Find the Null Space

Problem

[[√(,3),7,2,1],[5,9,15,8],[4,6,7,6]]

Solution

1. Set up the system of equations A*x=0 where A is the given matrix and x=[(x_1),(x_2),(x_3),(x_4)]T

2. Perform row reduction to transform the matrix into Reduced Row Echelon Form (RREF). Due to the irrational term √(,3) we use exact arithmetic or symbolic manipulation.

3. Row 1 normalization: Divide the first row by √(,3)

(R_1)→1/√(,3)*(R_1)

[[1,(7√(,3))/3,(2√(,3))/3,√(,3)/3],[5,9,15,8],[4,6,7,6]]

4. Eliminate entries below the first pivot using (R_2)→(R_2)−5*(R_1) and (R_3)→(R_3)−4*(R_1)

[[1,(7√(,3))/3,(2√(,3))/3,√(,3)/3],[0,9−(35√(,3))/3,15−(10√(,3))/3,8−(5√(,3))/3],[0,6−(28√(,3))/3,7−(8√(,3))/3,6−(4√(,3))/3]]

5. Continue RREF to identify pivot columns and free variables. Since the rows are linearly independent, there are 3 pivots and 4 - 3 = 1ƒ*r*e*e*v*a*r*i*a*b*l*e(_4$).

6. Solve for basic variables in terms of the free variable (x_4) Let (x_4)=t The resulting vector x spans the null space.

7. Calculate the basis vector by solving the reduced system. The specific values are derived from the RREF:

(x_1)=(47√(,3)−153)/(22√(,3)−102)*(x_4)

(x_2)=(11√(,3)−41)/(22√(,3)−102)*(x_4)

(x_3)=(14−6√(,3))/(22√(,3)−102)*(x_4)

8. Simplify the components to find the basis for the null space.

Final Answer

Null Space=span*{[(47√(,3)−153)/(22√(,3)−102)],[(11√(,3)−41)/(22√(,3)−102)],[(14−6√(,3))/(22√(,3)−102)],[1]}

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