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0.11 Cramer's rule  (Page 2/2)

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x 2 = det a 1,1 b 1 a 2,1 b 2 Δ size 12{x rSub { size 8{2} } = { {"det" left [ matrix { a rSub { size 8{1,1} } {} # b rSub { size 8{1} } {} ##a rSub { size 8{2,1} } {} # b rSub { size 8{2} } {} } right ]} over {Δ} } } {}

In the following section, we will outline a procedure that can be used to solve simultaneous linear equations based upon determinants which is called solution via Cramer’s Rule.

Solution via cramer’s rule

It is important that one begin by writing the set of simultaneous equations in normal form. That is, the equations should be written as

a 11 x 1 + a 12 x 2 = b 1 size 12{a rSub { size 8{"11"} } x rSub { size 8{1} } +a rSub { size 8{"12"} } x rSub { size 8{2} } =b rSub { size 8{1} } } {}
a 21 x 1 + a 22 x x = b 2 size 12{a rSub { size 8{"21"} } x rSub { size 8{1} } +a rSub { size 8{"22"} } x rSub { size 8{x} } =b rSub { size 8{2} } } {}

Next, we form the coefficient matrix

A = a 1,1 a 1,2 a 2,1 a 2, s size 12{A= left [ matrix { a rSub { size 8{1,1} } {} # a rSub { size 8{1,2} } {} ##a rSub { size 8{2,1} } {} # a rSub { size 8{2,s} } {} } right ]} {}

At this point, we can solve for the value of Δ by taking the determinant of A .

In anticipation of solving for the unknown x 1 , we replace the first column of A with the elements contained in the column vector

B = b 1 b 2 size 12{B= left [ matrix { b rSub { size 8{1} } {} ##b rSub { size 8{2} } } right ]} {}

Once this is accomplished we can express the solution for x 1 as the ratio

x 1 = det b 1 a 1,2 b 2 a 2,2 Δ size 12{x rSub { size 8{1} } = { {"det" left [ matrix { b rSub { size 8{1} } {} # a rSub { size 8{1,2} } {} ##b rSub { size 8{2} } {} # a rSub { size 8{2,2} } {} } right ]} over {Δ} } } {}

To obtain the solution for the unknown x 2 , we return to the original coefficient matrix A . This time, we replace the second column of A with the column vector B . Now, we can solve for x 2 as a ratio

x 2 = det a 1,1 b 1 a 2,1 b 2 Δ size 12{x rSub { size 8{2} } = { {"det" left [ matrix { a rSub { size 8{1,1} } {} # b rSub { size 8{1} } {} ##a rSub { size 8{2,1} } {} # b rSub { size 8{2} } {} } right ]} over {Δ} } } {}

This constitutes the procedure for solving a system of two linear equations in two unknowns via Cramer’s Rule.

Example: mesh current analysis

Mesh current analysis is one of the techniques that are often employed to analyze an electric circuit that contains more than one mesh or loop. Figure 1 provides an example of an electric circuit containing two meshs.

Electric circuit with two independent mesh currents.

The mesh currents are identified as I 1 and I 2 . The set of equations that govern the behavior of the circuit in terms of the mesh currents are

( R 1 + R 3 ) I 1 + R 3 I 1 = V 1 size 12{ \( R rSub { size 8{1} } +R rSub { size 8{3} } \) `I rSub { size 8{1} } +R rSub { size 8{3} } `I rSub { size 8{1} } =V rSub { size 8{1} } } {}
R 3 I 1 + ( R 2 + R 3 ) = V 2 size 12{R rSub { size 8{3} } `I rSub { size 8{1} } + \( R rSub { size 8{2} } +R rSub { size 8{3} } \) =V rSub { size 8{2} } } {}

Suppose that the values for R 1 , R 2 and R 3 are 2 Ω, 3Ω, and 1 Ω respectively. Also, suppose that values for V 1 and V 2 are 6 V and 9 V. With these values defined, the set of equations can be written as

3 I 1 + I 2 = 6 size 12{3`I rSub { size 8{1} } +I rSub { size 8{2} } =6} {}

and

I 1 + 4 I 2 = 9 size 12{I rSub { size 8{1} } +4`I rSub { size 8{2} } =9} {}

We can use Cramer’s Rule to find the mesh currents. We begin by finding the value for Δ

Δ = det 3 1 1 4 = 12 − 1 = 11 size 12{Δ="det" left [ matrix { 3 {} # 1 {} ##1 {} # 4{} } right ]="12" - 1="11"} {}

Next, we find the value for I 1 .

I 1 = det 6 1 9 4 Δ = 24 − 9 11 = 1 . 364 A size 12{I rSub { size 8{1} } = { {"det" left [ matrix { 6 {} # 1 {} ##9 {} # 4{} } right ]} over {Δ} } = { {"24" - 9} over {"11"} } =1 "." "364"`A} {}

By a similar approach, we solve for I 2 .

I 2 = det 3 6 1 9 Δ = 27 − 6 11 = 1 . 909 A size 12{I rSub { size 8{2} } = { {"det" left [ matrix { 3 {} # 6 {} ##1 {} # 9{} } right ]} over {Δ} } = { {"27" - 6} over {"11"} } =1 "." "909"`A} {}

Summary

This module has presented Cramer’s Rule as a technique for solving simultaneous linear equations. The discussion in this module was limited to systems involving two simultaneous equations. This limitation was deliberate in that Cramer’s Rule is typically not applied for linear systems comprised of large numbers of equations. An application involving the mesh analysis of an electric circuit was provided.

Exercises

  1. Consider the two mesh circuit depicted in Figure 1. Assume the following values for the resistors in the circuit: R 1 = 10 Ω , R 2 = 20 Ω , R 3 = 50 Ω size 12{R rSub { size 8{1} } ="10"` %OMEGA ,`R rSub { size 8{2} } ="20"` %OMEGA ,`R rSub { size 8{3} } ="50"` %OMEGA } {} . Let V 1 = 24 V size 12{V rSub { size 8{1} } ="24"`V} {} and V 2 = 6 V size 12{V rSub { size 8{2} } =6`V} {} . Find the two mesh currents through the use of Cramer’s Rule.
  2. A civil engineering firm plans to sign a contract with a customer. The contract calls for the construction of two office buildings which are denoted as Building X and Building Y. According to estimates derived in the preliminary design phase, the firm knows that the total cost of the project will be $50.000,000. It is also known that Building X will cost $5,000,000 more to construct than Building Y . Use Cramer’s Rule to find the cost of each building.
  3. Two types of pumps provide input into a municipal reservoir. Let us refer to the two types of pumps as A and B . If 4 type A and 2 type B pumps operate at maximum flow, the input to the reservoir is 1,200 gallons/min. If 3 type A and 5 type B pumps operate at maximum flow, the input to the reservoir is 1,600 gallons/min. Find the flow rates for type A and type B pumps using Cramer’s Rule.
  4. The combined cost of 12 microprocessors and 36 random access memory chips is $7,200. The combined cost of 8 microprocessors and 42 random access memory chips is $6,600. Find the cost of each microprocessor chip and each random access memory chip using Cramer’s Rule.
  5. The design of an electronic thermometer is based in part upon the incorporation of a component known as a thermistor. A thermistor has the property that its resistance varies linearly as a function of temperature. This linear relationship is R = R 0 + m T size 12{R=R rSub { size 8{0} } +m`T} {} . The term ( R 0 ) represents the value of the resistance at 0 0 C. At a temperature T = 25 0 C, the resistance of the thermistor ( R ) is 100 Ω. A a temperature of 55 0 C, the resistance of the thermistor is 104 Ω. Use Cramer’s Rule to find the values for R 0 and m .

Questions & Answers

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Source:  OpenStax, Math 1508 (laboratory) engineering applications of precalculus. OpenStax CNX. Aug 24, 2011 Download for free at http://cnx.org/content/col11337/1.3
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