Answer to Question #161418 in Differential Equations for Ojugbele Daniel

Question

Solve by finding the basis over R for the solution space.

(A) X + 3y -3z=0

2x - 3y + z=0

3x -2y + 2z=0

(B) X + Y + Z + W=0

2X + 3Y - Z +W=0

3X + 4Y +2W=0

Answer 1

(A)

(1) X + 3y -3z=0

(2) 2x - 3y + z=0

(3) 3x -2y + 2z=0

Adding (1) and (2) we have

(4) 3x-2z=0

Adding (3) and (4) we have

(5) 6x-2y=0

Let x=2t. Then from (4) z=3t and from (5) y=6t.

Applying this to the equation (1) we have:

2t +3(6t)-3(3t)=0

11t=0

Therefore, the system (1)-(3) has only solution x=y=z=0.

(B)

(1) X + Y + Z + W=0

(2) 2X + 3Y - Z +W=0

(3) 3X + 4Y +2W=0

Adding (1) and (2) we have (3), hence the equation (3) is linearly dependent with (1)-(2). From (3) we have

(4) W = - 3/2 X - 2Y

From (2) we have

(5) Z = 2X + 3Y + W = 2X + 3Y - 3/2 X - 2Y = 1/2X + Y

The equations (4) and (5) represent a common solution of the system.

The basis can be obtained with (X,Y)=(1,0) and (X,Y)=(0,1):

(X, Y, Z, W) = X(1, 0, 1/2, -3/2) + Y(0, 1, 1, -2)

(1, 0, 1/2, -3/2) with (0, 1, 1, -2) is a basis for the solution space.