Find the Null Space [[3,7,1,0],[5,2,3,1],[8,5,2,1]]

Problem

Null Space*[[3,7,1,0],[5,2,3,1],[8,5,2,1]]

Solution

1. Set up the augmented matrix for the homogeneous system A*x=0

[[3,7,1,0,0],[5,2,3,1,0],[8,5,2,1,0]]

2. Perform row operations to reach row echelon form. Subtract row 1 and row 2 from row 3 ((R_3)−(R_1)−(R_2)→(R_3).

[[3,7,1,0,0],[5,2,3,1,0],[0,−4,−2,0,0]]

3. Continue row reduction to find the reduced row echelon form (RREF). Divide (R_3) by −4

[[3,7,1,0,0],[5,2,3,1,0],[0,1,0.5,0,0]]

4. Eliminate the second column entries in (R_1) and (R_2)

[[3,0,−2.5,0,0],[5,0,2,1,0],[0,1,0.5,0,0]]

5. Normalize (R_1) and eliminate the first column entry in (R_2)

[[1,0,−5/6,0,0],[0,1,1/2,0,0],[0,0,37/6,1,0]]

6. Identify free variables and express the solution in parametric form. Let (x_3)=t

(x_1)=5/6*t

(x_2)=−1/2*t

(x_4)=−37/6*t

7. Write the basis for the null space by factoring out the parameter t

x=t*[[5/6],[−1/2],[1],[−37/6]]

8. Scale the vector to use integers for a cleaner basis. Multiply by 6.

v=[[5],[−3],[6],[−37]]

Final Answer

Null Space*([3,7,1,0],[5,2,3,1],[8,5,2,1])=span*{[5],[−3],[6],[−37]}

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