Find the Kernel

Problem

ker*(([−6,−3,−4],[7,5,7])+4*([1,6,−9],[4,−3,−3]))

Solution

1. Distribute the scalar 4 into the second matrix.

4*([1,6,−9],[4,−3,−3])=([4,24,−36],[16,−12,−12])

2. Add the two matrices together to find the matrix A

A=([−6+4,−3+24,−4−36],[7+16,5−12,7−12])=([−2,21,−40],[23,−7,−5])

3. Set up the homogeneous system A*x=0 to find the kernel.

([−2,21,−40],[23,−7,−5])*([x],[y],[z])=([0],[0])

4. Perform row reduction on the augmented matrix. Divide the first row by −2

([1,−10.5,20],[23,−7,−5])

5. Eliminate the first entry of the second row by subtracting 23 times the first row from the second row.

([1,−10.5,20],[0,234.5,−465])

6. Solve for the variables in terms of the free variable z From the second row:

234.5*y=465*z⇒y=465/234.5*z=930/469*z

7. Substitute y back into the first row equation to find x

x=10.5*y−20*z=21/2*(930/469)*z−20*z=9765/469*z−9380/469*z=385/469*z

8. Simplify the fractions. Both 385 and 469 are divisible by 7

x=55/67*z

y=(930/7)/(469/7)*z=(132.85...)/67*z* (re-evaluating: *465/234.5=930/469=132.85/67* is not integer)

y=930/469*z=(132.85...)/67*z

Let's re-calculate 234.5*y−465*z=0⇒469/2*y=465*z⇒y=930/469*z
Then x=21/2930/469*z−20*z=(9765−9380)/469*z=385/469*z
Dividing by 7 x=55/67*z and y=(930/7)/(469/7)*z=(132.85...)/67*z
Actually, 469=7×67 930 is not divisible by 7
The basis vector is ([385],[930],[469])

Final Answer

ker*(([−6,−3,−4],[7,5,7])+4*([1,6,−9],[4,−3,−3]))=span*{[385],[930],[469]}

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