**Sec. 3.1 (BDH)**
#5
(x') = (2 1) (x)
(y') (1 1) (y)

#17a Reducing to a first order system of equations by letting
v = y', we get that v=0, and y = anything.
#17b Using the same approach we get the same answer.
#24 Just plug and chug.
**Sec 3.2 (BDH)**
#17a eigenvalues 2, -3, corresponding eigenvectors
(0) (0.98058067569092)
(1) (0.19611613513818)

#19a eigenvalues -5, -2, corresponding eigenvectors
(-0.70710678118655) (-0.44721359549996)
(0.70710678118655 ) (-0.89442719099992)

#30 eigenvalues are **a** and **d** and corresponding eigenvectors
(1) (1 )
(0) ((d-a)/b)

#31 eigenvalues are 0.5*(a+d+ ( (a-d)^2+4b^2 )^(1/2) and
0.5*(a+d - ( (a-d)^2+4b^2 )^(1/2). Since the discrminant is positive
these eigenvalues are always real. Doing some algebra one can see that
the solutions are distinct so long as b is not zero.

Last modified: Thur Jan 30 16:44:48 1997