You need to find the kernel or null space of the matrix `A` , hence, you need to solve the equation `A*X = 0` , such that:

`((1,1,3,1),(2,1,5,4),(1,2,4,-1))*((x_1),(x_2),(x_3),(x_4))` = `((0),(0),(0))`

`((x_1 + x_2 + 3x_3 + x_4),(2x_1 + x_2 + 5x_3 + x_4),(x_1 + 2x_2 + 4x_3 - x_4)) = ((0),(0),(0))`

`{(x_1 + x_2 + 3x_3 + x_4 = 0),(2x_1 + x_2 + 5x_3 + x_4 = 0),(x_1 + 2x_2 + 4x_3 - x_4 = 0):}`

You need to evaluate the minor `delta` , such that:

`delta = [(1,1,3),(2,1,5),(1,2,4)]` => `delta = 4 + 13 + 5 - 3 - 10 - 8 = 1`

Since `delta != 0` you may consider `x_4` as free variable such that:

`{(x_1 + x_2 + 3x_3 = -x_4),(2x_1 + x_2 + 5x_3 = -x_4),(x_1 + 2x_2 + 4x_3 = x_4):}`

Considering `x_4 = alpha` , yields:

`delta_(x_1) = [(-alpha,1,3),(-alpha,1,5),(alpha,2,4)]` => `delta_(x_1) = -4alpha - 6alpha + 5alpha - 3alpha + 10alpha + 4alpha => delta_(x_1) = 6alpha`

`x_1 = (delta_(x_1))/(delta) => x_1 = (6alpha)/1 => x_1 = 6alpha`

`delta_(x_2) = [(1,-alpha,3),(2,-alpha,5),(1,alpha,4)] ` => `delta_(x_2) = -4alpha + 6alpha - 5alpha + 3alpha - 5alpha + 8alpha => delta_(x_2) = 3alpha`

`x_2 = 3alpha`

`x_1 + x_2 + 3x_3 = -alpha => 9alpha + 3x_3 = -alpha => 3x_3 = -10alpha => x_3 = (-10alpha)/3`

**Hence, evaluating the kernel of matrix A yields the set of vectors **`X = ((6alpha),(3alpha),((-10alpha)/3),(alpha)).`