Key Practice Exam 1 

>	with(linalg):

 

>	with(LinearAlgebra):

 

Problem 1 

a.  

>	b:=<-2,1,2>;

 

(1)

 

>	a:=(-6/Norm(b,2))*b;

 

(2)

 

>

 

b.  

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

 

(3)

 

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

 

(4)

 

>	AB:=<<Row(A,1).Column(B,1),Row(A,2).Column(B,1)>|<Row(A,1).Column(B,2),Row(A,2).Column(B,2)>>;

 

(5)

 

>

 

c.   

>	c:=<b1,1/sqrt(3),1/sqrt(3)>;

 

(6)

 

>	solve(Norm(c,2)-1=0);

 

(7)

 

2. 

>	C:=<<1,2,2>|<2,5,4>|<3,7,6>|<1,3,a1>>;

 

(8)

 

>	C1:=addrow(addrow(C,1,2,-2),1,3,-2);

 

(9)

 

a1=2 makes both matrices have rank 2. 

3. 

a   

>	F:=<<1,2,3>|<-1,-2,-5>|<3,7,5>>;

 

(10)

 

>	F1:=addrow(addrow(F,1,2,-2),1,3,-3);

 

(11)

 

>	F2:=swaprow(F1,2,3);

 

(12)

 

b.  

>

Df:=2;

 

(13)

 

4.  

>	A:=<<1,2,1>|<1,3,3>|<1,2,4>>;

 

(14)

 

>	b:=<1,3,4>;

 

(15)

 

>	Ab:=<A|b>;

 

(16)

 

>	Ab1:=addrow(addrow(Ab,1,2,-2),1,3,-1);

 

(17)

 

>	Ab2:=addrow(Ab1,2,3,-2);

 

(18)

 

>

x3:=1/3;

 

(19)

 

>

x2:=1;

 

(20)

 

>	x1:=1-1-1/3;

 

(21)

 

>	X:=<-1/3,1,1/3>;

 

(22)

 

>

A.X;

 

(23)

 

>

 

5. 

>	B:=<<1,2,-1>|<1,2,0>|<1,3,-1>>;

 

(24)

 

>	I3:=IdentityMatrix(3);

 

(25)

 

>	BI3:=<B|I3>;

 

(26)

 

>	BI31:=addrow(addrow(BI3,1,2,-2),1,3,1);

 

(27)

 

>	BI32:=swaprow(BI31,2,3);

 

(28)

 

>	BI33:=addrow(addrow(BI32,3,1,-1),2,1,-1);

 

(29)

 

>	InvB:=delcols(BI33,1..3);

 

(30)

 

>

 

6. 

>	A:=<<-4,-2>|<2,2>>;

 

(31)

 

>	B:=<<1,2>|<0,1>>;

 

(32)

 

>

 

a. 

>	AI:=1/(-4)*<<2,2>|<-2,-4>>;

 

(33)

 

>

A.AI;

 

(34)

 

>

 

b.  B is an elementary matrix so the inverse is found by changing the sign of 2. i.e 

>	BI:=<<1,-2>|<0,1>>;

 

(35)

 

>

B.BI;

 

(36)

 

>

 

c. 

>	ABI:=BI.AI;

 

(37)

 

>

A.B.ABI;

 

(38)

 

>

 

7. 

>	AA:=<<1,2,5>|<1,3,7>|<-1,-1,-3>>;

 

(39)

 

>	bb:=<2,4,10>;

 

(40)

 

>	AAbb:=<AA|bb>;

 

(41)

 

>	AAbb1:=addrow(addrow(AAbb,1,2,-2),1,3,-5);

 

(42)

 

>	AAbb2:=addrow(AAbb1,2,3,-2);

 

(43)

 

>	AAbb3:=addrow(AAbb2,2,1,-1);

 

(44)

 

>

x3:=t1;

 

(45)

 

>

x2:=-t1;

 

(46)

 

>	x1:=2+2*t1;

 

(47)

 

>	X:=<x1,x2,x3>;

 

(48)

 

>

AA.X;

 

(49)

 

>	xp:=<2,0,0>;

 

(50)

 

>	xh:=<2*t1,-t1,t1>;

 

(51)

 

>	L:=<<1,0,0,0>|<0,1,0,0>|<0,0,1,0>|<0,0,0,1>>;

 

(52)

 

>	L[2,1]:=-2;

 

(53)

 

>	L[3,1]:=-1;

 

(54)

 

>	L[4,1]:=-2;

 

(55)

 

Since the next step is to swap row 2,3 then we will swap the entries in L[2,1] and L[3,1]; 

>	L[2,1]:=-1;

 

(56)

 

>	L[3,1]:=-2;

 

(57)

 

We should now swap L[3,1l and L[4,1] but they are the same so that will not be necessary.  

>	L[3,2]:=-1;

 

(58)

 

>	U:=<<-1,0,0,0>|<-2,-1,0,0>|<3,2,-2,0>|<0,3,4,5>>;

 

(59)

 

>	P:=IdentityMatrix(4);

 

(60)

 

>	P:=swaprow(swaprow(P,2,3),3,4);

 

(61)

 

>	G:=<<-1,2,1,1>|<-2,4,1,5>|<3,-6,-1,-10>|<0,5,3,1>>;

 

(62)

 

>

P.G;

 

(63)

 

>

L.U;

 

(64)

 

c. 

>	b:=<1,0,-1,2>;

 

(65)

 

>	ForwardSubstitute(<L|P.b>);

 

(66)

 

9.   

>	H:=<<1,2,-1,0>|<-1,3,1,-2>|<1,1,1,1>|<5,6,-4,3>>;

 

(67)

 

>

 

We wand the element in coefactor matrix in the 3,2 position since the adjoint is the transpose of the matrix hence putting it in the 2,3 position.  Must also consider the sign of this subdeterminate.  

>	C32:=delrows(delcols(H,2..2),3..3);

 

(68)

 

>	Adj23:=-det(C32);

 

(69)

 

>

 

10. 

a.  T. 

b.  F  The Zero matrix is upper triangular 

c.  T 

d.  F  To be perpendicular then the inner product should be zero. 

e.  T