Find the Null Space A=[[1,-1,3],[5,-4,-2],[7,-6,4]]

Problem

A=[[1,−1,3],[5,−4,−2],[7,−6,4]]

Solution

1. Set up the equation for the null space by solving the homogeneous system A*x=0 where x=[[(x_1)],[(x_2)],[(x_3)]]

2. Perform row operations to transform the matrix into row-echelon form. Subtract 5 times the first row from the second row ((R_2)−5*(R_1)→(R_2).

[[1,−1,3],[0,1,−17],[7,−6,4]]

3. Continue row reduction by subtracting 7 times the first row from the third row ((R_3)−7*(R_1)→(R_3).

[[1,−1,3],[0,1,−17],[0,1,−17]]

4. Eliminate the third row by subtracting the second row from the third row ((R_3)−(R_2)→(R_3).

[[1,−1,3],[0,1,−17],[0,0,0]]

5. Find the reduced row echelon form by adding the second row to the first row ((R_1)+(R_2)→(R_1).

[[1,0,−14],[0,1,−17],[0,0,0]]

6. Identify the free variable and express the pivot variables in terms of it. Here, (x_3) is free.

(x_1)−14*(x_3)=0⇒(x_1)=14*(x_3)

(x_2)−17*(x_3)=0⇒(x_2)=17*(x_3)

7. Write the solution vector in terms of the parameter (x_3)

x=[[14*(x_3)],[17*(x_3)],[(x_3)]]=(x_3)*[[14],[17],[1]]

Final Answer

Null(A)=span*{[14],[17],[1]}

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