# The nullspace of a 3 by 4 matrix A is the line through (2,3,1,0)? a. what is the rank of A and the complete solution to Ax=0? b. what is the exact row-reduced echelon form R of A?

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The nullspace is `{ [[2],[3],[1],[0]] }`

That is, the complete solution to `Ax=0` is all the vectors that are multiples of `[[2],[3],[1],[0]]`

The rank + nullity is n (the number of columns)

So A must have rank 3, which means we must have 3 leading 1s in the reduced row eschelon form

`[ [. , . , . , . ], [. , . , . , . ], [. , . , . , . ] ] [[2],[3],[1],[0]] = [[0],[0],[0]]`

Starting with the bottom row:

We're looking for a linear combination that adds to 0, and using as few of 0,1,3,2 as possible.

We just use the 0, and get:

`1*0=0`

So the bottom row should be `[ [0 , 0 , 0 , 1 ] ]`

We have a "leading 1" in the fourth component, so we want an equation that doesn't use the 0. We can't make an equation with just the 1. We can't make a nonzero linear combination from just the number 1, and still get 0. But if we use both 1 and 3, we can make a linear combination:

1*3+(-3)*1=0`

The next row from the bottom should be `[ [0 , 1 , -3 , 0 ] ]`

For the top row, we must have 0s in the 2nd and 4th components because that corresponds to columns with leading 1s. That means we can't use the 3 or the 0 in our equation, and we must use the 2 and the 1. The equation for these is:

`1*2+(02)*1=0`

So the top row should be `[ [1 , 0 , -2 , 0 ] ]`

So our reduced row eschelon matrix is:

`[[1 , 0 , -2 , 0 ], [0 , 1 , -3 , 0 ], [0 , 0 , 0 , 1 ]]`