1 Find the eigenvalues of A 2 Find a matrix P such that P1AP

1) Find the eigenvalues of A;

2) Find a matrix P such that P^-1AP is diagonal.

3) Find the solution x(t) of the equations with x(0) = (-1,1,0)

dt

Solution

A = [ 5 -28 -18

-1 5 3

3 16 10]

Now, to find eigen value | A - I | = 0

[ 5 -28 -18 [ 1 0 0

-1 5 3 -    0 1 0 = 0

3 16 10] 0 0 1]

[ 5 - -28 -18

-1 5 - 3

3 16 -10 - ]

| A - I| = 0

(5 - )[ (5-) (-10 - ) - 48] + 28 [ -1 ( -10-) - 9] - 18[ -16 - 3(5-)]

(5-) [ ² + 5 - 98] + 28 [ + 1] -18 [ 3 - 31]

-³ + 97 + 96 = 0

³ - 97 - 96 = 0

= -1 , 10.3 , -9.3

1 = -1

2 = 10.3

3 = -9.3

2) to show, P^-1A P = D

[I - A] X = 0

For = -1

[ 6 -28 -18   

-1 6 3

3 16 -9 ] X = 0

Using Gaussian elimination method,

R1 -> R1/6

[ 1 -14/3 -3

-1 6 3

3 16 -9]

R2 - ( -R1) * R1 ->R2

[ 1 -14/3 -3

0 4/3 0

3 16 -9]

R3 3×R1R3

[1 -14/3 -3

0 4/3 0

0 30 0]

R2 * 4/3 ->R2

[ 1 -14/3 -3

0 1 0

0 30 0]

R3 - 30* R2---> R3

[ 1 -14/3 -3

0 1 0

0 0 0]

R1 - (-14/3) × R2R1

[1 0 -3

0 1 0

0 0 0]

Hence ,

x1 - 3x3 = 0

x2 = 0

Hence X1 = [ 3x3

0

x3]

Similarly,

X2 = [-881 x3

166.64 x3

x3]

X3 = [0.98 x3

-0.14 x3

x3]

Hence,

P = [3 -881.98 0.98

0 166.4 -0.14

1 1 1]

3)

x(t) = e^(-t) c1v1 + c2 v2 e^(-9.3t) + c3(tv2 + v3)e^(10.3t)

A x[0] = [3 e^-t -881.98e^10.3 t (t + 0.98) e^(-9.3)t

0 166.64 e^(10.3)t -e^(-9.3)t

1e^(-t) e^(10.3)t (t+1) e^(-9.3t)]

Put t = 0

= [3 -881.98 0.98

0 166.64 -0.14

1 1 1]