Answer:
To find the basis for the null space of A, we need to find all solutions to the equation Ax = 0, where 0 is a vector of zeros and x is a vector of unknowns. We can solve this equation using Gaussian elimination and back-substitution.
After performing Gaussian elimination, we get the row-reduced echelon form of A:
⎡⎣
1 0 -1 0
0 1 -2 0
0 0 0 1
0 0 0 0
0 0 0 0
⎤⎦
The last two rows of the row-reduced echelon form of A correspond to the equation 0x = 0, which is always true. The first three rows correspond to the equation x1 - x3 = 0, x2 - 2x3 = 0, and x4 = 0.
Solving for the variables in terms of the free variable x3, we get:
x1 = x3
x2 = 2x3
x3 = x3
x4 = 0
Thus, the general solution to Ax = 0 is:
x = t1 * ⎡⎣
1
2
1
0
⎤⎦
+ t2 * ⎡⎣
0
0
1
0
⎤⎦
where t1 and t2 are any constants.
Therefore, the basis for the null space of A is:
{⎡⎣
1
2
1
0
⎤⎦, ⎡⎣
0
0
1
0
⎤⎦}
To find the basis for the row space of A, we need to find the row echelon form of A. After performing Gaussian elimination, we get:
⎡⎣
1 2 -1 2
0 1 -2 0
0 0 4 2
0 0 0 0
0 0 0 0
⎤⎦
The non-zero rows of the row echelon form of A correspond to the rows of A that contain the pivots. Therefore, the basis for the row space of A is:
{⎡⎣
1
2
-1
2
⎤