Let the columns of A be vectors named C1 C2 C3 C4 and C5 Giv

Let the columns of A be vectors named C_1, C_2, C_3, C_4 and C_5. Given that [A|b_1|b_2] reduces to the matrix give below, which of the b_i vectors is in col(A)? For the b_i in col(A) express it as a linear combination of the C_i vectors. [1 0 3 -2 0 7 -2 0 1 -1 6 0 -4 0 0 0 0 0 1 2 -3 0 0 0 0 0 1 0] If {(1234)^T (243 -1)^T} is a basis for null (M), and if b = 3C_1-7C_3 where the C_i are the columns of M, then what is the general solution for M x= b?

Solution

3. (a) The 1^st , 2^nd and the 5^th columns of A i.e. c₁, c₂ and c₅ are linearly idependent and c₃ and c₄ are linear combinations of c₁ and c₂. Further, b₁ is not in Col(A) as it has 1 in the last row. Also, since b₂= -2c₁ -3c₅ , hence b₂ is in Col (A).

(b) Let A =

1

2

2

4

3

3

4

-1

Then the RREF of A is

1

0

0

1

0

0

0

0

We know that the vector b is a linear combination of the columns of a matrix M if and only if the equation Mx = b has at least one solution , say x_p . The general solution to Mx = b is given by x = x_p + x_n, where x_p is a particular solution of the equation Mx = b and x_n is a generic vector in the nullspace of M. Here, x_n is a linear combination of (1,2,3,4)^T and (2,4,3,-1)^T which is the same as a linear combination of ( 1,0,0,0)^T and (0,1,0,0)^T. Thus, x_n is (a,b,0,0)^T, wher a,b are arbitrary real numbers.Then, the general solution to Mx = b is x = x_p +(a,b,0,0)^T where x_p is a particular solution of the equation Mx = b.

1	2
2	4
3	3
4	-1