Find the Null Space [[2,x,5,2],[x,7,3,0]]=5

Problem

A=[[2,x,5,2],[x,7,3,0]]

Solution

1. Identify the goal, which is to find the null space of the matrix A The null space consists of all vectors v such that A*v=0

2. Set up the homogeneous system of linear equations A*v=0 where v=[(v_1),(v_2),(v_3),(v_4)]^T

3. Write the system of equations based on the matrix rows:

2*(v_1)+x*(v_2)+5*(v_3)+2*(v_4)=0

x*(v_1)+7*(v_2)+3*(v_3)=0

4. Solve for the pivot variables (v_1) and (v_2) in terms of the free variables (v_3) and (v_4) From the second equation:

x*(v_1)=−7*(v_2)−3*(v_3)

(v_1)=(−7*(v_2)−3*(v_3))/x

5. Substitute (v_1) into the first equation:

2*((−7*(v_2)−3*(v_3))/x)+x*(v_2)+5*(v_3)+2*(v_4)=0

(−14*(v_2)−6*(v_3))/x+x*(v_2)+5*(v_3)+2*(v_4)=0

6. Multiply by x to clear the denominator:

−14*(v_2)−6*(v_3)+x^2*(v_2)+5*x*(v_3)+2*x*(v_4)=0

(x^2−14)*(v_2)+(5*x−6)*(v_3)+2*x*(v_4)=0

7. Isolate (v_2)

(x^2−14)*(v_2)=(6−5*x)*(v_3)−2*x*(v_4)

(v_2)=(6−5*x)/(x^2−14)*(v_3)−(2*x)/(x^2−14)*(v_4)

8. Substitute (v_2) back into the expression for (v_1)

(v_1)=(−7*((6−5*x)/(x^2−14)*(v_3)−(2*x)/(x^2−14)*(v_4))−3*(v_3))/x

(v_1)=(35*x−42−3*x^2+42)/(x*(x^2−14))*(v_3)+(14*x)/(x*(x^2−14))*(v_4)

(v_1)=(35−3*x)/(x^2−14)*(v_3)+14/(x^2−14)*(v_4)

9. Express the solution in vector form v=(v_3)*(n_1)+(v_4)*(n_2)

v=(v_3)*[[(35−3*x)/(x^2−14)],[(6−5*x)/(x^2−14)],[1],[0]]+(v_4)*[[14/(x^2−14)],[(−2*x)/(x^2−14)],[0],[1]]

Final Answer

Null(A)=span*([[35−3*x],[6−5*x],[x^2−14],[0]],[[14],[−2*x],[0],[x^2−14]])

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