Find the Nullity [[b+c,c+a,a+b],[a,b,c],[1,1,1]]=0

Problem

nullity*[[b+c,c+a,a+b],[a,b,c],[1,1,1]]

Solution

1. Set up the matrix A and identify that the nullity is the dimension of the solution space to A*x=0

A=[[b+c,c+a,a+b],[a,b,c],[1,1,1]]

2. Perform row operations to simplify the matrix. Add the second row to the first row.

(R_1)←(R_1)+(R_2)

[[a+b+c,a+b+c,a+b+c],[a,b,c],[1,1,1]]

3. Factor out the common term (a+b+c) from the first row.

(a+b+c)*[[1,1,1],[a,b,c],[1,1,1]]

4. Eliminate the third row by subtracting the first row from it.

(R_3)←(R_3)−(R_1)

(a+b+c)*[[1,1,1],[a,b,c],[0,0,0]]

5. Analyze the rank based on the variables a,b,c If a,b,c are not all equal, the first two rows are linearly independent, so the rank is 2.

rank(A)=2

6. Apply the Rank-Nullity Theorem which states that rank(A)+nullity(A)=n where n is the number of columns.

2+nullity(A)=3

7. Solve for the nullity.

nullity(A)=1

Final Answer

nullity*[[b+c,c+a,a+b],[a,b,c],[1,1,1]]=1

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