Find the Null Space

Problem

null*[[63,−36,45,−27,54],[−81,54,−63,45,−72],[−36,63,−54,72,−45],[45,−18,45,−9,36],[27,−54,−54,−63,36],[−27,−45,18,−36,81],[−63,45,72,72,27]]

Solution

1. Set up the matrix A and divide each row by its greatest common divisor to simplify the arithmetic.

A=[[63,−36,45,−27,54],[−81,54,−63,45,−72],[−36,63,−54,72,−45],[45,−18,45,−9,36],[27,−54,−54,−63,36],[−27,−45,18,−36,81],[−63,45,72,72,27]]

2. Perform row reduction to transform the matrix into Reduced Row Echelon Form (RREF).

[[1,0,0,0,1],[0,1,0,0,1],[0,0,1,0,1],[0,0,0,1,1],[0,0,0,0,0],[0,0,0,0,0],[0,0,0,0,0]]

3. Identify the free variables by looking for columns without pivots. The fifth column corresponds to the free variable (x_5)

(x_1)+(x_5)=0

(x_2)+(x_5)=0

(x_3)+(x_5)=0

(x_4)+(x_5)=0

4. Express the pivot variables in terms of the free variable (x_5)

(x_1)=−(x_5)

(x_2)=−(x_5)

(x_3)=−(x_5)

(x_4)=−(x_5)

5. Write the solution vector in parametric form to find the basis for the null space.

[[(x_1)],[(x_2)],[(x_3)],[(x_4)],[(x_5)]]=(x_5)*[[−1],[−1],[−1],[−1],[1]]

Final Answer

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

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