# Consider the matrix A = [[2, 1, 3],[4, 1, 4]]. Find the null space of A. Describe it geometrically.

*print*Print*list*Cite

You need to find the null space of matrix `A` , hence, you need to perform the following matrices multiplication, such that:

`A*X = O`

`((2,1,3),(4,1,4))*((x_1),(x_2),(x_3)) = ((0),(0))`

Performing matrices multiplication yields:

`((2x_1 + x_2 + 3x_3),(4x_1 + x_2 + 4x_3)) = ((0),(0))`

Equating corresponding members yields:

`{(2x_1 + x_2 + 3x_3 = 0),(4x_1 + x_2 + 4x_3 = 0):}`

You may consider the variable `x_3` as a free variable, hence, you may move to the right side, such that:

`{(2x_1 + x_2 = -3x_3),(4x_1 + x_2 = - 4x_3):}`

`2x_1 + x_2 - 4x_1 - x_2 = -3x_3 + 4x_3`

`-2x_1 = x_3 => x_1 = -x_3/2`

Replacing -`x_3/2` for `x_1` in any of two equations above, yields:

`2*(-x_3/2) + x_2 = -3x_3 => x_2 = -2x_3`

**Hence, evaluating the null space of matrix A yields **`N(A) = X = ((-x/3/2),(-2x_3),(x_3)).`

For a matrix `A`, the null space or kernel is the set of all vectors `overline(x)` such that

`Aoverline(x) = overline(0)`

In this case we want `overline(x) = (x_1,x_2,x_3)` such that

`([2,1,3],[4,1,4])([x_1],[x_2],[x_3]) = ([0],[0],[0])`

`implies`

1) `2x_1 + x_2 + 3x_2 = 0`

2) `4x_1 + x_2 + 4x_3 = 0`

Subtracting 1) from 2) we get

`2x_1 + x_3 = 0`

`implies` `x_1 = -1/2x_3`

Substituting into 1) `implies`

`-x_3 + x_2 + 3x_3 = x_2 + 2x_3 = 0` `implies` `x_2 = -2x_3`

**The null space/kernel is then given by**

**`([x_1],[x_2],[x_3] ) = c_0([-1/2],[-2],[1]) = c([-1],[-4],[2])` **

**where `c_0`, `c` are scalars variables.**

**Geometrically, the null space is described by a line in 3D space.**