Math 310, Lesieutre Problem set # September 9, 015 Problems for M 8/31: 1.3.11 Determine if b is a linear combination of a 1, a, and a 3, where 1 a 1 =, 0 a = 1, 5 a 3 = 6, b = 1. 0 8 6 We need to form the matrix 1 0 5 1 6 1 0 8 6 and put it into echelon form to see whether there are any solutions. 1 0 5 1 0 5 1 0 5 1 6 1 0 1 4 3 0 1 4 3 0 8 6 0 8 6 0 0 0 0 If all we re interested in is whether there are solutions are not (and not specifically what they are), reaching echelon form is enough we don t need to go all the way to reduced echelon form. In this case we have no rows like [000 b], so there are solutions, and b is indeed a linear combination of the a i. 1.3.18 Define 1 h v 1 = 0, v = 1, y = 5. 8 For what values of h is y in the plane generated by v 1 and v? To figure this out, we ll find the reduced echelon form of the matrix, but in terms of h. Then we can see which values of h make us end up with a row [00 b]. 1 h 1 h 1 h 0 1 5 0 1 5 0 1 5 8 0 14 + h 0 + h 1 h 1 h 0 1 5 0 1 5. 0 + h 0 0 7 + h We ve again reached echelon form, which is enough to check whether or not there are solutions. In this case, if 7 + h isn t 0, we have one of the infamous [00 b] rows, which
means there are no solutions, and y is not in the span. On the other hand, if 7+h = 0, then the system is consistent. This gives us our answer: if h = 7/, then y is in the span. If h 7/, then y is not in the span. 1.4.1 Compute the product Ax using (a) the definition and (b) the row-vector method. If the product is undefined, say why. 4 3 1 6 0 1 7 The matrix A has columns, whereas the vector x has three entries. There is a size mismatch: we can t multiple these two things. 1.4.3 Compute the product Ax using (a) the definition and (b) the row-vector method. If the product is undefined, say why. 6 5 [ ] 4 7 6 This sizes here match, so this one we can do. (a) First we compute it using the definition, in terms of linear combinations of the columns. 6 5 [ ] 6 5 1 15 4 = () 4 + () = 8 + 9 = 1 7 6 7 6 14 18 4 (b) Using the other method, 6 5 4 7 6 [ ] = ()(6) + ()(5) ()( 4) + ()() ()(7) + ()(6) We get the same answer either way, whcih is a relief. = 1 4 1.4.10 Write the system first as a vector equation and then as a matrix equation. 8x 1 x = 4 5x 1 + 4x = 1 x 1 3x =
As a vector equation, this is: x 1 8 1 4 5 + x 4 = 1 1 In matrix form, the same system is: 8 1 [ ] 4 5 4 x1 = 1. x 1 The book doesn t ask us to solve it, so I won t. Problems for W 9/: 1.5.5 Write the solution set of the given homogeneous system in parametric vector form. x 1 + 3x + x 3 = 0 4x 1 9x + x 3 = 0 x 6x 3 = 0 First we need to find the solution set, using row reduction. 1 3 1 0 1 3 1 0 1 3 1 0 4 9 0 0 3 6 0 0 3 6 0 0 6 0 0 6 0 0 0 0 0 1 3 1 0 1 0 5 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 The general solution, using a parameter s for the free variable x 3, is: x 1 = 5s x = s x 3 = s In parametric vector form, this is x 1 5 x = s x 3 1
1.5.7 Describe all solutions of Ax = 0 in parametric vector form, where A is row equivalent to the given matrix. [ ] 1 3 7 A =. 0 1 4 5 We have to form the augmented matrix and put it in rref, but that s easy: just add times row to row 1 and we get: [ ] [ ] 1 3 7 0 1 0 9 8 0 0 1 4 5 0 0 1 4 5 0 The variables x 3 and x 4 are free; let s assign them parameters s and t respectively. The general solution is x 1 = 9s + 8t x = 4s 5t x 3 = s x 4 = t. In parametric vector form, we get x 1 9 8 x x 3 = s 4 1 + t 5 0 x 4 0 1 1.5.9 Same as above, but with A = [ ] 3 9 6. 1 3 First, rref: [ ] 3 9 6 0 1 3 0 [ ] 1 0 1 3 0 [ 1 ] 0 0 0 0 0 The variables x and x 3 are free. Parametrize them by s and t. x 1 = 3s t x = s = t x 3 In parametric vector form, x 1 3 x = s 1 + t 0 x 3 0 1
1.5.15 Follow the method of Example 3 to describe the solutions of the following system in parametric vector form. Also give a geometric description of the solution set and compare it to that in Exercise 1.5.5. x 1 + 3x + x 3 = 1 4x 1 9x + x 3 = 1 x 6x 3 = First step, as usual, is to find the general solution using row reduction. 1 3 1 1 1 3 1 1 1 3 1 1 4 9 1 0 3 6 3 0 3 6 3 0 6 0 6 0 0 0 0 1 3 1 1 1 0 5 0 1 1 0 1 1 0 0 0 0 0 0 0 0 So we get x 1 = + 5s x = 1 s x 3 = s, which in parametric vector form is x 1 5 x = 1 + s. x 3 0 1 This is the same thing we got way back in 1.5.5, but translated by a particular solution v p = (, 1, 0). Problems for F 9/4: 1.6.1 Suppose an economy has only two sectors, Goods and Services. Each year, Goods sells 80% of its output to Services and keeps the rest, while Services sells 70% of its output to Goods and retains the rest. Find the equilibrium prices for the annual outputs of the Goods and Services sectors that make each sector s income match its expenses. The matrix corresponding to this economy (as in the example on page 51) is [ 0. ] 0.7 0.8 0.3 Goods expenses are 0.p G + 0.7p S : this to buy 0% of its own output, and 70% of Services output. This should be equal to p G, its total income, so 0.p G + 0.7p S = p G. Likewise for Services we should have 0.8p G + 0.3p S = p S. This is [ 0.8 0.7 0 0.8 0.7 0 ].
To get rref, add the first row to the second, which makes it 0. Then p S is free, and p G = 7 8 p S. 1.6.5 Balance the unbalanced chemical equation: B S 3 + H O H 3 BO 3 + H S. Let s give variable names to the missing coefficients: x 1 B S 3 + x H O x 3 H 3 BO 3 + x 4 H S. Each of the elements B, S, H, O gives us an equation involving this variables. In this order, we have x 1 = x 3 3x 1 = x 4 x = 3x 3 + x 4 x = 3x 3 Writing down the system and running row reduction (please forgive me for not writing out all the steps this time), 0 1 0 0 1 0 0 1/3 0 3 0 0 1 0 0 0 0 1 0 0 0 0 1 /3 0 0 1 0 0 0 0 0 0 0 The variable x 4 is free, which gives a general solution x 1 = 1 x 3 4 x = x 4 x 3 = x 3 4 x 4 is free. Let s plug in 3 for x 4 to make everything integers: x 1 = 1, x = 6, x 3 =, x 4 = 3. So the balanced equation is B S 3 + 6 H O H 3 BO 3 + 3 H S. 1.6.6 Balance the unbalanced chemical equation: Na 3 PO 4 + Ba(NO 3 ) Ba 3 (PO 4 ) + NaNO 3.
This works a lot like the previous problem. The elements this time are Na, P, O, and Ba. Give names to the unknowns: x 1 Na 3 PO 4 + x Ba(NO 3 ) x 3 Ba 3 (PO 4 ) + x 4 NaNO 3. Let s write the equations straight into the matrix: Na: P: O: Ba: Rref for this matrix is 3 0 0 1 0 1 0 0 0 4 6 8 0 0 1 0 0 1 0 0 1/3 0 0 1 0 1/ 0 0 0 1 1/6 0 0 0 0 0 0 This means that the general solution, in terms of the free variable x 4 is x 1 = 1 x 3 4 x = 1x 4 x 3 = 1x 6 4 x 4 is free. Plugging in 6 for x 4, we obtain the balanced equation:. Na 3 PO 4 + 3Ba(NO 3 ) Ba 3 (PO 4 ) + 6NaNO 3. 1.6.11 Find the general flow pattern of the network shown in the figure. Assuming that the flows are all nonnegative, what is the largest possible value for x 3? Correction: the branch going out of C and labeled 80 should instead go in to C. We get equations from each of the three nodes, plus one equation for the total: A : x 1 + x 3 = 0 B : x = x 3 + x 4 C : x 1 + x = 80 T : x 4 + 0 = 80
In matrix form, this is the equation: 1 0 1 0 0 0 1 1 1 0 1 1 0 0 80. 0 0 0 1 60 Putting this in rref, we obtain 1 0 1 0 0 0 1 1 0 60 0 0 0 1 60. 0 0 0 0 0 The variable x 3 is free, and the general solution is x 1 = 0 x 3 x = 60 + x 3 x 3 is free x 4 = 60. The maximum possible value of x 3 is 0: if x 3 were any larger than this, then x 1 would be negative.