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Which of the following descriptions are correct? The solutions x of

Ax = $\begin{bmatrix} 1 & 1 & 1\\ 1 & 0 & 2 \end{bmatrix}$ $\begin{bmatrix} x1\\ x2\\ x3 \end{bmatrix}$ = $\begin{bmatrix} 0\\ 0\\ \end{bmatrix}$

form
(a) a plane.
(b) a line.
(c) a point.
(d) a subspace.
(e) the nullspace of A.
(f) the column space of A

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Systematic Procedure of Finding Null Space $N(A)$ of any matrix $A$ -  

Given any equation $AX = 0$ where Amxn  is a matrix having m rows and n columns. 
Null Space of A = N(A) is - 

All solutions $X = \begin{bmatrix} x1\\ x2\\ x3 \end{bmatrix}$ in Rn that solves $AX = 0$ .
Note- Here $N(A)$ is a subspace of Rn because X has n components where n is number of columns in A. 

Given Equation -

$Ax = \begin{bmatrix} 1 & 1 & 1\\ 1 & 0 & 2 \end{bmatrix} \begin{bmatrix} x1\\ x2\\ x3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $

$A = \begin{bmatrix} 1 & 1 & 1\\ 1 & 0 & 2 \end{bmatrix}$

Step 1 - 
Convert the given matrix A into Echelon form by applying row operations - 

$A = \begin{bmatrix} 1 & 1 & 1\\ 1 & 0 & 2 \end{bmatrix}$ $ \overset{{R2 \rightarrow R2-R1 }}{\large \rightarrow}$ $U = \begin{bmatrix} 1 & 1 & 1\\ 0 & -1 & 1 \end{bmatrix}$

Now we have to solve $UX = 0$ where U is upper triangular matrix. 
x1 + x2 + x3 = 0    --(1)
-x2 + x3 = 0          --(2)

Number of pivot columns = 2 
Rank of the matrix (r) = Number of pivots = 2

No. of free Columns = n-r = 3-2 = 1

So Dimension of N(A) i.e null space of A = n-r = 3-2 =   1
i.e it's a Straight Line in R3

Step 2 -
Find Special Solutions.

Number of Special Solutions = Number of free variables
Here it is 1 i.e x3.

 Put x3 = 1 and do Back Substitution.
-x2+1 = 0 $\Rightarrow$ x2 = 1
x1+ 1 +1 = 0 $\Rightarrow$ x1 = -2

Special Solution = $\begin{bmatrix} x1\\ x2\\ x3 \end{bmatrix} = \begin{bmatrix} -2\\ 1\\ 1 \end{bmatrix}$ 

Null Space - All linear combinations of special solutions.

So Null Space of $A \rightarrow N(A)$ is $C. \begin{bmatrix} -2\\ 1\\ 1 \end{bmatrix}$ where $C$ is any non zero real number.

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$A_{x}=\begin{bmatrix} 1& 1 &1 \\ 1& 0 &2 \end{bmatrix}$$\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3} \end{bmatrix}$$=\begin{bmatrix} 0\\ 0 \end{bmatrix}$

If we form simple equation, it will be

$x_{1}+x_{2}+x_{3}=0$

$x_{1}+2x_{3}=0$

Solving this we get

$x_{1}=-2x_{3}$

$x_{2}=x_{3}$

Now if we put $x_{3}=K$

$x_{2}=K$ too

and $x_{1}=-2K$

So, eigen vector will be $\begin{bmatrix} -2\\ 1\\ 1 \end{bmatrix}$

that is also the null space of A

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