00:01
We are asked to find the null space of a.
00:05
The definition of the null space is shown up top.
00:08
We have that the null space of an n -by -n matrix a is equal to the solution set with a vector x with elements in rn, which satisfies the homogenous linear system, a -x, equals zero.
00:27
And the problem gives a as a 1 by 2 matrix 1 -4.
00:31
So we can see from that our n is equal to 2.
00:37
So we're in r2 space, which means our x vector will be a 2x1 matrix, which will look like this, x1, x2.
00:56
So now we'll write out our a x equals 0.
01:01
So we have our matrix a comes our vector x, which is equal to 1, 4 times x1, and x2, or x vector.
01:20
Multiplying that out, we have 1 times x1 plus 4 times x2 is equal to 0.
01:34
Now solving for x1, we can get x1 is equal to negative 4 times x2...