Find the Null Space

Problem

Null Space*[[5,−10,25,−20],[1,−2,5,−4],[3,−6,15,12]]

Solution

1. Set up the homogeneous system A*x=0 by representing the matrix as an augmented matrix.

[[5,−10,25,−20,0],[1,−2,5,−4,0],[3,−6,15,12,0]]

2. Perform row operations to reach row echelon form. Swap (R_1) and (R_2)

[[1,−2,5,−4,0],[5,−10,25,−20,0],[3,−6,15,12,0]]

3. Eliminate entries below the first pivot by performing (R_2)−5*(R_1)→(R_2) and (R_3)−3*(R_1)→(R_3)

[[1,−2,5,−4,0],[0,0,0,0,0],[0,0,0,24,0]]

4. Simplify the matrix by swapping (R_2) and (R_3) and dividing the new (R_2) by 24

[[1,−2,5,−4,0],[0,0,0,1,0],[0,0,0,0,0]]

5. Identify the pivot columns and free variables. The pivots are in columns 1 and 4. The free variables are (x_2) and (x_3)

(x_1)−2*(x_2)+5*(x_3)−4*(x_4)=0

(x_4)=0

6. Solve for the basic variables in terms of the free variables.

(x_4)=0

(x_1)=2*(x_2)−5*(x_3)

7. Express the general solution in vector form.

[[(x_1)],[(x_2)],[(x_3)],[(x_4)]]=[[2*(x_2)−5*(x_3)],[(x_2)],[(x_3)],[0]]

[[(x_1)],[(x_2)],[(x_3)],[(x_4)]]=(x_2)*[[2],[1],[0],[0]]+(x_3)*[[−5],[0],[1],[0]]

Final Answer

Null Space=span*{[[2],[1],[0],[0]],[[−5],[0],[1],[0]]}

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