Answer to Question #309019 in Linear Algebra for stuti

Question

Determine the subspace of ℝ^3 spanned by the vectors α = 1, 2, 3 , β = 3, 1, 0 . Examine whether

γ = 2, 1, 3 , δ = (−1, 3, 6) are in the subspace or not.

Answer 1

$\alpha=(1, 2, 3), \beta=(3,1,0),$

$V=\begin{Bmatrix} (1, 2, 3); (3,1,0) \end{Bmatrix}$ - the set of vectors

Let's write the augmented matrix of the system of linear equations where the coefficient matrix is composed by the vectors of V as columns, and a generic vector of the space specified by means of variables as the additional column used to compose the augmented matrix:

$\begin{bmatrix} 1 & 3 & | & x \\ 2 & 1 & | & y \\ 3 & 0 & | & z \end{bmatrix}$

Now we'll transform the augmented matrix to row echelon form taking into account the last column composed of expressions depending on variables.

$R_2\gets2R_1-R_2$

$R_3\gets3R_1-R_3$

$\begin{bmatrix} 1 & 3 & | & x \\ 0 & 5 & | & 2x-y \\ 0 & 9 & | & 3x-z \end{bmatrix}$

$R_3\gets5R_3-9R_2$

$\begin{bmatrix} 1 & 3 & | & x \\ 0 & 5 & | & 2x-y \\ 0 & 0 & | & -3x+9y-5z \end{bmatrix}$

For the system to have solution is necessary that the entries in the last column, corresponding to null rows in the coefficient matrix be zero (equal ranks). The equations defined by those expressions, are the implicit equations of the vector subspace spanning for the set of vectors.

$\begin{bmatrix} V \end{bmatrix}=\begin{Bmatrix} (x, y,z)\in R^3 |-3x+9y-5z=0 \end{Bmatrix}-$ the desired subspace of $R^3.$

Let's examine whether vectors γ =( 2, 1, 3) , δ = (−1, 3, 6) are in the subspace or not.

$-3\cdot2+9\cdot1-5\cdot3=-12\not =0, \gamma$ is not in the subspace;

$-3\cdot(-1)+9\cdot3-5\cdot6=0, \delta$ is in the subspace.