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 SIAM Activity Group on Linear Algebra::
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Is it possible to expand on #'s 4 and 5 a bit?
Thanks
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 !
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:
> PLAPACK::
Jun 12, 2007 Look how pretty Parallel Linear Algebra code can be! FLAME: Formal Linear Algebra Methods Environment · ITXGEMM: The high-performance DGEMM
http://www.cs.utexas.edu/~plapack/HOME |
> 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.
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Linear Algebra