Answer to Question #203389 in Linear Algebra for umme habiba sumi

Question

prove that the vectors (1,0,0) ( 0,1,0) (1,1,0) is linearly dependent .

Answer 1

The set S = {v₁, v₂, v₃} of vectors in R³ is linearly independent if the only solution of

  c₁v₁ + c₂v₂ + c₃v₃ = 0

is c₁, c₂, c₃ = 0

Otherwise (i.e., if a solution with at least some nonzero values exists), S is linearly dependent.

With our vectors v₁, v₂, v₃, becomes:

c₁(1,0,0) + c₂ (0,1,0) + c₃(1,1,0) =(0,0,0)

Rearranging the left hand side yields

1 c₁ +0 c₂ +1 c₃ =0

0 c₁ +1 c₂ +1 c₃ =0

0 c₁ +0 c₂ +0 c₃ = 0

The matrix equation above is equivalent to the following homogeneous system of equations

1 c₁ +0 c₂ +1 c₃ =0

0 c₁ +1 c₂ +1 c₃ =0

0 c₁ +0 c₂ +0 c₃ = 0

Step 2: Transform the coefficient matrix of the system to the reduced row echelon form  

We now transform the coefficient matrix of the homogeneous system above to the reduced row echelon form to determine whether the system has

• the trivial solution only (meaning that S is linearly independent), or
• the trivial solution as well as nontrivial ones (S is linearly dependent).

$\begin{bmatrix} 1 &0 &1 \\ 0 &1 & 1 \\ 0 &0& 0\\ \end{bmatrix}$

Matrix is already in the reduced row echelon form.

 Step 3: Interpret the reduced row echelon form

The reduced row echelon form of the coefficient matrix of the homogeneous system is

$\begin{bmatrix} \boldsymbol{1} &0 &1 \\ 0 &\boldsymbol{1} & 1 \\ 0 &0& 0\\ \end{bmatrix}$

which corresponds to the system

1 c₁   +1 c₃=0 

1 c₂ +1 c₃ =0  

0=0

The leading entries have been highlighted in bold.

Since some columns do not contain leading entries, then the system has nontrivial solutions, so that some of the values c₁, c₂, c₃ solving may be nonzero.

Therefore the set S = {v₁, v₂, v₃} is linearly dependent.