JUST THE MATHS SLIDES NUMBER 7.4 DETERMINANTS 4 (Homogeneous - PDF document

“JUST THE MATHS” SLIDES NUMBER 7.4 DETERMINANTS 4 (Homogeneous linear equations) by A.J.Hobson 7.4.1 Trivial and non-trivial solutions

UNIT 7.4 - DETERMINANTS 4 HOMOGENEOUS LINEAR EQUATIONS 7.4.1 TRIVIAL AND NON-TRIVIAL SOLUTIONS Consider three “homogeneous” linear equations of the form a 1 x + b 1 y + c 1 z = 0 , a 2 x + b 2 y + c 2 z = 0 , a 3 x + b 3 y + c 3 z = 0 . Observations 1. In Cramer’s Rule, if ∆ 0 � = 0, there will exist a unique solution, namely x = 0, y = 0, z = 0. Each of ∆ 1 , ∆ 2 and ∆ 3 will contain a column of zeros. But this solution is obvious and we call it the “trivial solution” . 2. Question: are there any “non-trivial” solutions. 1

3. Non-trivial solutions occur when the equations are not linearly independent. (a) If one of the equations is redundant, we could solve the remaining two in an infinite number of ways by choosing one of the variables at random. (b) If two of the equations are redundant, we could solve the remaining equation in an infinite number of ways by choosing two of the variables at random. 4. There will be non-trivial solutions if a 1 b 1 c 1 � � � � � � � � ∆ 0 = = 0 . � a 2 b 2 c 2 � � � � � � � � a 3 b 3 c 3 � � � � � 5. If non-trivial solutions exist, any solution x = α, y = β, z = γ will imply other solutions of the form x = λα, y = λβ, z = λγ, where λ is any non-zero number. This is because the right-hand-sides are all zero. 2

TYPE 1 - One of the three equations is redundant The solution will be of the form x : y : z = α : β : γ. That is, x y = α z = β y γ and x z = α β, γ. Method (a) First eliminate z between two equations in order to find the ratio x : y Then eliminate y between two equations in order to find the ratio x : z Note: For a slightly simpler method, see Example 1 later. 3

TYPE 2 - Two of the three equations are redundant This case arises when the three homogeneous linear equa- tions are multiples of one another. If the only equation remaining is ax + by + cz = 0 , we could choose any two of the variables at random and solve for the remaining variable. For example, putting y = 0 we obtain x : y : z = − c a : 0 : 1; Similarly, putting z = 0 we obtain x : y : z = − b a : 1 : 0 Now we can generate solutions with any values, y = β , z = γ and a suitable x . In fact x = − ( β.b a + γ.c a ) , y = β, z = γ. Note: It may be shown that, for homogeneous linear simultane- ous equations, no other types of solution exist. 4

EXAMPLES 1. Show that the homogeneous linear equations 2 x + y − z = 0 , x − 3 y + 2 z = 0 , x + 4 y − 3 z = 0 have solutions other than x = 0, y = 0, z = 0 and determine the ratios x : y : z for these non-trivial solutions. Solution (a) Using the Rule of Sarrus, ∆ 0 =. 2 1 − 1 2 1 � � � � � � � � 1 − 3 2 1 − 3 = (18+2 − 4) − (3+16 − 3) = 0 . � � � � � � � � 1 4 − 3 1 4 � � � � � � Thus, the equations are linearly dependent and, hence, have non-trivial solutions. Note: We could, alternatively, have noticed that the first equation is the sum of the second and third equations. 5

(b) Slightly simpler method than before In a set of ratios α : β : γ , any of the three quantities ( � = 0) may be replaced by 1. For example, α : β : γ = α γ : β γ : 1 as long as γ � = 0. Let us now suppose that z = 1, giving 2 x + y − 1 = 0 , x − 3 y + 2 = 0 , x + 4 y − 3 = 0 . These give x = 1 7 and y = 5 7 , which means that x : y : z = 1 7 : 5 7 : 1 That is, x : y : z = 1 : 5 : 7 Any three numbers in these ratios form a solution. 6

2. Obtain the values of λ for which the homogeneous linear equations (1 − λ ) x + y − 2 z = 0 , − x + (2 − λ ) y + z = 0 , y − (1 − λ ) z = 0 have non-trivial solutions. Solution First we solve the equation 1 − λ 1 − 2 � � � � � � � � 0 = − 1 2 − λ 1 R 1 − → R 1 − R 3 � � � � � � � � � 0 1 − 1 − λ ) � � � � � 1 − λ 0 − 1 + λ � � � � � � � � = − 1 2 − λ 1 R 1 − → R 1 ÷ (1 − λ ) � � � � � � � � 0 1 − 1 − λ � � � � � � 1 0 − 1 � � � � � � � � = (1 − λ ) − 1 2 − λ 1 − → C 3 + C 1 � � C 3 � � � � � � 0 1 − 1 − λ � � � � � � 1 0 0 � � � � � � � � = (1 − λ ) − 1 2 − λ 0 = − (1 − λ )(2 − λ )(1+ λ ) � � � � � � � � � 0 1 − 1 − λ � � � � � Hence, λ = 1 , − 1 or 2. 7

3. Determine the general solution of the homogeneous linear equation 3 x − 7 y + z = 0 . Solution Substituting y = 0, we obtain 3 x + z = 0 and hence x : y : z = − 1 3 : 0 : 1. Substituting z = 0, we obtain 3 x − 7 y = 0 and hence x : y : z = 7 3 : 1 : 0 The general solution may thus be given by x = 7 β 3 − γ 3 , y = β, z = γ, where β and γ are arbitrary numbers. Note: Other equivalent versions are possible according to which of the three variables are chosen to have ar- bitrary values. 8