View Question
 Question
 Subject: Linear Algebra Category: Science > Math Asked by: bildy-ga List Price: \$35.00 Posted: 05 Dec 2002 14:43 PST Expires: 04 Jan 2003 14:43 PST Question ID: 119961
 ```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.```
 bildy-ga rated this answer: ```Went above and beyond what was necessary ! Thanks ! Would love to work with you again !```