# Define null space . Find the basis for the null space of the matrix $$A = \begin{bmatrix}1 & 2 & 3\\ 2 & 3 & 4\end{bmatrix}$$

Let A be a m x n matrix. Then, the null space of A, denoted by nulA, is the set of all vectors x ∈ Rn, such that Ax = 0 i.e.

Nul A = {x ∈ Rn : Ax = 0}

Note: Nul A ≠ Φ

Given, A = $$\begin{bmatrix}1 & 2 & 3\\ 2 & 3 & 4\end{bmatrix}$$

Let the homogeneous equation be

Ax = 0, x = $$\begin{bmatrix}x1\\ x2\\ x3\end{bmatrix}$$

Now, the augmented matrix is

$$\begin{bmatrix}1 & 2 & 3 & : & 0\\ 2 & 3 & 4 & : & 0\end{bmatrix}$$

Apply R2 → R2 – 2R1

$$~ \begin{bmatrix}1 & 2 & 3 & : & 0\\ 0 & -1 & -5 & : & 0\end{bmatrix}$$

Here, x1 and x2 are basic variable and x3 is free variable

Now, Apply R2 → (-1) R2

$$~ \begin{bmatrix}1 & 2 & 3 & : & 0\\ 0 & 1 & 5 & : & 0\end{bmatrix}$$

Apply R1 → R1 – 2R2

$$~ \begin{bmatrix}1 & 0 & -7 & : & 0\\ 0 & 1 & 5 & : & 0\end{bmatrix}$$

From 1st row,

x1 – 7×3 = 0

∴ x1 = 7×3

From 2nd row,

x2 + 5×3 = 0

∴ x2 = -5×3

The solution is

x = $$\begin{bmatrix}x1\\ x2\\ x3\end{bmatrix}$$ = $$\begin{bmatrix}7×3\\ -5×3\\ x3\end{bmatrix}$$ = x3 $$\begin{bmatrix}7\\ -5\\ 1\end{bmatrix}$$

∴ Basis for the null space = $$\left \{ \begin{bmatrix}7\\ -5\\ 1\end{bmatrix} \right \}$$