Find the Nullity

Problem

nullity*[[1,−3,2,0],[−1,x,0,0],[0,−1,x,0],[0,0,−1,x]]

Solution

1. Identify the matrix A and the definition of nullity. The nullity is the dimension of the solution space to A*v=0 which is calculated as n−rank(A) where n is the number of columns.

2. Calculate the determinant of the matrix to determine when the rank is less than 4. We expand along the first row:

det(A)=1*|[x,0,0],[−1,x,0],[0,−1,x]|−(−3)*|[−1,0,0],[0,x,0],[0,−1,x]|+2*|[−1,x,0],[0,−1,0],[0,0,x]|−0

3. Simplify the determinants of the 3×3 submatrices:

det(A)=1*(x3)+3*(−x2)+2*(x)

det(A)=x3−3*x2+2*x

4. Factor the determinant to find the values of x where the matrix is singular:

det(A)=x*(x2−3*x+2)

det(A)=x*(x−1)*(x−2)

5. Determine the nullity based on the value of x If det(A)≠0 the rank is 4 and the nullity is 0. If x∈{0,1,2} the rank is less than 4.

6. Evaluate the rank for x=0

(A_x=0)=[[1,−3,2,0],[−1,0,0,0],[0,−1,0,0],[0,0,−1,0]]

The first three columns are linearly independent, so rank(A)=3 and nullity=4−3=1

7. Evaluate the rank for x=1

(A_x=1)=[[1,−3,2,0],[−1,1,0,0],[0,−1,1,0],[0,0,−1,1]]

The sum of the first three columns is the zero vector in the first three entries, but the fourth row prevents a rank drop below 3. Thus, rank(A)=3 and nullity=1

8. Evaluate the rank for x=2

(A_x=2)=[[1,−3,2,0],[−1,2,0,0],[0,−1,2,0],[0,0,−1,2]]

Similarly, the rank is 3, so nullity=1

Final Answer

nullity(A)={[1,if *x∈{0,1,2}],[0,if *x∉{0,1,2}])

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