Answer to Question #285331 in Linear Algebra for Sabelo Xulu

Question

8. 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) $\isin$ R^5: x base 1 = 3x base 2 and x base 3 =7x base 4}. Then the basis of W is

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

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

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

9. The basis of a solution space of given homogeneous 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 =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

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

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

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

(iv) None

10. For a given matrix A $\begin{bmatrix} 1 & 0 & 2 & - 3 \\ 2 & 0 & 4 & - 6 \\ - 3 & 0 & - 6 & 9 \end{bmatrix}$. Which of the following is true

(i) rank (A) =3, nullity (A) =1

(ii) rank (A) =2, nullity (A) =2

(iii) rank (A) =1, nullity (A) =3

(iv) None

Answer 1

8.

$x_1=3x_2$ and $x_3=7x_4$

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

Basis of W is $(3,1,0,0,0),(0,0,7,1,0),(0,0,0,0,1)$

9.

$x_1+x_2-x_3=0.....(i)\\ x_1-x_3=0......(ii)\\ -2x_1-x_2+2x_3=0....(iii)\\ -2x_1+2x_3=0....(iv)$

$-2(ii)=(iv),$ therefore ignoring $(iv)$

$\begin{pmatrix} 1&&1&&-1 \\ 1&&0&&-1\\ -2&&-1&&2 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$

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

$x_1=x_3\\ x_2=0$

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

Basis is $[1,0,1]$

10.

$\begin{pmatrix} 1&&0&&2&& -3 \\ 2&&0&&4&&-6 \\ -3&&0&&-6&&9 \end{pmatrix}$

$\frac{1}{2}R_2\to\>R_2$

$\frac{-1}{3}R_3\to\>R_3$

$\begin{pmatrix} 1&&0&& 2&&-3 \\ 1&&0&&2& & -3\\ 1&&0&&2&&-3 \end{pmatrix}$

$R_2-R_1\to\>R_2\\ R_3-R_1\to\>R_3$

$\begin{pmatrix} 1&&0&&2&&-3 \\ 0&&0& & 0&&0\\ 0&&0&&0&&0 \end{pmatrix}$

Rank $=1$

Nullity $=3$