1. b = b = b = b = 5

Transcription

1 Version 001 Minterm 1 tsishchanka (54615) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. FinM4a points Panditisanagingdogwhohastobekepton a strict diet containing, among other things, 2.5gramsofproteinand1.8gramsoffat. Two dog foods are available to Pandit s owner. Food A has 8% protein and 6% fat, while Food B has 3% protein and 2% fat. Howmany gramsoffoodashould Pandit s owner use in his diet? 1. # grams food A = 20 correct 2. # grams food A = # grams food A = # grams food A = # grams food A = 23 Let x and y be the amounts (in grams) of foods A and B respectively used in Pandit s diet. Then 0.08x+0.03y = 2.5, 0.06x+0.02y = 1.8. Solving for x, y we see that so x = 20, y = 30, # grams food A = 20. FitParabola01b points When the graph of the function y = ax 2 +bx+c is the unique parabola passing through the points determine b. 1. b = 2 2. b = 6 3. b = 4 4. b = 5 (1, 7), ( 1, 1), ( 3, 3), 5. b = 3 correct Since the parabola passes through the points (1, 7), ( 1, 1), ( 3, 3), the coefficients a, b and c must satisfy the equations a+b+c =7 a b+c =1 9a 3b+c =3 To solve these equations for c we reduce the augmented matrix to echelon form by successive row operations: R 2 R 1 R 3 9R 1 R 3 6R

2 Version 001 Minterm 1 tsishchanka (54615) 2 Thus b = TRUE LinearSystemT/F01a points Elementary row operations on an augmented matrix never change the the solution set of the associated linear system. 1. TRUE correct 2. FALSE Elementary row operations on an augmented matrix (i) replace one row by the sum of itself and the multiple of another row, (ii) interchange rows, (iii) multiply all entries in a row by a nonzero constant. For the linear system associated with the original augmented matrix these respective operations (i) add a constant times one equation to another, (ii) interchange two equations, (iii) multiply an equation through by by a non-zero constant. But none of these will affect the solutions of the equations. LinSysUniqueTF points If a system of linear equations has no free variables, then it has a unique solution. 2. FALSE correct A linear system must have no solutions, unique solutions, or infinitely many solutions. Having no free variables indicates that the system does not have infinitely many solutions, but it does not determine which of the other two cases it might be. Consider the following counterexample: x 1 + x 2 = 1 x 2 = 5 x 1 + x 2 = 2 Subtracting the first row from the third row obtains the equation 0 = 1, so this is clearly not consistent and hence has no solution. You canalsoshowthesystemhasnofreevariables. Henceithasno free variablesandnosolution. FALSE. RowReduceMan02a points Theaugmentedmatrixofalinearsystemof equations has been reduced by row operations to (a) Continue row operations to write the matrix in reduced row echelon form. (b) Then determine the solution set of the original system. 1. correct (a) (b) x 1 = 1, x 2 = 1, x 3 = 2

3 Version 001 Minterm 1 tsishchanka (54615) 3 (a) Using row operations we see that The augmented matrix is now in reduced row echelon form. (b) By row reduction the original system is equivalent to the system x 1 = 1 x 2 = 1 x 3 = 2, so the solution set of the original system is x 1 = 1, x 2 = 1, x 3 = 2. AxisIntersect01a points When P is the plane in R 3 given in vector form by x = 1 2 +s 2 2 +t 3 4, determine where P intersects the z-axis. 1. z = 7 correct 2. z = 8 3. z = 9 4. z = 5 5. z = 6 Since the z-axis consists of points in R 3 with x = y = 0, P intersects the z-axis when 0 0 z = 1 2 +s 2 2 +t for some choice of s and t, i.e., when s 2 2 +t 3 4 = z 1 is consistent as a vector equation in s, t. This will be true if and only if the associated augmented matrix A = z 1 is row equivalent to an echelon matrix whose entries in the last row are all zero. Now byrowreduction inthefirst columnof A, we obtain A , 0 2 z +1 and after row reduction in the second column A z +7 Consequently, P intersects the z-axis at Given z = 7. M340LSpanM points v 1 = 1 1, v 2 = 2 4, v 3 = 1 0, 0 2 1

4 Version 001 Minterm 1 tsishchanka (54615) 4 determine all values of λ for which w = 2 1 λ is a vector in Span{v 1, v 2, v 3 }? 1. λ = 3 2. λ = 1, 3 3. λ = 1, 1 4. λ = 1 correct 5. λ = 1 6. λ = 1, 3 ThevectorwisinSpan{v 1, v 2, v 3 }ifthere exist weights x 1, x 2, x 3 such that x 1 v 1 +x 2 v 2 +x 3 v 3 = w. Such weights exist when the rightmost column in the augmented matrix [v 1 v 2 v 3 w]= λ is not a pivot column. But λ λ λ+1 Thus the rightmost column is not a pivot column when λ+1 = 0. Consequently, w lies in Span{v 1, v 2, v 3 } when λ = 1. VectorEquTF01e points If u, v are vectors in R 3, when can Span{u, v} be visualized as a plane through the origin in R ALWAYS 2. SOMETIMES correct 3. NEVER If u, v are linearly independent, then Span{u, v} can be visualized as a plane through the origin in R 3. But if v is a nonzero scalar multiple of u, or when v = 0, then Span{u, v} consists of all scalar multiples of u, and hence can be identified with a line through the origin. Consequently, Span{u, v} can SOMETIMES be visualized as a plane through the origin in R 3, but not always. Consistent01d points Describe geometrically the conditions on a vector b in R 2 under which the equation has a solution in R 2. [ ] 2 3 x = b b lies on line y +3x = 0 2. b lies on line y 3x = 0 correct 3. arbitrary b in R 2 4. any b not on line y +3x = 0 5. any b not on line y 3x = 0

5 Version 001 Minterm 1 tsishchanka (54615) 5 A matrix equation Ax = b has a solution when the last column of the associated augmented matrix [A b] is not a pivot column. Now for the given equation, [ ] 2 3 b1 [A b] = 6 9 b 2 [ ] 2 3 b b 2 3b 1 Thus the equation has a solution when b 2 3b 1 = 0, i.e., (b 1, b 2 ) satisfies the linear equation y 3x = 0. Consequently, Ax = b has a solution if and only if b lies on the line y 3x = 0. MatEquTF02b points IfthematrixequationAx = bisconsistent, thenbisintheset spanned by the columnsof A. 1. TRUE correct 2. FALSE Since Ax = [a 1 a 2... a n ] x 1 x 2. x n = x 1 a 1 +x 2 a x n a n, the equation Ax = b is consistent, i.e., has a solution, if and only if b is a linear combination of the columns of A. On the other hand, Span{a 1 a 2... a n } consists of all linear combination of the columns of A. Hence, Ax = b is consistent if and only if b is in the span of the columns of A. SolSetsLinSysTF points If the equation Ax = b has more than one solution, then so does the homogeneous equation Ax = FALSE 2. TRUE correct The homogeneous equation Ax = 0 always has the trivial solution x = 0. Now suppose p 1, p 2 are solutions of Ax = b with p 1 p 2, i.e., w = p 1 p 2 0 and But then Ap 1 = b, Ap 2 = b. Aw = Ap 1 Ap 2 = b b = 0. So x = 0 and x = w are two different solutions of Ax = 0. ThreePoints01a points Determine the linear equation of the unique plane in R 3 containing the points and P(1, 2, 1), Q( 1, 1, 0), R( 1, 4, 2).

6 Version 001 Minterm 1 tsishchanka (54615) x 4y +6z = 1 correct 2. 3x+4y 6z +1 = x 4y 6z +1 = x+4y +6z = x+4y 6z = x 4y +6z +1 = 0 A linear equation in x, y, z has the form ax+by +cz +d = 0. Its graph will contain the points P, Q, and R when a, b, c, and d satisfy the homogeneous system a+2b+z +d = 0, a b+0z +d = 0, x 4y 2z +d = 0, of 3 linear equations in 4 variables. Now the associated augmented matrix is A = , and rref(a) = So d is a free variable, say d = s, while a = 3s, b = 4s, c = 6s. Although there are infinitely many solutions, the parameter s is common to each of a, b, c, and d. Thus there is a unique plane passing through P, Q, and R, namely the graph of 3x 4y +6z = 1, obtained by cancelling the common factor s. BalChemEqt01a points When butane C 4 H 10 burns in the presence of oxygen O 2 it produces carbon dioxide CO 2 and water H 2 O, represented chemically by C 4 H 10 + O 2 CO 2 + H 2 O. If 60 molecules of water were produced in one particular reaction, how many molecules of butane were burned in that reaction? 1. # molecules = # molecules = # molecules = # molecules = # molecules = 12 correct We need to solve first for the relative numbers x 1,..., x 4 of molecules in the balanced chemical equation x 1 C 4 H 10 +x 2 O 2 x 3 CO 2 +x 4 H 2 O. Now the fundamental rule governing this reaction is that the left and right hand sides contain the same number of the respective carbon, hydrogen and oxygen atoms. Thus 4x 1 +0x 2 = x 3 +0x 4, 10x 1 +0x 2 = 0x 3 +2x 4, 0x 1 +2x 2 = 2x 3 +x 4, which as a homogeneous system can be written in augmented matrix form [A 0] = But rref([a 0]) =

7 Version 001 Minterm 1 tsishchanka (54615) 7 So x 4 is a free variable, say x 4 = s, and x 1 = 1 5 s, x 2 = s, x 3 = 4 5 s, give the respective proportions of the other molecules in the reaction with respect to x 4. Consequently, if 60 molecules of water were produced, then of butane were burned. 12 molecules form: h h h , 0 0 h+4 0 LinIndependMan01a points Find all values h for which the vectors 2 2 4, are linearly independent. 1. h 4 correct The vectors a 1 = 2 2 4, a 2 =, h, a 3 = 2 2, h arelinearlyindependentifandonlyiftheonly solutions of x 1 a 1 +x 2 a 2 +x 3 a 3 = 0 are x 1 = x 2 = x 3 = 0. To determine the solutions of this vector equation we use row operations to reduce the corresponding augmented matrix to echelon showing that the solutions of the original vector equation are x 3 (h+4) = 0, x 2 = 0, x 1 = x 3. But then if h+4 0, the only solutions are x 3 = 0, x 2 = 0, x 1 = 0, whileifh+4 = 0, x 3 isafreevariable. Consequently, the vectors are linearly independent if and only if h 4. LinIndepTF01c points Thecolumnsofany4 5matrixarelinearly dependent. 1. FALSE 2. TRUE correct Ifasetcontainsmorevectorsthanthereare entries in each vector, then the set is linearly dependent. That is, any set {v 1, v 2,..., v p }

8 Version 001 Minterm 1 tsishchanka (54615) 8 in R n is linearly dependent if p > n. Now when A = [a 1 a 2... a 5 ] isa4 5matrix,thereare5columnsandeach is a vector in R 4. Because 5 > 4, the columns must therefore be linearly dependent. ( sinθ, cosθ) θ (0, 1) θ (cosθ, sinθ) (1, 0) LinTransform01e points A transformation T : R n R m is linear if and only if T(c 1 v 1 +c 2 v 2 ) = c 1 T(v 1 )+c 2 T(v 2 ) for all vectors v 1, v 2 in R n and all scalars c 1, c FALSE 2. TRUE correct A transformation T : R n R m by definition is linear when: (i) T(u+v) = T(u)+T(v), (ii) T(cu) = ct(u), for all u, v in R n and scalars c. The property T(c 1 v 1 +c 2 v 2 ) = c 1 T(v 1 )+c 2 T(v 2 ) combines (i) and (ii) into a single condition. MatrixTrans01a points Determine the Standard Matrix for the transformation rotating the plane counterclockwise about the origin through 60. [ ] A = 1 3 [ ] 2. A = [ ] A = [ ] 4. A = [ ] A = A = 1 2 [ ] 1 3 correct 3 1 When T is rotation counter-clockwise through θ about the origin in the plane, then by right-angle trig in the graphic above, its associated Standard Matrix is [T(1, 0) T(0, 1)] = [ cosθ sinθ sinθ cosθ When θ = 60, therefore, this is the matrix [ ] ].