Answer to Question #101022 in Linear Algebra for Joshua

Question

1. Determine the dimension of and a basis for the solution space of the homogeneous linear system.

2x subscript 1 + x subscript 2 + 2x subscript 3 = 0
x subscript 1 + 6x subscript 3 = 0
x subscript 2 + x subscript 3 = 0

2. Find a basis for the given subspace of R Superscript 3, and state its dimension for the plane 2 x - 3 y + 5 z = 0.

3. Let v1 = (1,-1,3,-6) and v2 = (-3,4,-12,24).
Find standard basis vectors for R4 that can be added to the set {v1, v2} to produce a basis for R4.

A. v3 = (1,0,0,0) and v4 = (0,1,0,0)
B. v3 = (0,0,1,0) and v4 = (1,0,0,0)
C. v3 = (1,0,0,0) and v4 = (0,0,0,1)
D. v3 = (0,1,0,0) and v4 = (0,0,0,1)

Answer 1

$2x_1+x_2+2x_3=0,$

$x_1+6x_3=0,$

$x_2+x_3=0.$

$(R_1 \gets R_1-2R_2)$

$x_2+2x_3-12x_3=0,$

$x_1+6x_3=0,$

$x_2+x_3=0.$

$(R_1 \gets R_1-R_3)$

$-11x_3=0,$

$x_1+6x_3=0,$

$x_2+x_3=0.$

Thus, $x_1=0, x_2=0, x_3=0.$

A basis for the solution space is $(0,0,0),$ the dimension of this space in $R^3$ is 0.

2.$2x=3y-5z$

$x=3y/2-5z/2$

$\begin{pmatrix} ( 3/2) y-(5/2)z \\ y\\ z \end{pmatrix}=\begin{pmatrix} (3/2) \\ 1 \\ 0 \end{pmatrix}y+\begin{pmatrix} - 5/2 \\ 0\\ 1 \end{pmatrix}z$

{$v_1,v_2$ }={(3/2,1,0),(5/2,0,1)} spans R³

We just have to prove that $a_1,a_2=0$ in $a_1v_1+a_2v_2=0$

$a_1(3/2,1,0)+a_2(-5/2,0,1)=0$

$(3/2)a_1-(5/2)a_2=0$

$a_1=0$

$a_2=0$

So, { $v_1,v_2$ } is the basis of the subspace in R³, the dimension of the subspace in $R^3$ is 2.

3.(D)

$v_1=(1,-1,3,-6)$

$v_2=(-3,4,-12,24)$

$v_3=(x_1,y_1,z_1,w_1)$

$v_4=(x_2,y_2,z_2,w_2)$

$a_1v_1+a_2v_2+a_3v_3+a_4v_4=0$

$a_1 (1,-1,3,-6) +a_2(-3,4,-12,24)+a_3(x_1,y_1,z_1,w_1)+a_4 (x_2,y_2,z_2,w_2) =0$

From above we obtain the following equation:

$a_1-3a_2+x_1a_3+x_2a_4=0$

$-a_1+4a_2+y_1a_3+y_2a_4=0$

$3a_1-12a_2+z_1a_3+z_2a_4=0$

$-6a_1+24a_2+w_1a_3+w_2a_4=0$

From the options we put the values of v₃ and v₄ in the above equations.

$a_1=a_2=a_3=a_4=0$ only for D option.