# Let A = [[-4,-7,6],[5,10,-10],[-1,-1,0]] , vector v =[[4],[-1],[5]] , vector w = [[-2],[2],[1]] , vector x = [[4],[-4],[3]] Type yes or no: Is vector x in NulA?

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You need to test if X is the null space of matrix A, hence, you need to solve the following equation, such that:

`A*X = 0`

`((-4,-7,6),(5,10,-10),(-1,-1,0))*((x_1),(x_2),(x_3)) = ((0),(0),(0))`

Performing the multiplication of matrices yields:

`((-4x_1 - 7x_2 + 6x_3),(5x_1 + 10x_2 - 10x_3),(-x_1 - x_2)) =` `((0),(0),(0))`

You need to solve for `x_1,x_2,x_3` the following system of equations, such that:

`{(-4x_1 - 7x_2 + 6x_3 = 0),(5x_1 + 10x_2 - 10x_3 = 0),(-x_1 - x_2 = 0):}`

Replacing -`x_2` for `x_1` yields:

`{(4x_2 - 7x_2 + 6x_3 = 0),(-5x_2 + 10x_2 - 10x_3 = 0):}`

`{(-3x_2 + 6x_3 = 0),(5x_2 - 10x_3 = 0):} => {(-3x_2 =- 6x_3),(5x_2 - 10x_3 = 0):} => {(x_2 = 2x_3),(5x_2 - 10x_3 = 0):} `

Replacing 2`x_3` for `x_2` yields:

`10x_3 - 10x_3 = 0 => x_3 = 0, x_2 = 0, x_1 = 0`

**You may notice that `X = ((0),(0),(0))` does not coincide with the given `X = ((4),(-4),(3))` , hence, X is not the kernel of matrix A.**