Given the following linear system Ax b A 2 4 6 4 2 5 7 6 2

Given the following linear system Ax = b: A= [2 4 6 4 2 5 7 6 2 3 5 2] b = [b_1 b_2 b_3] = [4 3 5] Find the condition on (b_1, b_2, b_3) for Ax = b to have a solution. Describe C(A). Find an equation for the plane C(A). Describe N(A). Find the complete solution (i.e. general solution) to Ax = b.

Solution

a) b must be in the column space of A

b) find the rank of A.

If A has a rank 3 then for any b in R^3 Ax=b has a solution

calculate the basis for column space, C_i is the ith column vector

column 1 , column 2 are Linearly independent

C_3 = C_2 + C_1

C_4 = 2(C_2 - C_1)

C_3 and C_4 are not independent

So column space has dimension 2

and basis ={C_1,C_2}

Now b = 4C_1 - C_2

is in the column space of A

So Ax=b has solutions

x is in Null space of A

if A(x) =0

if x = [a_1,a_2,a_3,a_4]

then a_1C_1 + a_2C_2 + a_3C_3 + a_4C_4 = 0

a_1C_1 + a_2C_2 + a_3(C_1 + C_2) + a_4*2(C_2 - C_1) = 0

=> a_1 +a_3 - 2a_4 = 0

a_2 +a_3 +2a_4 = 0

a_3 = -(a_1 + a_2)/2

a_4 =(a_1 - a_2)/4

So

[1,1, -1,0] and [1,-1,0,1/2] form basis for N(A)

complete solution for Ax =b

x = c_1 [1,1, -1,0] + c_2 [1,-1,0,1/2] + particuar solution

particular solution = [4,-1,0,0]

since b= 4C_1 -C_2