Determine the null space of each of the following1 1 -1 22 2 -3 1-1 -1 0 -5

To find the null space, we need to solve the equation Ax = 0, where A is the matrix and x is a vector in the null space. We can set up the augmented matrix and row reduce as follows:

[ 1 1 -1 2 | 0 ] [ 2 2 -3 1 | 0 ] [-1 -1 0 -5 | 0 ]

R2 - 2R1 -> R2: [ 1 1 -1 2 | 0 ] [ 0 0 -1 -3 | 0 ] [-1 -1 0 -5 | 0 ]

R1 + R3 -> R3: [ 1 1 -1 2 | 0 ] [ 0 0 -1 -3 | 0 ] [ 0 0 -1 -3 | 0 ]

R3 - R2 -> R2: [ 1 1 -1 2 | 0 ] [ 0 0 -1 -3 | 0 ] [ 0 0 0 0 | 0 ]

Now we can solve for the variables in terms of the free variable, say x4 = t:

x3 = -t x2 = 3t x1 = -t - 2x4

So the null space is the set of all vectors of the form:

[ -t - 2s ] [ 3t ] [ -t ] [ t ]

where t and s are any scalars. In other words, the null space is the span of the following vector:

[ -1 ] [ 3 ] [ -1 ] [ 1