3 Find the general solution of each of the following equatio

3. Find the general solution of each of the following equations (presented, as in the text, as a combination of an appropriate number of vectors a. x1 2x2 3x3 4 (in Rs) d. 2x2 3x3 4 (in R4) 0 in R4) 2 in R4) b. x1 x2 x3 2x4 3x4 e. x2 x3 c. x1 x2 x3 2x4 5 (in R4)

Solution

a. 3 variables and 1 equation hence, 3-1=2 free variables.

Free variables are variables which can take arbitrary values

WE choose:x2 and x3 to be the free variables

So, x1=2x2-3x3

General solution is

(x1,x2,x3)=(2x2-3x3,x2,x3)=x2(2,1,0)+x3(-3,0,1)

Hence general solution is a linear combination of vectors

(2,1,0) and (-3,0,1)

d.

Here we have the same equation as in a. but now in R4 so we have 4 variables and 1 equation so number of free variables is equal to 4-1=3. So one more free variables which is the fourth coordinate x4

We choose:x2,x3,x4 to be the free variables

So, x1=2x2-3x3

So general solution is

(x1,x2,x3,x4)=(2x2-3x3,x2,x3,x4)=x2(2,1,0,0)+x3(-3,0,1,0)+x4(0,0,0,1)

e.

Space is R4 so 4 variables and one equation. So 3 free variables

We choose, x1,x3,x4 to be the free variables

So, x2=-x3+3x4

General solution is

(x1,x2,x3,x4)=(x1,-x3+3x4,x3,x4)+x1(1,0,0,0)+x3(0,-1,1,0)+x4(0,3,0,1)