Test 2 Practice Exam Key 

>	with(LinearAlgebra):

 

1. 

>	A:=<<1,-1,2>|<1,0,4>|<1,2,8>>;

 

(1)

 

>

AT:=A^%T;

 

(2)

 

a. 

>	A1:=GaussianElimination(A);

 

(3)

 

>	A2:=ReducedRowEchelonForm(A1);

 

(4)

 

>	r1:=Row(A1,1);

 

(5)

 

>	r2:=Row(A1,2);

 

(6)

 

>	r11:=Row(A2,1);

 

(7)

 

>	f21:=Row(A2,2);

 

(8)

 

rowspace=span{r1,r2}=span{r11,r21} 

b. 

>	AT1:=GaussianElimination(AT);

 

(9)

 

>	AT2:=ReducedRowEchelonForm(AT1);

 

(10)

 

>	c1:=Row(AT1,1);

 

(11)

 

>	c2:=Row(AT1,2);

 

(12)

 

>	c11:=Row(AT2,1);

 

(13)

 

>	c21:=Row(AT2,2);

 

(14)

 

c. 

>	z:=<0,0,0>;

 

(15)

 

>	Az:=<A|z>;

 

(16)

 

>	NT:=BackwardSubstitute(GaussianElimination(Az));

 

(17)

 

>	n1:=<2,-3,1>;

 

(18)

 

d.  

>	RN:=<r1^%T|r2^%T|n1>;

 

(19)

 

>	GaussianElimination(RN);

 

(20)

 

The matrix is if rank 3 so they are linearly independent.   

2. 

a. 

>	v1:=<1,0,1>;

 

(21)

 

>	v2:=<1,0,0>;

 

(22)

 

>	v3:=<0,1,1>;

 

(23)

 

>	V:=<v1|v2|v3>;

 

(24)

 

>	GaussianElimination(V);

 

(25)

 

Three piviots so they are linearly independent. 

b. 

>	alpha:=DotProduct(v1,v2)/DotProduct(v1,v1);

 

(26)

 

>	p:=alpha*v1;

 

(27)

 

c. 

>

p1:=v1;

 

(28)

 

>

p2:=v2-p;

 

(29)

 

>	p3:=v3-DotProduct(v3,p2)/DotProduct(p2,p2)*p2-DotProduct(v3,p1)/DotProduct(p1,p1)*p1;

 

(30)

 

d.   

>	q1:=Normalize(p1,Euclidean);

 

(31)

 

>	q2:=Normalize(p2,Euclidean);

 

(32)

 

>

q3:=p3;

 

(33)

 

>	Q:=<q1|q2|q3>;

 

(34)

 

>	R:=Q^%T.<v1|v2|v3>;

 

(35)

 

e. 

>	e:=<0,1,0>;

 

(36)

 

>	QTe:=Q^%T.e;

 

(37)

 

>	x:=BackwardSubstitute(<R|QTe>);

 

(38)

 

>	<v1|v2|v3>.x;

 

(39)

 

3. 

>	E:=<<1,1,1>|<-1,0,1>>;

 

(40)

 

>	b:=<-1,1,3.1>;

 

(41)

 

>	ETE:=E^%T.E;

 

(42)

 

>	ETb:=E^%T.b;

 

(43)

 

>	s:=BackwardSubstitute(<ETE|ETb>);

 

(44)

 

>

a:=s[1];

 

(45)

 

>

b:=s[2];

 

(46)

 

y=1.033333333+2.05x.   

4a 

>	<x+y,x-y,x,y>:=x*<1,1,1,0>+y*<1,-1,0,1>;

 

(47)

 

>	a:=<1,1,1,0>;

 

(48)

 

>	b:=<1,-1,0,1>;

 

(49)

 

=span(a,b) so a vector space.   

b. 

Y in U then YA=0 and Z in U then ZA=0    Look at (a*Y+b*Z)A=a*YA+b*Za=a*0+b*0 =0 so is closed under linear combinations which tell us we have a subspace. 

5.  Notice that the following is true. 

>	B1:=<<1,0>|<0,-1>>;

 

(50)

 

>	B2:=<<1,1>|<-1,1>>;

 

(51)

 

>	B3:=<<0,1>|<-1,0>>;

 

(52)

 

>

B2=B1+B3;

 

(53)

 

Also, B1 and B3 are linearly independent so a basis is B1,B3.  The dimension is 2.  You could look at a linear combination and set it equal to zero 

>	a1*B1+b1*B2+c1*B3;

 

(54)

 

We then have a1+b1=0 and b1+c1=0 . The other two equations -b1-c1=0 and -a1-b1=0 are the same as the first tow.  This tell us that a1=c1and b1=-1 is a possiblity so they are dependent.  But the first and the last one are definitely independent so the dimension is two.   

6.   

>	w1:=<1,-1,1>;

 

(55)

 

>	w2:=<0,1,-1>;

 

(56)

 

>	X:=<x,y,z>;

 

(57)

 

>	CE:=<w1|w2|X>;

 

(58)

 

>	GaussianElimination(CE);

 

(59)

 

b. 

We must have z+y=0 or z=-y which tell us the vectors look like <x,y,-y>   Try <0,0,1>; 

>	w3:=<0,0,1>;

 

(60)

 

>	GaussianElimination(<w1|w2|w3>);

 

(61)

 

They are linearly independent so they form a basis for three space.   

7.   0:=c1*v1+c2*v2+...+ck*vk 

Want to show that ci for any i is zero 

Next multiply both sides by  vi^T 

The gives 

vi^T.0= vi^Tc1v1+.....+vi^Tck*vk but everything is zero since orthogonal except when k=i and we have 

vi^T.civi =0 or civi^T.vi=0  Since viT.vi is the square of the lenght then ci=0 and the vectors are linearly independent.    

8.  a. T 

    b.  F  Must be linearly independent. 

    c.  T  your text defines a orthogonal matrix as a square matrix so the answer must be true.   

    d.  T  Theorem in you book 

    e.  T  If not then the dimension would be bigger than n which is impossible.