Find the Null Space

Problem

null*[[−42,54,24,−30,−18,18,42],[24,−36,−42,12,36,30,−30],[−30,42,36,−30,36,−12,−48],[18,−30,−48,6,42,24,−48],[−36,48,30,−24,−24,−54,−18]]

Solution

1. Set up the augmented matrix [A|0] to solve the homogeneous system A*x=0

2. Perform Gaussian elimination to transform the matrix into row-reduced echelon form (RREF).

3. Divide the first row by −42 and use row operations to eliminate entries below the first pivot.

4. Continue row reductions for subsequent columns to identify pivot columns and free variables.

5. Identify the RREF of the matrix:

[[1,0,0,0,0,−1,−1],[0,1,0,0,0,−1,1],[0,0,1,0,0,1,1],[0,0,0,1,0,1,2],[0,0,0,0,1,1,−1]]

6. Express the pivot variables (x_1),(x_2),(x_3),(x_4),(x_5) in terms of the free variables (x_6) and (x_7)

(x_1)=(x_6)+(x_7)

(x_2)=(x_6)−(x_7)

(x_3)=−(x_6)−(x_7)

(x_4)=−(x_6)−2*(x_7)

(x_5)=−(x_6)+(x_7)

7. Write the general solution in vector form by factoring out the free variables:

x=(x_6)*[[1],[1],[−1],[−1],[−1],[1],[0]]+(x_7)*[[1],[−1],[−1],[−2],[1],[0],[1]]

Final Answer

null(A)=span*{[[1],[1],[−1],[−1],[−1],[1],[0]],[[1],[−1],[−1],[−2],[1],[0],[1]]}

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