Problems for M 8/31: and put it into echelon form to see whether there are any solutions.

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1 Math 310, Lesieutre Problem set # September 9, 015 Problems for M 8/31: 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 = We need to form the matrix and put it into echelon form to see whether there are any solutions 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 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 h 0 + h 1 h 1 h h 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

2 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 Compute the product Ax using (a) the definition and (b) the row-vector method. If the product is undefined, say why The matrix A has columns, whereas the vector x has three entries. There is a size mismatch: we can t multiple these two things Compute the product Ax using (a) the definition and (b) the row-vector method. If the product is undefined, say why. 6 5 [ ] 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 [ ] = () 4 + () = = (b) Using the other method, [ ] = ()(6) + ()(5) ()( 4) + ()() ()(7) + ()(6) We get the same answer either way, whcih is a relief. = 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 =

3 As a vector equation, this is: x x 4 = 1 1 In matrix form, the same system is: 8 1 [ ] x1 = 1. x 1 The book doesn t ask us to solve it, so I won t. Problems for W 9/: 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 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

4 1.5.7 Describe all solutions of Ax = 0 in parametric vector form, where A is row equivalent to the given matrix. [ ] A = 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: [ ] [ ] 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 x x 3 = s t 5 0 x Same as above, but with A = [ ] First, rref: [ ] [ ] [ 1 ] 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

5 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 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 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 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: 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. ] 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 [ ].

6 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 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), / / 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 H O H 3 BO H S Balance the unbalanced chemical equation: Na 3 PO 4 + Ba(NO 3 ) Ba 3 (PO 4 ) + NaNO 3.

7 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 / / / 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 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 = 80

8 In matrix form, this is the equation: Putting this in rref, we obtain 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.