A = 2x4 matrix:
( row 1= -6, 2, -4, 4)
(row 2 = 6, -2, 4, -4)
Find a spanning set for the null space of A (4x3)
To find a spanning set for the null space of a matrix A, we need to find all the solutions to the equation Ax = 0, where x is a vector.
In this case, A is a 2x4 matrix:
A = ((-6, 2, -4, 4),
(6, -2, 4, -4))
We want to find all vectors x such that Ax = 0, which means finding vectors x that satisfy the following system of linear equations:
-6x1 + 2x2 - 4x3 + 4x4 = 0
6x1 - 2x2 + 4x3 - 4x4 = 0
To find the spanning set of the null space of A, we need to find the special solutions of this system. We can do that by setting up an augmented matrix with the original matrix A and the column vector 0:
((-6, 2, -4, 4, 0),
(6, -2, 4, -4, 0))
Now, let's perform row reduction on this augmented matrix to find the solutions.
First, add the first row to the second row to eliminate the 6 in the second row:
((-6, 2, -4, 4, 0),
(0, 0, 0, 0, 0))
Now, we can see that this matrix represents the following equations:
-6x1 + 2x2 - 4x3 + 4x4 = 0
0 = 0
From the second equation, we can see that x2, x3, and x4 are free variables. We can choose any values for them. Let's choose x2 = a, x3 = b, and x4 = c.
Now, we can express x1 in terms of the free variables:
-6x1 + 2x2 - 4x3 + 4x4 = 0
-6x1 + 2a - 4b + 4c = 0
-6x1 = -2a + 4b - 4c
x1 = (2a - 4b + 4c) / -6
x1 = (-a/3) + (2b/3) - (2c/3)
Therefore, the general solution to the system of equations is:
x = (-a/3, a, b, c)
This means the spanning set of the null space of A is:
{((-a/3, a, b, c) | a, b, c ∈ ℝ)}
In other words, any vector in this form is a solution to the equation Ax = 0 and therefore spans the null space of A.