solution of system

[bhuvaneshnick]

B column must be linearly dependent with A so only we can make the B column zero to further move to find solution of the system.am i thinking right?

 

[stapel]

A set of linear equations is represented by the matrix equation Ax = b. The necessary condition for the existence for this system is:

a) A must be invertible.
b) b must be linearly dependent on the columns of A.
c) b must be linearly independent of the columns of A.
d) None of these.

B column must be linearly dependent with A so only we can make the B column zero to further move to find solution of the system.am i thinking right?

Since when does the b-column "have" to be all zeroes in order for there to be a solution to the system of equations? (Think back to the systems you've solved by hand.)

If you're working with the matrix equation, what would you multiply by (on the left) in order to get x by itself, with the solution vector on the other side of the equation? What must be true in order for that multiplication to be able to take place?

 

[bhuvaneshnick]

Since when does the b-column "have" to be all zeroes in order for there to be a solution to the system of equations? (Think back to the systems you've solved by hand.)

If you're working with the matrix equation, what would you multiply by (on the left) in order to get x by itself, with the solution vector on the other side of the equation? What must be true in order for that multiplication to be able to take place?

For multiplication to take place B should be dependent
okay we get to that question later, before that i have few question,let A be matrix of

1 0
3 0
1)now say are the column vectors dependant or not? what i was thinking is multiplying anything with column 1 dont make the column 2, isn't it .so the vectors are independent right? .though we multiply zero,but rule of elementary matrix operation is only about non zero multiplication

 

[HallsofIvy]

For multiplication to take place B should be dependent
okay we get to that question later, before that i have few question,let A be matrix of

1 0
3 0
1)now say are the column vectors dependant or not? what i was thinking is multiplying anything with column 1 dont make the column 2, isn't it .so the vectors are independent right? .though we multiply zero,but rule of elementary matrix operation is only about non zero multiplication

The "rule of elementary matrix operations" is to get to "echelon form" or "reduced echelon form", it does not have any thing to do with whether or not the columns, thought of as vectors, are dependent or independent.

The definition of "dependent" is this: "a set of vectors, \(\displaystyle \{v_1, v_2, \cdot\cdot\cdot, v_n\}\) is dependent if and only if there exist a set of numbers, \(\displaystyle \{a_1, a_2, \cdot\cdot\cdot, a_n\}\), not all 0, such that \(\displaystyle a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n= 0\).

Here, \(\displaystyle v_1= <1, 3>\) and \(\displaystyle v_2= <0, 0>\) with \(\displaystyle a_1= 0\), \(\displaystyle a_2= 1\), not both 0, \(\displaystyle a_1v_1+ a_2v_2= 0<1, 3>+ 1<0 , 0>= <0, 0>\). Any set of vetors, containing the 0 vector, is dependent.

Again, you are, as so many students, unfortunately, do, focusing on "formulas" and "methods" rather than the basic concepts and definition of the subject.

 

[Ishuda]

Let x = [x₁ x₂ x₃]^T where the superscript T stands for transpose and let column c_i of A be given [a_i1 a_i2 a_i3]^T then it is easy to show that
A x = x₁ c₁ + x₂ c₂ + x₃ c₂
Thus, if the system of equations
A x = b
is to have a solution then there must exist an x such that
b = x₁ c₁ + x₂ c₂ + x₃ c₂
and, unless the trivial solution is the only solution and b is the zero column vector, b is linearly dependent on the column vectors of A.

So, to answer your question, the b column must be linearly dependent with A unless b is zero and the only solution is the trivial solution.
EDIT: Most texts count the zero vector as being linearly dependent with any other set of vectors so the answer would still be 'the b column must be linearly dependent with A' even if b were the zero vector and the only solution were the trivial solution.