Linear Algebra Quiz: Modeling With Linear Systems
7 questions · exam conditions
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Modeling With Linear SystemsQuestion 1 of 7

A water treatment facility has three treatment stages connected in series with recycling loops. Fresh water enters at 1000 L/min, and treated water exits at 800 L/min. Stage 2 recycles 150 L/min back to stage 1, and stage 3 recycles 100 L/min back to stage 2. If f1,f2,f3f_1, f_2, f_3 represent the total flow rates through stages 1, 2, and 3 respectively, what is f1f_1?

950 L/min
1050 L/min
1100 L/min
1150 L/min
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Linear Algebra Quiz

Linear Algebra Quiz: Modeling With Linear Systems

Practice Modeling With Linear Systems in Linear Algebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Modeling With Linear Systems, giving you a quick way to practice the rules, question types, and explanations that matter most for Linear Algebra.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A water treatment facility has three treatment stages connected in series with recycling loops. Fresh water enters at 1000 L/min, and treated water exits at 800 L/min. Stage 2 recycles 150 L/min back to stage 1, and stage 3 recycles 100 L/min back to stage 2. If f1,f2,f3f_1, f_2, f_3 represent the total flow rates through stages 1, 2, and 3 respectively, what is f1f_1?

  1. 950 L/min
  2. 1050 L/min
  3. 1100 L/min
  4. 1150 L/min (correct answer)
Explanation: Working backwards from stage 3: f3=800+100=900f_3 = 800 + 100 = 900 L/min (output plus recycled amount). For stage 2: f2=f3+150=900+150=1050f_2 = f_3 + 150 = 900 + 150 = 1050 L/min (flow to stage 3 plus recycling to stage 1). For stage 1: f1=1000+150=1150f_1 = 1000 + 150 = 1150 L/min (fresh water input plus recycled flow from stage 2). Choice A incorrectly subtracts the recycle flow. Choice B gives the flow through stage 2. Choice C results from adding only half the recycle flows.

Question 2

A traffic network has four intersections connected by one-way streets. Traffic flows are measured in vehicles per hour. At intersection 2, the conservation equation is x1x3+x4=50x_1 - x_3 + x_4 = 50, where x1,x3,x4x_1, x_3, x_4 represent flows on streets 1, 3, and 4 respectively, and 50 represents net traffic generation at that intersection. If street 1 carries 200 vehicles/hour toward intersection 2, and street 4 carries 120 vehicles/hour away from intersection 2, what is the flow on street 3?

  1. 30 vehicles/hour toward intersection 2
  2. 30 vehicles/hour away from intersection 2
  3. 270 vehicles/hour toward intersection 2 (correct answer)
  4. 270 vehicles/hour away from intersection 2
Explanation: Using the conservation equation x1x3+x4=50x_1 - x_3 + x_4 = 50 with x1=200x_1 = 200 (toward intersection 2) and x4=120x_4 = 120 (away from intersection 2, so positive in the equation): 200x3+120=50200 - x_3 + 120 = 50, which gives 320x3=50320 - x_3 = 50, so x3=270x_3 = 270. Since x3x_3 appears with a negative sign in the equation, positive x3x_3 represents flow toward the intersection. Choice A uses incorrect arithmetic (200+12050=270200 + 120 - 50 = 270, not 30). Choice B has the wrong magnitude and direction. Choice D has correct magnitude but wrong direction interpretation.

Question 3

A supply chain network distributes products from 2 warehouses to 3 retail stores. The cost matrix CC has entry cijc_{ij} representing the cost per unit to ship from warehouse ii to store jj. If warehouse 1 ships quantities x11,x12,x13x_{11}, x_{12}, x_{13} to stores 1, 2, 3 respectively, and warehouse 2 ships quantities x21,x22,x23x_{21}, x_{22}, x_{23}, which constraint ensures that store 2 receives exactly its demand d2d_2?

  1. x12+x22=d2x_{12} + x_{22} = d_2 (correct answer)
  2. x21+x22+x23=d2x_{21} + x_{22} + x_{23} = d_2
  3. c12x12+c22x22=d2c_{12}x_{12} + c_{22}x_{22} = d_2
  4. x11+x12+x13=d2x_{11} + x_{12} + x_{13} = d_2
Explanation: Store 2 receives shipments from both warehouses: x12x_{12} units from warehouse 1 and x22x_{22} units from warehouse 2. The total supply to store 2 must equal its demand, giving x12+x22=d2x_{12} + x_{22} = d_2. Choice B represents the total shipment from warehouse 2 to all stores, not the total received by store 2. Choice C incorrectly includes costs rather than quantities. Choice D represents total shipment from warehouse 1 to all stores.

Question 4

A metabolic pathway involves three enzymes (E1, E2, E3) that process substrates at rates v1,v2,v3v_1, v_2, v_3 respectively. At steady state, the concentration of each intermediate remains constant. The pathway can be represented as: Sv1Av2Bv3PS \xrightarrow{v_1} A \xrightarrow{v_2} B \xrightarrow{v_3} P, where enzyme E2 also has a side reaction that removes intermediate A at rate 0.2v20.2v_2. What is the correct steady-state balance equation for intermediate A?

  1. v1=v2v_1 = v_2
  2. v1=v2+0.2v2v_1 = v_2 + 0.2v_2
  3. v1=1.2v2v_1 = 1.2v_2 (correct answer)
  4. v1+0.2v2=v2v_1 + 0.2v_2 = v_2
Explanation: At steady state, the rate of production of intermediate A must equal its rate of consumption. Intermediate A is produced at rate v1v_1 and consumed at rate v2v_2 (main pathway) plus 0.2v20.2v_2 (side reaction). Therefore: v1=v2+0.2v2=1.2v2v_1 = v_2 + 0.2v_2 = 1.2v_2. Choice A ignores the side reaction. Choice B shows the correct relationship but isn't simplified. Choice D incorrectly places the side reaction term on the production side rather than consumption side.

Question 5

A logistics network ships products between 4 cities using a hub-and-spoke model with city 1 as the hub. Direct shipments between non-hub cities are not allowed. The network is represented by a flow matrix where positive entries indicate outbound flows and negative entries indicate inbound flows. At city 2 (a non-hub city), the flow balance equation is x12+x21=d2-x_{12} + x_{21} = d_2, where d2=50d_2 = -50 represents net outbound demand. If x21=200x_{21} = 200 units/day, what does this imply about the logistics operation at city 2?

  1. City 2 receives 200 units from hub and ships 150 units to hub
  2. City 2 receives 250 units from hub and ships 200 units to hub (correct answer)
  3. City 2 receives 200 units from hub and ships 250 units to hub
  4. City 2 receives 150 units from hub and ships 200 units to hub
Explanation: From the balance equation x12+x21=d2-x_{12} + x_{21} = d_2 with x21=200x_{21} = 200 and d2=50d_2 = -50: x12+200=50-x_{12} + 200 = -50, so x12=250x_{12} = 250. Here, x12x_{12} represents flow from hub (city 1) to city 2 (inbound to city 2), and x21x_{21} represents flow from city 2 to hub (outbound from city 2). Therefore, city 2 receives 250 units from the hub and ships 200 units back to the hub. The negative demand indicates net outbound flow from city 2. Choice A uses incorrect calculation. Choice C reverses the flow interpretation. Choice D uses the wrong value for inbound flow.

Question 6

A manufacturing company produces three types of electronic components (A, B, C) using four different machines (M1, M2, M3, M4). Each machine can produce multiple component types but with different efficiency rates. The production requirements are 500 units of A, 300 units of B, and 200 units of C per day. The company wants to minimize total machine hours while meeting all production demands.

If the efficiency matrix EE has entry eije_{ij} representing units of component ii produced per hour on machine jj, and h=[h1,h2,h3,h4]T\mathbf{h} = [h_1, h_2, h_3, h_4]^T represents hours each machine operates, which matrix equation correctly models this optimization problem's constraints?

  1. Eh=[500,300,200]TE\mathbf{h} = [500, 300, 200]^T (correct answer)
  2. ETh=[500,300,200]TE^T\mathbf{h} = [500, 300, 200]^T
  3. hTE=[500,300,200]\mathbf{h}^TE = [500, 300, 200]
  4. hE=[500,300,200]T\mathbf{h}E = [500, 300, 200]^T
Explanation: The efficiency matrix EE is 3×4 (3 components, 4 machines) where eije_{ij} is units of component ii per hour on machine jj. When we multiply EE by the hours vector h\mathbf{h} (4×1), we get total production of each component: (Eh)i=j=14eijhj(E\mathbf{h})_i = \sum_{j=1}^4 e_{ij}h_j = total units of component ii produced. This must equal demand [500,300,200]T[500, 300, 200]^T. Choice B uses ETE^T which would have wrong dimensions (4×3 matrix times 4×1 vector is invalid). Choice C creates a row vector output instead of column vector. Choice D has dimension mismatch (1×4 times 3×4 is invalid).

Question 7

An economic input-output model for a three-sector economy (agriculture, manufacturing, services) uses the equation (IA)x=d(I - A)\mathbf{x} = \mathbf{d}, where AA is the technology matrix, x\mathbf{x} is total output, and d\mathbf{d} is final demand. If a21=0.3a_{21} = 0.3 and the manufacturing sector increases its output by 1000 units, how much additional input does manufacturing require from agriculture?

  1. 300 units (correct answer)
  2. 700 units
  3. 1000 units
  4. 1300 units
Explanation: In the technology matrix AA, entry a21=0.3a_{21} = 0.3 represents the amount of agriculture output needed per unit of manufacturing output. If manufacturing increases output by 1000 units, the additional agricultural input required is a21×1000=0.3×1000=300a_{21} \times 1000 = 0.3 \times 1000 = 300 units. Choice B incorrectly uses (1a21)×1000(1 - a_{21}) \times 1000. Choice C assumes a one-to-one relationship. Choice D incorrectly adds the coefficient to the output change.