Solving simultaneous equations using the inverse matrix
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1 Solving simultaneous equations using the inverse matrix 8.2 Introduction The power of matrix algebra is seen in the representation of a system of simultaneous linear equations as a matrix equation. Matrix algebra allows us to write the solution of the system using the inverse matrix of the coefficients. In practice the method is suitable only for small systems. Its main use is the theoretical insight into such problems which it provides. Prerequisites Before starting this Block you should... Learning Outcomes After completing this Block you should be able to... use the inverse matrix of coefficients to solve a system of two linear simultaneous equations use the inverse matrix of coefficients to solve a system of three linear simultaneous equations Recognise and identify cases where there is no unique solution be familiar with the basic rules of matrix algebra be able to evaluate 2 2and 3 3 determinants be able to find the inverse of 2 2and 3 3 matrices Learning Style To achieve what is expected of you... allocate sufficient study time briefly revise the prerequisite material attempt every guided exercise and most of the other exercises
2 1. Using the inverse matrix on a system of two equations If we have one linear equation ax = b in which the unknown is x and a and b are constants then there are just three possibilities a 0 then x = b a a 1 b. The equation ax = b has a unique solution for x. a =0,b = 0 then the equation ax = b becomes 0 = 0 and any value of x will do. There are infinitely many solutions to the equation ax = b. a = 0 and b 0 then ax = b becomes 0 = b which is a contradiction. In this case the equation ax = b has no solution for x. What happens if we have more than one equation and more than one unknown? In this block we copy the algebraic solution x = a 1 b used for a single equation to solve a system of linear equations. As we shall see, this will be a very natural way of solving the system if it is first written in matrix form. Consider the system 2x 1 +3x 2 = 5 x 1 2x 2 = 1. In matrix form this becomes [ ][ x1 x 2 ] [ = 5 1 ] which is of the form AX = B. If A 1 exists then the solution is X = A 1 B. (compare the solution x = a 1 b above) Try each part of this exercise Part (a) Given the matrix A = A 1? [ ] find its determinant. What does this tell you about Part (b) Now find A 1 To solve the system AX = B we use X = A 1 B Now do this exercise [ 2 3 Solve the system AX = B where A = 1 2 [ ] [ ] [ ] (i) (ii) (iii) ] and B is Engineering Mathematics: Open Learning Unit Level 1 2
3 2. Non-unique solutions The key to obtaining a unique solution of the system AX = B is to find A 1. What happens when A 1 does not exist? Consider the system 2x 1 +3x 2 = 5 (i) 4x 2 +6x 2 = 10. (ii) In matrix form this becomes [ ][ x1 x 2 ] = [ 5 10 ]. Now do this exercise Identify the matrix A and hence find A 1. Looking at the original system we see that equation (ii) is simply equation (i) with all coefficients doubled. In effect we have only one equation for the two unknowns x 1 and x 2. The equations are consistent but have infinitely many solutions. If we let x 2 assume a particular value, t say, then x 1 must take the value x 1 = 1 (5 3t) i.e. 2 the solution to the given equations is: x 2 = t, x 1 = 1 (5 3t) 2 For each value of t there are values for x 1 and x 2 which satisfy the original system of equations. For example, if t = 1, then x 2 =1,x 1 =1,ift = 3 then x 2 = 3, x 1 = 7 and so on. Now consider the system 2x 1 +3x 2 = 5 (i) 4x 1 +6x 2 = 9 (ii) Since the left-hand sides are the same as those in the previous system then A is the same and again A 1 does not exist. There is no unique solution to the equations (i) and (ii). However, if we double equation (i) we obtain 4x 1 +6x 2 =10, which conflicts with equation (ii). There are thus no solutions to (i) and (ii) and the equations are said to be inconsistent. Try each part of this exercise What can you conclude about the solutions of the systems (i) x 1 2x 2 = 1 3x 1 6x 2 = 3 (ii) 3x 1 + 2x 2 = 7 6x 1 4x 2 = 5 Part (a) First write the systems in matrix form and find A. 3 Engineering Mathematics: Open Learning Unit Level 1
4 Part (b) Now compare the equations in each system. 3. Three equations in three unknowns It is much more tedious to use the inverse matrix to solve a system of three equations although in principle, the method is the same as for two equations. Consider the system x 1 2x 2 + x 3 = 3 2x 1 + x 2 x 3 = 5 3x 1 x 2 +2x 3 = 12. We met this system in Block 8.1 where we found that Hence A 1 exists. Try each part of this exercise Find A 1 by the method of determinants. A =10. Part (a) First form the matrix where each element of A is replaced by its minor. Part (b) Now use the 3 3 array of signs to obtain the matrix of cofactors. Part (c) Now transpose this matrix and divide by A to obtain A 1. Part (d) Now use X = A 1 B to solve the system of linear equations. Comparing this approach to the use of Cramer s rule for three equations in section 2of Block 8.1 we can say that the two methods are both rather tedious. Engineering Mathematics: Open Learning Unit Level 1 4
5 Equations with no unique solution If A =0,A 1 does not exist and therefore, early on, it is easy to see that the system of equations has no unique solution. But it is not obvious whether this is because the equations are inconsistent or whether they have infinitely many solutions. Try each part of this exercise Consider the systems (i) x 1 x 2 + x 3 = 4 2x 1 +3x 2 2x 3 = 3 3x 1 +2x 2 x 3 = 7 (ii) x 1 x 2 + x 3 = 4 2x 1 +3x 2 2x 3 = 3 x 1 11x 2 +9x 3 = 13 Part (a) In system (i) add the first equation to the second. What does this tell you about the system? Part (b) In system (ii) take the combination 5 times the first equation minus 2times the second equation. What does this tell you about the system? In practice, systems containing three or more linear equations are best solved by the method which we shall introduce in Block 8.3. More exercises for you to try 1. Solve the following using the inverse matrix approach: (a) 2x 3y = 1 4x + 4y = 2 (b) 2x 5y = 2 4x + 10y = 1 (c) 6x y = 0 2x 4y = 1 2. Solve the following equations using matrix methods: (a) 2x 1 + x 2 x 3 = 0 x 1 + x 3 = 4 x 1 + x 2 + x 3 = 0 (b) x 1 x 2 + x 3 = 1 x 1 + x 3 = 1 x 1 + x 2 x 3 = 0 5 Engineering Mathematics: Open Learning Unit Level 1
6 4. Computer Exercise or Activity For this exercise it will be necessary for you to access the computer package DERIVE. Derive is capable of doing matrix algebra and can be used to obtain matrix inverses, and so can be used to solve systems of linear equations using the inverse matrix approach. For example to solve: x 1 2x 2 + x 3 = 3 2x 1 + x 2 x 3 = 5 3x 1 x 2 +2x 3 = 12. we would key in the coefficient matrix A and the matrix of constants B on the right-hand-sides. If these are labelled as A and B so that DERIVE responds: a := b := Now key Author:Expression a ( 1)b. DERIVE responds with a 1 b Now hit Simplify:Basic and DERIVE responds with: which are the required solutions. It will be a useful exercise to solve the problems in this block using the matrix facilities of DERIVE. Engineering Mathematics: Open Learning Unit Level 1 6
7 End of Block Engineering Mathematics: Open Learning Unit Level 1
8 A =2 ( 2) 1 3= 7 since A 0 then A 1 exists. Engineering Mathematics: Open Learning Unit Level 1 8
9 [ 2 3 A 1 = 1 ( 7) 1 2 ] = 1 7 [ ] 9 Engineering Mathematics: Open Learning Unit Level 1
10 [ ][ ] 2 3 (i) X = [ ][ ] (ii) X = [ ][ ] (iii) X = expected. = 1 7 = 1 7 = 1 7 [ ] [ ] 7 1 =. Hence x =1,x 2 =1. [ ] [ ] 14 2 =. Hence x =2,x 2 = 1. [ ] [ ] 0 0 =. Hence x =0,x 2 =0, as might have been Engineering Mathematics: Open Learning Unit Level 1 10
![Hence x 7 1 1 =1,x 2 =1. [ ] [ ] 14 2 =. Hence x 7 1 1 =2,x 2 = 1.](/docs-images/46/21588107/images/page_10.jpg "[ ] [ ] 0 0 =.")
11 [ 2 3 A = 4 6 ] and A = =0. Hence A 1 does not exist. 11 Engineering Mathematics: Open Learning Unit Level 1
12 (i) (ii) [ [ ][ x1 x 2 ][ x1 ] [ ] 1 = 3 ] [ 7 = 5 x 2 ] A = 6+6=0; A = 12+12=0. Engineering Mathematics: Open Learning Unit Level 1 12
13 (i) The second equation is 3 times the first equation. There are infinitely many solutions of the form x 2 = t, x 1 =1+2t where t is arbitrary. (ii) If we multiply the first equation by ( 2) we obtain 6x 1 4x 2 = 14 which is in conflict with the second equation. The equations are inconsistent and have no solution. 13 Engineering Mathematics: Open Learning Unit Level 1
14 = Engineering Mathematics: Open Learning Unit Level 1 14
15 The array of signs is so that we obtain Engineering Mathematics: Open Learning Unit Level 1
16 Transposing gives Finally, A 1 = Engineering Mathematics: Open Learning Unit Level 1 16
17 X = = = Then x 1 =3,x 2 =1,x 3 =2. 17 Engineering Mathematics: Open Learning Unit Level 1
18 The sum is 3x 1 +2x 2 x 3 = 7, which is identical to the third equation. Thus, essentially, there are only two equations x 1 x 2 + x 3 = 4 and 3x 1 +2x 2 x 3 = 7. Now adding these two gives 4x 1 + x 2 =11orx 2 =11 4x 1 and then x 3 =4 x 1 + x 2 =4 x x 1 =15 5x 1 Hence if we give x 1 a value, t say, then x 2 =11 4t and x 3 =15 5t. Thus there are an infinite number of solutions (one for each value of t). Engineering Mathematics: Open Learning Unit Level 1 18
19 The combination is x 1 11x 2 +9x 3 =14, which conflicts with the third equation. There are thus no solutions. 19 Engineering Mathematics: Open Learning Unit Level 1
20 1. (a) x = 1 2,y= 0 (b) A 1 does not exist. (c) x = 1 22,y= (a) x 1 = 8 3,x 2 = 4, x 3 = 4 3 (b) x 1 = 1 2,x 2 = 1 2,x 3 =1 Engineering Mathematics: Open Learning Unit Level 1 20
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