MATH 307 Test 1 Study Guide

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1 MATH 37 Test 1 Study Guide Theoretical Portion: No calculators Note: It is essential for you to distinguish between an entire matrix C = (c i j ) and a single element c i j of the matrix. For example, consider the following lines: n Let A B = C = (c i j ), where c i j = a i k b k j. Then (AB) T = C T = ( c i j ) is p m, k =1 n where c i j = c ji = a jk b k i. Here (c i j ) is the entire matrix, and c i j is a single term. k =1 1. Let A = (a i j ) be m n and let B = (b i j ) be n p. (a) Explain why AB is defined and give the formal definition of the matrix product A B = C. (b) Give formal definitions of the matrices A T, B T, (A B) T, and give their dimensions. (c) Prove that B T A T = (AB) T. (See More Theorems, Theorem 1.11) 2. (a) Give the definition of trace of an n n matrix A = (a i j ). (b) Use the definition to prove that trace is a linear operator on the set of n n matrices (i.e., prove Tr(c A) = c Tr(A) for all scalars c, and Tr(A + B) = Tr(A) + Tr(B) ). (See More Theorems, Theorem 1.9) 3. Prove the following results about inverses of n n matrices: (a) If A is invertible, then A 1 is unique. (b) If A and B are invertible, then A B is invertible and (AB) 1 = B 1 A 1. (c) If A is invertible, then A T is invertible and (A T ) 1 = (A 1 ) T. (d) If A 1 exists, then det (A) and det (A 1 1 ) = det (A). (See Inverses file, Theorem 1.5 for (a), (b), (d), See More Theorems, Theorem 1.12 for (c)) 4. The characterization of n n matrices: Know five statements that are equivalent to det (A) for an n n matrix A (and those for det (A) = ). (See n n Part 1.) Computational Part: Calculator Allowed 5. Given a system of n equations and n unknowns, write the matrix of coefficients A and compute det (A). If det (A), then find A 1 (exact fraction values) and solve a given system A X = B by X = A 1 B or by using the rref form of the augmented matrix A B. If det (A) =, then use the reduced row echelon form of the augmented matrix A B to determine if A X = B has infinitely many solutions or no solutions. If there are infinitely many solutions, then give the general form of all solutions.

2 Example Problems for 5: For each, (a) Give the matrix of coefficients A and the matrix of constants B. (b) Give the exact values of det(a), A 1, and det(a 1 ). (c) Give the exact solution to the system AX = B. (i) 2x y + 2z 3w = 4 x + y w = 2 x 2y + 3z = x 2y + z 4w = 4 x 1 x 2 + 2x 3 x 4 = 1 2x 1 + x 2 2x 3 2x 4 = x 1 + 2x 2 4x 3 + x 4 = 1 3x 1 3x 4 = x 1 + x 2 3x 3 = x 1 + 8x 2 + 3x 3 = x 1 + 2x 2 + 3x 3 = 8 (iv) x + y + z = 2x + 3y + 4z = 3x + 4y + 5z =. Given the rref form of a system of equations, write the general form of all solutions and give a specific solution. (i) Analyze a Markov Chain with Fixed Transitions. In particular, determine the limiting state and prove the limiting state does not depend on the initial state. Example. At one large firm, 5% use a Mac desktop (M), 7% use a PC desktop (P), and 25% do not use a desktop computer (N). Each quarter, 1% of Mac users switch to PC, 9% stop using a desktop, and the rest continue using a Mac. Each quarter, 3% of PC users switch to Mac, 12% of them stop using a desktop, and the rest continue using a PC. Each quarter, 2% of non-desktop users start using a Mac, % of them start using a PC, and the rest continue without a desktop. (a) Give the initial state matrix I and the matrix of transition probabilities T. (b) Assuming fixed transitions, give the percentages of each type of computer user in 5 quarters, and the percentages in the far future. (c) Prove that the far-future percentages of each type of computer user do not depend on the initial state.

3 Solutions to Example Problems for Test 1 Review (i) A = B = 4 det (A) = 32, so A 1 exists. det(a 1 ) = 1 det(a) = / 2 1 / 2 1 / 4 1 / 4 5 / x = 5 / 2 A 1 11/ 1 23 / 1 13 / 32 5 / 32 = X = A 1 1 / 1 y = 1 / 1 B = > 5 / 8 9 / 8 11 / 1 3 / 1 19 / 8 z = 19 / 8 / 1 1 / 1 5 / 32 / / 1 w = 13 / A = and B =. Then det (A) = ; so A 1 does not exist and A X = B has either no solutions or infinitely many solutions. From the reduced row echelon form of A B, we see that there are infinitely many solutions. rref (A B) = Let x 3 = s and x 4 = t. Then 2nd eq implies x 2 = 2 x 3 = 2s and 1st equation implies x 1 = 1 + x 4 = 1 + t. General solution: x 1 = 1+ t, x 2 = 2s, x 3 = s, x 4 = t, where s, t can be any real numbers. A Specific Solution: x 1 = 1, x 2 =, x 3 =, x 4 = 1 A = and B =. Because det (A) =, A 1 does not exist and A X = B has either no solutions or infinitely many solutions. rref (A B) = ; the last row implies that there are no solutions to the system (i.e., it is 1 inconsistent).

4 (iv) det (A) = ; so the homogeneous system has infinitely many solutions. rref (A ) = General Solution: x 1 = t, x 2 = t, x 3 = t where t can be any real number (i) From the leading 1's, we see that the variables x 1, x 2, and x 4 are dependent. The others are independent. (So let x 3 = r x 5 = s x = t x 7 = u) General Solution: x 1 = 2 r 5s + u x 2 = x 3 = r x 4 = + 2s 4u x 5 = s x = t x 7 = u, where r, s, t, u can be any real numbers. One Specific Solution: (2,,, 3,,, ) (from letting all ind. variables be ) The last row is inconsistent. There are no solutions From the leading 1's, we see that the variables x 1, x 3, x 4, and x 5 are dependent. The others are independent. So let x 2 = a x = b x 7 = c General Solution: x 1 = 8 5b + 2c x 2 = a x 3 = x 4 = 4 3b 7c x 5 = + 2b 4c x = b x 7 = c, where a, b, c can be any real numbers. One Specific Solution: ( 8,, 2, 4,,, ) (from letting all ind. variables be ) 7. (a) I = ( ) M P N M P = T N.2..92

5 (b) I T 5 ( ) (which are the proportions of Mac, PC, and nondesktop users, respectively, after 5 quarters) So about 12.% use Mac, 39.8% use a PC, and 47.% do not use a desktop. For large k, I T k ( ) So in the far future, about 18.7% will use a Mac, 24.1% will use a PC, and 57.2% will not use a desktop. (c) For large k, T k = c 1 c 2 c 3 c 1 c 2 c 3 c 1 c 2 c 3 where c c c Then for any initial state I = (p 1 p 2 p 3 ) with p 1 + p 2 + p 3 = 1, and large k, we have I T k ((p 1 + p 2 + p 3 )c 1 (p 1 + p 2 + p 3 )c 2 ( p 1 + p 2 + p 3 )c 3 ) = (c 1 c 2 c 3 ). So regardless of the initial state I, the far-future proportions are (c 1 c 2 c 3 ) ( ).