + 3a On comparing we get a + 2a + ax = 3, 2a, - a = 2,-2a, +az + 3a, = 1 (1 2 1 13 The matrix form is Ax=v= 2 0 -1 az = 2 (-2 1 3 ) a (1) (1 2 13 (10 09/5 Consider (4 v) = 2 0 -1 | 2 the rref of (4|v) is 0 1 0 1-1/5 -2 1 3 1) OO 18/5) Therefore, v=B**

(c)Find the orthogonal basis using gram schmidth process: W = V1 = 2 ,= 12 - 2 1 w = 0 - W-W 1 (-2) (4/9 (1) (2) (-2/9 So, the orthogonal basis is {wy, W2,w;}={ 2,0 519 (-2) (1) 4/9) * chuo mw-(2,067 #0340- 2 W2 = 1 2 ,W2 =- 0 W = w,-W, V5 So, the orth onormal basis of W is {w, W2, W;}.