Test 3 practice exam 

1. 

>	with(linalg):

 

>	I3:=diag(1,1,1);

 

(1)

 

>	A:=matrix([[1,0,1],[0,2,0],[1,0,1]]);

 

(2)

 

>	p:=det(evalm(A-lambda*I3));

 

(3)

 

>	solve(p=0,lambda);

 

(4)

 

>	v1:=nullspace(evalm(A));

 

(5)

 

>	v2:=nullspace(evalm(A-2*I3));

 

(6)

 

a.  spectrum ={0,2} spectrial radius=2 

b.{ λ=0,[-1,0,1]}, {λ=2,[1,0,1],[0,1,0]}  

c.  

>	q1:=normalize(v1[1]);

 

(7)

 

>	q2:=normalize(v2[1]);

 

(8)

 

>	q3:=normalize(v2[2]);

 

(9)

 

>	Q:=augment(q1,q2,q3);

 

(10)

 

>	evalm(transpose(Q)&*A&*Q);

 

(11)

 

2l   

>	A:=matrix([[2,1],[1,2]]);

 

(12)

 

>	X:=matrix([[x],[y]]);

 

(13)

 

>	simplify(evalm(transpose(X)&*A&*X));

 

(14)

 

>	I2:=diag(1,1);

 

(15)

 

>	p:=det(evalm(A-lambda*I2));

 

(16)

 

>	solve(p=0,lambda);

 

(17)

 

b.  both eigenvalues are postive.  

c.  2>0 and det(A)=4-1=3 so postive definite.  

>	gausselim(A);

 

(18)

 

>

det(%);

 

Both piviots are postivie.   

(19)

 

>	p1:=nullspace(evalm(A-3*I2))[1];

 

(20)

 

>

 

>	p2:=nullspace(evalm(A-I2))[1];

 

(21)

 

3.   

>	H:=matrix([[1,1],[0,3]]);

 

(22)

 

>	P:=det(evalm(H-lambda*I2));

 

(23)

 

>	p1a:=nullspace(evalm(H-3*I2))[1];

 

(24)

 

>	p2a:=nullspace(evalm(H-I2))[1];

 

(25)

 

a.   

For F.  [1,1],[-1,1]; for H [1,2],[1,0] 

b. 

>	Q:=augment(normalize(p1),normalize(p2));

 

(26)

 

>	F:=evalm(A);

 

(27)

 

>	Lambda:=evalm(transpose(Q)&*F&*Q);

 

(28)

 

c.  

>	P1:=augment(p1a,p2a);

 

(29)

 

>	Lambda:=evalm(inverse(P1)&*H&*P1);

 

(30)

 

d. 

>	P:=evalm(Q&*inverse(P1));

 

(31)

 

>	evalm(inverse(P)&*F&*P);

 

(32)

 

It is always good to check your answers.   

4.   

a.  

>	A:=matrix([[2,3,1],[0,-1,1],[0,0,1]]);

 

(33)

 

Since an upper triangular then eigenvalues are on the diagonal and are 2,-1,1 so there are three linearly independent eigenvectors so it is diagonlizable.   

b.  

>	B:=matrix([[2,1,-1],[1,-4,1],[-1,-1,9]]);

 

(34)

 

1. circle centered at 2 radius 2, 2.  circle centered at -4 radius 2, 3.  circle centered at 9 radius 2.   

in the interval [0,4], in the interval [-6,-2] and in the interval [7,11] .  There are three real eigenvalues so three distinct eigenvectors.  This means that the matrix can be made similar to a diagonl matrix.     

The interval for the largest eigenvlaue is [7,11] 

5 

a.  

>	C:=matrix([[7/8,1/8],[1/8,7/8]]);

 

(35)

 

>	x0:=matrix([[2],[1]]);

 

(36)

 

>	eigenvects(C);

 

(37)

 

Ther other eigenvalue is 3/4. 

b.  { 1,[1,1]}, {3/4,[-1,1]} 

c.    

>	q1:=normalize([1,1]); q2:=normalize([-1,1]);

 

 

	(38)

 

>	Q:=augment(q1,q2);

 

(39)

 

>	Lambda:=evalm(transpose(Q)&*C&*Q);

 

(40)

 

d.   

>	steadysc:=diag(1,0);

 

(41)

 

>	steadystate:=evalm(Q&*steadysc&*transpose(Q)&*x0);

 

(42)

 

>

 

6. 

>	B:=matrix([[1,3],[3,1]]);

 

(43)

 

>	simplify(evalm(transpose(X)&*B&*X));

 

(44)

 

>	EB:=eigenvects(B);

 

(45)

 

b product of eigenvalues is -2*4 =-8 so we have a hyperbola 

c. 

>	q1:=normalize(EB[2][3][1]);

 

(46)

 

>	q2:=normalize(EB[1][3][1]);

 

(47)

 

>	Q:=augment(q1,q2);

 

(48)

 

>	Y:=evalm([[u],[v]]);

 

(49)

 

>	Y=evalm(transpose(Q)&*X);

 

(50)

 

>	Lambda:=evalm(transpose(Q)&*B&*Q);

 

(51)

 

So the new equatyion will be  

>	evalm(transpose(Y)&*Lambda&*Y)[1,1]=1;

 

(52)

 

>

 

Angle of rotation is cos(θ)= 

7. 

>	L:=matrix([[1,1,1],[0,2,0],[1,0,1]]);

 

(53)

 

>	EL:=eigenvects(L);

 

(54)

 

a  Î»=2 a.m.=2 g.m.=1,  Î»=0, a.m =1 and g.m=1.    

b.  

>	q1:=normalize(EL[1][3][1]);

 

(55)

 

>	q2:=normalize(EL[2][3][1]);

 

(56)

 

>	q3:=vector([0,1,0]);

 

(57)

 

>	Q:=augment(q1,q2,q3);

 

(58)

 

>	R:=evalm(transpose(Q)&*L&*Q);

 

(59)

 

c.  

 

8. 

a.  T. 

b.  F  A must be diagonalizable 

c.  F.. 

d.  F  must be greater than zero 

e.  T.