Answer to Question #288345 in Linear Algebra for Sabelo Xulu

Question

Q7.

The vectors (1, 2, 0), (0, -4, 2), (1, 0, 1) are

(1) Span R^3

(2) linearly dependent

(3) linearly independent

(4) None of the given answers is true.

Q8.

Let W be the subspace of R^5 defined by

W = {(x base 1; x base 2; x base 3; x base 4; x base 5) 2 R^5 : x base 1 = 3x base 2 and x base 3 = 7x base 4}

Then the basis of W is

(1) (3,1,0,0,1), (3,1,3,0,0), (3,1,0,0,1)

(2) (3,1,0,1,1), (0,0,3,0,1), (0,0,1,3,1)

(3) (3,1,1,0,1), (0,1,1,0,3), (0,0,1,0,1)

(4) None of the given answers is true.

Q9.

The basis for a solution space of given homogenous linear system

x base 1 + x base 2 - x base 3 = 0

x base 1 + x base 2 - x base 3 = 0

x base 1 - x base 3=0

- x base 1 + x base 3=0 $\implies$x base 2 = 0 $\implies$x base 2 = 0

- 2x base 1 - x base 2 + 2x base 3 = 0

- 2x base 1 - x base 2 + 2x base 3= 0

- 2x base 1 + 2x base 3 = 0 is

(1) {(1, 0, 1)}

(2) {(1, 0, 1), (0, 1, 0)}

(3) f(1; 1; 1);(1; 0; 1), (2, 1, 2)}

(4) None of the given answers is true.

Answer 1

Q7

Let $A=\begin{pmatrix} 1&0 &1 \\ 2&-4&0\\0&2&1 \end{pmatrix}$

Reduced row echelon form:

rref $A=\begin{pmatrix} 1&0&1 \\ 0&1&½\\0&0&0 \end{pmatrix}$

The vectors are linearly dependent and they don't span $\Reals ^3$

Q8

w={($x_1;x_2;x_3;x_4;x_5)\in\Reals^5:x_1=3x_2\>,x_3=7x_4$

}

Let $x_5=1$ and $x_4=1$

$\begin{pmatrix} x_1\\ x_2\\x_3\\x_4\\x_5 \end{pmatrix}=\begin{pmatrix} 3x_2 \\ x_2\\7x_4\\x_4\\x_5 \end{pmatrix}=x_2\begin{pmatrix} 3\\ 1\\0\\0\\0 \end{pmatrix}+x_4\begin{pmatrix} 0 \\ 0\\7\\1\\0 \end{pmatrix}+x_5\begin{pmatrix} 0 \\ 0\\0\\0\\1 \end{pmatrix}$

None of the given answer is true

Q9

$x_1+x_2-x_3=0\\x_2=0$ $\implies x_1=x_3$

$\begin{pmatrix} x_1 \\x_2\\x_3 \end{pmatrix}=\begin{pmatrix} x_3 \\0\\x_3 \end{pmatrix}=x_3\begin{pmatrix} 1 \\0\\1 \end{pmatrix}$

The base is {(1,0,1)}