Find the Null Space x=1/(2[[5,4,8],[3,3,7]])

Problem

null*(1/2*[[5,4,8],[3,3,7]])

Solution

1. Identify the matrix A Scaling a matrix by a non-zero constant does not change its null space, so we can find the null space of the simpler matrix M

M=[[5,4,8],[3,3,7]]

2. Set up the homogeneous system M*x=0 to find the vectors in the null space:

[[5,4,8],[3,3,7]]*[[(x_1)],[(x_2)],[(x_3)]]=[[0],[0]]

3. Perform row reduction to find the row-echelon form. Divide the first row by 5

[[1,0.8,1.6],[3,3,7]]

4. Eliminate the first entry of the second row by subtracting 3 times the first row from the second:

[[1,0.8,1.6],[0,0.6,2.2]]

5. Normalize the second row by dividing by 0.6

[[1,0.8,1.6],[0,1,11/3]]

6. Back-substitute to reach reduced row-echelon form by subtracting 0.8 times the second row from the first:

[[1,0,−4/3],[0,1,11/3]]

7. Express the variables in terms of the free variable (x_3)

(x_1)=4/3*(x_3)

(x_2)=−11/3*(x_3)

8. Write the solution in vector form by choosing (x_3)=3 to clear denominators:

x=[[4],[−11],[3]]

Final Answer

null*(1/2*[[5,4,8],[3,3,7]])=span*{[4],[−11],[3]}

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