Find the Kernel [[2,3,4],[1,2,1],[3,1,2]]=[[x],[y],[z]]

Problem

[[2,3,4],[1,2,1],[3,1,2]]*[[x],[y],[z]]=[[0],[0],[0]]

Solution

1. Set up the homogeneous system of linear equations A*x=0 to find the kernel (null space) of the matrix.

2*x+3*y+4*z=0

x+2*y+z=0

3*x+y+2*z=0

2. Form the augmented matrix and perform row operations to reach row-echelon form.

[[2,3,4,0],[1,2,1,0],[3,1,2,0]]

3. Swap the first and second rows to create a leading 1 in the first row.

[[1,2,1,0],[2,3,4,0],[3,1,2,0]]

4. Eliminate the entries below the leading 1 by performing (R_2)−2*(R_1)⇒(R_2) and (R_3)−3*(R_1)⇒(R_3)

[[1,2,1,0],[0,−1,2,0],[0,−5,−1,0]]

5. Eliminate the entry in the third row, second column by performing (R_3)−5*(R_2)⇒(R_3)

[[1,2,1,0],[0,−1,2,0],[0,0,−11,0]]

6. Solve the resulting system using back-substitution.

−11*z=0⇒z=0

−y+2*(0)=0⇒y=0

x+2*(0)+0=0⇒x=0

7. Conclude that the only solution is the trivial solution, meaning the kernel contains only the zero vector.

Final Answer

ker*[[2,3,4],[1,2,1],[3,1,2]]={[0],[0],[0]}

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