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Problems B and F are graded.
Problem B: Find an eigenvector corresponding to the eigenvalue λ=4 of the matrix
A=[30−1231−345].
Solution: We must find a vector →x such that A→x=4→x, i.e.
[30−1231−345][x1x2x3]=4[x1x2x3].
or in other words
[3x1−x32x1+3x2+x3−3x1+4x2+5x3]=[4x14x24x3].
Subtracting the right-hand-side from both sides of the equation yields
[−x1−x33x1−x2+x3−3x1+4x2+x3]=→0,
yielding the system of equations
{−x1−x3=03x1−x2+x3=0−3x1+4x2+x3=0.
The reduced echelon form of the assocaited augmented matrix is
[101001100000].
which is equivalent to the system
{x1+x3=0x2+x3=0
and so we see that x1=−x3 and x2=−x3 with free variable x3. Thus we see that an eigenvector →x is any vector of the form
→x=[x1x2x3]=[−x3−x3x3]=x3[−1−11].
So pick the simplest constant, x3=1, to get an eigenvector [−1−11].
Problem F: Use integration by parts to calculate the antiderivative of f(x)=log(x). (Hint: use u=logx and dv=1. Also recall that ddxlogx=1x)
Solution: Using u=logx yields du=1xdx and using dv=1dx yields v=x. Therefore by integration by parts,
∫log(x)dx=∫udv=uv−∫vdu=xlogx−∫1dx=xlogx−∫x1xdx=xlogx−x+C,
where C is a constant introduced when finding the anti-derivative of −1.
