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.