Answer to Question #101514 in Linear Algebra for Joseph Cielo

Question

C02

1. Determine whether or not the set W = {(a + 2, a, 0) : a is a real number } is a subspace of R^3 with the standars operations.

2. Let V= R^2, and define addition and scalar multiplication as follows:
u+v= (u1, u2) + (v1, v2) = (u1+v1+1, u2+v2+2)
ku= k(u1,u2)=(ku1,-ku2)
Let u=(2,1) and v=(4,-1). Compute the value of 2u+v under the given operations.
(Note that V is not a vector space.)

3. Use a determinant to prove that the vectors (6,3,2), (3, 6, 0) and (0, 0, 2) form a basis of R^3.

4. Find the basis of the row space of the matrix.

1 -2 3 9
A= 0 1 3 5
0 -1 4 9

Answer 1

1

$\forall x=(a+2,a,0), y=(b+2,b,0) , z=(c+2,c,0)$

Check the axioms subspace

$1) x+y=y+x\\ x+y=(a+2,a,0)+(b+2,b,0)=\\ =(a+b+4,a+b,0)\\ y+x=(b+2,b,0)+(a+2,a,0)=\\ =(b+a+4,b+a,0)\\$

Performed

$2) (x+y)+z=x+(y+z)\\ (x+y)+z=(a+b+4,a+b,0)+(c+2,c,0)=\\ =(a+b+c+6,a+b+c,0)\\ x+(y+z)=(a+2,a,0)+(b+c+4,b+c,0)=\\ =(a+b+c+6,a+b+c,0)\\$

Performed

3)

$\exist z=(t+2,t,0): x+z=x\\ x+z=(a+2,a,0)+(t+2,t,0)=\\ =(a+t+4,a+t,0)=(a+2,a,0)$

$\left \{\begin{matrix} a+t+4=a+2 \\ a+t=a\\ 0=0 \end{matrix}\right.\\ \left \{\begin{matrix} t=-2 \\ t=0 \end{matrix}\right.\\$

The system has no solution.

W is not a subspace of $R^3$

2

$\forall u=(u_1,u_2), v=(v_1,v_2) \\ u+v=(u_1+v_1+1,u_2+v_2+2)\\ ku=(ku_1,-ku_2)$

Then

$u=(2,1),\\ 2u=(2\cdot2,-2\cdot1)=(4,-2)\\ v=(4,-1)\\ 2u+v=(4+4+1,-2+(-1)+2)=(9,-1)$

3

Find the determinant

$\begin{vmatrix} 6 & 3&2 \\ 3& 6&0\\ 0&0&2 \end{vmatrix}=\\ =6\cdot6\cdot2+3\cdot0\cdot0+3\cdot0\cdot2-\\ -2\cdot6\cdot0-3\cdot3\cdot2-0\cdot0\cdot6=72-18=54\not=0$

The vectors $(6,3,2), (3,6,0),(0,0,2)$ is a basis of $R^3$.

4

Find the rank of the matrix A.

$rk(A)=rk\begin{pmatrix} 1 & -2&3&9 \\ 0 & 1&3&5\\ 0&-1&4&9 \end{pmatrix}=\\ =rk\begin{pmatrix} 1 & -2&3&9 \\ 0 & 1&3&5\\ 0&0&7&14 \end{pmatrix}=3$

The basis of the row space consists of 3 vectors, for example,

$a_1=(1,-2,3,9)\\ a_2=(0,1,3,5)\\ a_3=(0,-1,4,9)$