1.  
A=[1,-2;-1,0]  ,  x=[x1,x2]  ,  b=[1,-3]
Given the matrices A, x, and b above, find the values of T for which
the linear system (TI2-A)x=b has a unique solution.
I2 = Identity Matrix
T = Theta

2.  Use the following procedure to find the steady-state vector for
the matrix [1/2, 2/7; 1/2, 5/7]
The first procedure for computing the steady-state vector u of a
regular transition matrix T is as follows.
step 1:  Compute the powers (T^n)x, where x is any probability vector.
Step 2:  u is the limit of the powers (T^n)x.
or
The second procedure for computing the steady-state vector u of a
regular transition matrix T is as follows.
Step 1:  Solve the homogeneous system (In-T)u=0
Step 2:  From the infinitely many solutions obtained in Step 1,
determine a unique u by requiring that its components satisfy equation
(4) u1+u2+...+un = 1

3.  Use theorem 3.9 to evaluate the determinant [1,0,1,8; 2,2,0,0;
3,4,-1,5; 0,7,11,5]
Theorem 3.9:  Let A = [aij] be an nxn matrix.  Then for each 1<=i<=n
det(A)=ai1Ai1 + ai2Ai2+......+ainAin
(expansion of det(A) along the ith row)
and for each 1<=j<=n,
det(A)=a1jA1j+a2jA2j+...+anjAnj
(expansion of det(A) along the jth column)

4.  Let L:  (R^3) -> (R^3) be the linear transformation defined by
L([x, y, z]) = [2x, 0, 5y].  Is [0, 0, 1] in range L.

5.  Let S = {v1, v2, v3, v4} consist of the vectors below.  Does S
form a basis for R^4?
v1=(1,0,1,0) , v2=(0,1,0,0) , v3=(0,1,1,0) , v4=(1,0,0,0)

6.  Suppose x=(3,-1,1),v=(2,1,-1), and w=(a,b,c).  Answer the
following:
(a)  Find a,b, and c so that the vector w is orthogonal to both x and
v.
(b)  Is it possible for the vector w to be a unit vector

7.  Let S=[v1,v2,v3,v4,v5], where
v1=(1,2,3), v2=(1,3,5), v3=(1,1,0), v4(1,0,1), v5=(0,1,1)
Find a basis for the subspace V = span S of R^3

Thank you for your question; here are the answers.

1. We have

A=

1 -2
-1 0


b=
1
-3

Let R(T) be the matrix

T 0
0 T


so that R(T)=T*I_2, for real T, where I_2 is the identity matrix of
order 2.

We wish to find those T for which the equation:

(R(T)-A) x = b

has a unique solution. There is a unique solution of the matrix
equation

B x = b

for any matrix B if and only if B is nonsingular, which is true if and
only if B has nonzero determinant.

R(T)-A=

T-1  2
1    T

The determinant of R(T)-A is thus

T^2-T - 2

= (T-2) * (T+1)

Hence, the determinant is zero if and only if T equals 2 or T= -1 .
Thus, R(T)-A is nonsingular if and only if T is not equal to 2 or -1 .

Therefore, the required set of values is the set of all T except 2 and
-1 .

----
2. Let the matrix T be

1/2 2/7
1/2 5/7

A steady-state vector for T is a vector u such that 

T u = u .
u_1 + u_2 = 1 .

We will use the second procedure to determine u.

Step 1:

We solve 

 (I_2-T) u = 0

I_2-T = A, where A=

1-1/2   -2/7
-1/2    1-5/7

=

1/2  -2/7
-1/2  2/7

We note that A is nonsingular, as it has zero determinant. We want to
find a u such that

A u = 0

Let u=

u_1
u_2


Then we have the two equations:

1/2 u_1 - 2/7 u_2 = 0
-1/2 u_1 +2/7 u_2 =0

From the first equation, we see that:

u_1 = 4/7 u_2

setting 

u_1 + u_2 =1 , we get

4/7 u_2 + u_2 =1,

11/7 u_2 =1
u_2 = 7/11
u_1=4/11

So u=

4/11
7/11


Let us check this:

T u = 

1/2 2/7       *    4/11
1/2 5/7            7/11

=

4/22+14/77
4/22+35/77

=
2/11 + 2/11
2/11 + 5/11

= 
4/11
7/11

= u 

And we are done.

---
3.

Let A be

1 0 1 8
2 2 0 0
3 4 -1 5
0 7 11 5

We let the cofactor 

A_{ij}
         i+j
be the -1    times the determinant of the matrix formed by removing
row i and colum j from A.

To get det(A) we expand A along the 2nd row, to get

det(A) =

 2 A_{2,1} + 2 A_{2,2}


A_{21} is the negative of the determinant of:

0 1 8
4 -1 5
7 11 5

expanding along the 1st row, this is then

-(
-det(
4 5
7 5
)

+ 8 det (

4 -1
7 11 )
)
which is then

-(( 20-35) + 8 (44+7) )
= -(15 + 8(51))
= -(15+408)
=-423

Similarly, A_{2,2} is the determinant of

1 1 8
3 -1 5
0 11 5

Expanding across the 3rd row, this is then:

-11 det(
1 8
3 5
)

+ 5 det(
1 1
3 -1)

=
=-11 (5-24) + 5 (-1-3)
=11 (19) -20
=189.

So the full determinant is:
2*A_{1,2} + 2*A_{2,2} = 
2*-423 + 2*189
= -468 .

which is the answer.

I checked this with Matlab:

>> A

A =

     1     0     1     8
     2     2     0     0
     3     4    -1     5
     0     7    11     5

>> det(A)

ans =

  -468

>> 

Really, though, linear algebra is about avoiding computations of this
kind, or at least using computers for them.

--
4. Since L([0, 1/5, 0])=[0,0,1] , it follows that [0,0,1] is in the
range of L.

---
5. No. By inspection, each of the 4 vectors has zero 4th coordinate.
Hence, no vector that has nonzero 4th coordinate can be a linear
combination of the given vectors.

--
6. I will write "." for the . product.

Two vectors s,t are orthogonal if s.t=0 .

There are two ways to do this. One is to take the cross-product of x
and v. In this case, though, I think it is simpler just to write it
out:

We want to find w so that

x.w=0 and
v.w=0 

written out, this is:

3a-b+c=0
2a+b-c=0

5a=0
a=0
b=c

so that orthogonal vectors are of the form:

[0,b,b]

(b) 
A sample unit vector is just 

u=[0, 1/sqrt(2), 1/sqrt(2)]

since the square of the norm is 1/2+1/2=1 .

7. The vectors v3,v4, and v5 are linearly independent, as can be
verified by computing the determinant of the matrix

v3
v4
v5

which turns out to be -2 .

alternatively, note that v3-v4=(0,1,-1), so

v5-(v3-v4)= 0,0,2.

Hence, (0,0,2) is in the span, and so also (0,0,1) is in the span.

Similarly or by symmetry, (1,0,0) is in the span as is (0,1,0), and so
all vectors of R^3 are in the span.

The required basis is therefore just

{v3,v4,v5}

I hope that these answers are helpful.

---

Request for Answer Clarification by bildy-ga on 05 Dec 2002 16:47 PST

Is it possible to expand on #'s 4 and 5 a bit?

Thanks

---

Clarification of Answer by rbnn-ga on 05 Dec 2002 17:15 PST

4. A vector v is in the range of L if and only if for some [x,y,z] we
have L([x,y,z])=v .

This is the definition of "range".
 
Choose x=0, y=1/5, z=0.

Then L([x,y,z])
=[2x, 0, 5y]
=[0,0,5*1/5]
=[0,0,1]

Hence, [0,0,1] is in the range of L.

5. 
Let u be the vector [0,0,0,1].

The span of S is the set of all sums of the form

a*v1+b*v2+c*v3+d*v4

where a,b,c,d are scalars.

Now, the 4th coordinate of v1,v2,v3, and v4 are each 0. 

So the 4th coordinate of a*v1 is 0, as is b*v2, as is c*v2, as is d*v3
.

Therefore, the sum of these values also has 4th coordinate of 0.

Hence, any vector in the span of S has 4th coordinate 0. 

But u has 4th coordinate 1.

Therefore, u is not in the span of S. Therefore, S is not a basis.

Went above and beyond what was necessary !  Thanks ! Would love to
work with you again !