Question

Show if b is in the null space of A and why

Answer 1

\[A= \left[\begin{matrix}1 & 2 \\ 3 & 4 \\ 5 & 6\end{matrix}\right] b = \left(\begin{matrix}2 \\ -1\end{matrix}\right)\]

Answer 2

In my notes he used gaussian elimination to find the null space of A but I don't think I'm doing it right

Answer 3

i got \[ \left[\begin{matrix}1 & 2 \\ 0 & 1 \\ 0 & 0\end{matrix}\right]\]

Answer 4

would I let it equal to b???

Answer 5

do you know what the null space of a matrix is?

Answer 6

No I don't really understand it, honestly

Answer 7

Ok, the null space of a matrix is all values of x, where Ax = 0

Answer 8

So, in general, you need to see if A is invertible. If it is invertible, then, it means the null space of A would just be \[\left(\begin{matrix}0 \\ 0\end{matrix}\right)\]

Answer 9

If it is not invertible, then you need to pull out the general solution, and see if b fits into the general solution.

Answer 10

However, in the case of your question A is not invertible. Hence, b is not in the null space of A.

Answer 11

How do you tell if it is invertible?

Answer 12

Firstly, if the matrix is not a square matrix, it is not invertible. So, in your case, the matrix is a 3x2 matrix. Therefore, it is automatically not invertible.

Answer 13

For those matrices that are square, you need to use Gaussian Elimination to get it down to this form:
\[\left[\begin{matrix}1 & a \\ 0 & 1\end{matrix}\right]\]

All the diagonal entries should be 1, everything below the 1's should be zero. That is called the row echelon form. You get this form through Gaussian Elimination.
There should be no rows of 0. As long as these conditions are met, the square matrix will be invertible.

Answer 14

Oh okay!!! Thanks. So where it's not a square matrix it's not invertible. But you said if it's not then you can pull out a general solution to see if b fits in?