let u1 336 u2 444 and u3 633SolutionGiven that u1 336 u2

let u1 (3,3,6) u2 = (4,4,4) and u3 (-6,3,3)

Given that u1 = (3,3,6) , u2 = (4,4,4) and u3 =(-6,3,3)

v = (41,5,12)

Let v = c1 u1 + c2 u2 + c3u3 where c1,c2,c3 are constants.

Hence,

(41,5,12) = c1 (3,3,6)+ c2 (4,4,4) + c3 (-6,3,3)

(41,5,12) = (3c1 +4c2 -6c3, 3c1+4c2+3c3, 6c1+4c2+3c3)

Then,

3c1 +4c2 -6c3 = 41 --------- Eq (1)

3c1+4c2+3c3 = 5 ------------ Eq(2)

6c1+4c2+3c3 = 12 ----------- Eq (3)

Calucation of c3:

Eq (1) - Eq(2),gives c3 .

3c1 +4c2 -6c3 = 41 --------- Eq (1)

3c1+4c2+3c3 = 5 ------------ Eq(2)

---------------------------------------------

-9c3 = 36

c3 = -4

Calculation of c1:

Eq (2)- Eq(3) gives c1.

3c1+4c2+3c3 = 5 ------------ Eq(2)

6c1+4c2+3c3 = 12 ----------- Eq (3)

----------------------------------------

-3c1 = -7

c1 = 7/3

Calculaton of c2:

Substitute values of c1 and c3 in Eq (1) to get c2.

3c1 +4c2 -6c3 = 41 --------- Eq (1)

3.(7/3) + 4c2 - 6(-4) = 41

7 + 4c2 +24 = 41

c2 = 10/4 = 5/2

Therefore,

c1 = 7/3, c2 = 5/2 , c3 =-4

Hence,

v = (7/3) u1 + (5/2) u2 -4 u3