practiceFinal

# practiceFinal - Math 310 Final Exam(Solutions Prof S Smith...

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Unformatted text preview: Math 310: Final Exam (Solutions) Prof. S. Smith: Wed 6 Dec 1995 Problem 1: (a) Find the eigenvalues of A = OMH OHM 0 0 3 We see det(A—:rI) : (3—:r)[(1—:r)2—2-2] : (3—50 (x2—2r—3) : (3—Jc)(x—3)(Jc+1). So eigenvalues —1, 3, 3. (b) Find the eigenspaces for each eigenvalue. Is A diagonalizable? (Why/Why not?) For —1 get (—1,1,0)T; for 3, span of (1, 1, 0)T and (0,0,1)T. So YES diagonalizable (in particular, 2 lin.indep. eigenvectors for (0) Note A is symmetric—is it positive deﬁnite? Give two methods of deciding. No. From above, not all eigenvalues are positive. Alternatively, the three “upper left” determinants are 1, —3, —9, not all positive. I _ Problem 2: (a) Give the general solution of the differential system y} _ yl + m 1/2 = —2y1 + 43/2 Find eigenvalues 2 and 3, and corresponding eigenvectors (1,1)T and (1, 2)T. 1 l (:1 ) = c1< 1 >e2t + Cg ( 2 > 6:” , so y1 = 016% + 0263” and 3/2 = 616% + 2026:”. 2 (b) Give the particular solution when y1(0) : 3 and 112(0) : 1. 1 2 1 Then Chapter 1 methods give solution 61 = 5, 02 = —2,' so y1 = 56% — 26375, y2 2 5e” — 46375. (a) Find the “steady state77 eigenvector for M. (Components of vector should add to 1). —2 2 - _ 1 For eigenvalue 1, M — 1] _ §( 2 _2 >, so eigenvector is span of (1, 1)T. Hence (g, is steady-state vector. (b) Diagonalize M and use this to get an expression for the power M ”. Compute det(M — 3:1) : (a: — g)? _ g2 2 \$2 _ g3; _ g 2 (a; _ 1W; + a 22 22 Putting in t = 0 gives system With augmented matrix ( 1 1 ‘ 3 ) , Problem 3: Let M be the Markov matrix ( 03m:le bah—teal» So other eigenvalue is —%; from M + %I : % < > get eigenvector (1, —1)T. UsingS 2(1 _i ), get S’lMS = A so M 2 SAS’1 hence M“ = S(A")S’1. 3—i><1:(_;n>%(1_1> ThusMn=< M1 Jamil 133:) :(i—i:> Problem 4: Let A : . I give you that the eigenvalues are 3, 0, 0. +—|+—|>—\ I—|I—|>—\ r—tr—I>—\ (a) Find eigenvectors for these eigenvalues. l 0 —1 A — 3I row-reduces to 0 l —l , so a 3—eigenvector is (1,1,1)T. 0 0 0 l 1 l A — OI = A row-reduces to 0 0 0 , 0 0 0 so independent O-eigenvectors are (1,—1,0)T and (1,07 —1)T. (b) Note A is symmetric. So ﬁnd orthogonal S with S‘lAS diagonal. (Remember this means the columns of 3 must be orthonormal). We need to apply Gram—Schmidt to each eigenspace. For 3, just divide by length to get i(1,1,1)T. \/§ For 0, ﬁrst is (1, —1,0)T and second is (column form of) (1307—1) — (1:_1JO)W(17_170)(170:_1)T = (1,0,—1)— §(1,—1,0) = §(1,1,—2). Divide by lengths to get %(1, —l,0)T and %(1,1,—2)T. So use 5 : | OEIHHH seals swash ...
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practiceFinal - Math 310 Final Exam(Solutions Prof S Smith...

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