Lecture 4: Gaussian Elimination and Homogeneous Equations

Lecture 4: Gaussian Elimination and Homogeneous Equations

Reduced Row Echelon Form An augmented matrix associated to a system of linear equations is said to be in Reduced Row Echelon Form (RREF) if the following properties hold: The first non-zero entry from the left in each non-zero row is, and is called the leading for that row. 2 All entries in any column containing a leading are zero, except the leading itself. 3 Each leading is to the right of the leading s in the rows above it. 4 All rows consisting entirely of zeros are at the bottom of the matrix. The following matrices are in RREF: 2 3 2 4 3 4, 3, 2.

Definition Recalling that each column of the coefficient matrix corresponds to a variable in the system of equations, we call each variable associated to a leading in the RREF a leading variable. A variable (if any) that is not a leading variable is called a nonleading variable or a free variable. Thus, if the original system of equations has n variables, then n equals the number of leading variables plus the number of free variables. Extremely Important Point. When describing the solution set, each free variable is replaced by an independent parameter. Thus, the number of independent parameters describing the solution set equals the number of free variables. Equivalently: The number of parameters describing the solution set to a system of m equations in n unknowns equals n minus the number of leading s in the RREF of the augmented matrix.

Example Solve the given system of equations by rendering the associated augmented matrix into RREF. x 2x 2 x 3 + 3x 4 = 2x 4x 2 + x 3 = 5 x 2x 2 + 2x 3 3x 4 = 4 Solution: 2 3 2 3 2 4 5 2 R+R2 3 6 3 R 2 2 3 4 +R 3 3 6 3 3 R2 2 3 2 2 2 3 R2+R3 2 R 3 6 3 2+R Note that x, x 3 are leading variables and x 2, x 4 are free variables.

Example continued Thus, using independent parameters t and s, we have x = 2 + 2t s, x 2 = t, x 3 = + 2s, x 4 = s. In terms of a solution set, we have {(2 + 2t s, t, + 2s, t) s, t R}.

Algorithm for Gaussian elimination The following steps lead effectively to the RREF of the augmented matrix: Find the first column from the left containing a non-zero entry, say a, and interchange the row containing a with the first row.in this first step, a will more often than not be in the first row, first column of the augmented matrix. 2 Divide R by a, so that the leading entry of R is now. 3 Subtract multiples of R from the rows below R so that every entry in the matrix below the in R is. 4 Find the next column from the left containing a non-zero entry, say b, and interchange the row containing b with R 2. Now divide R 2 by b to get leading entry. 5 Use the leading in R 2 to get s above and below it. 6 Continue in this fashion until arriving at the RREF. If at any point in the process, we have a row consisting entirely of s, such a row should be moved to the bottom of the matrix.

Example Solve the system by reducing the augmented matrix to RREF. 3x + 7x 2 x 3 = x + 3x 2 + x 3 = x 2x 2 + x 3 =. 3 7 3 Solution: 3 R R2 3 7 2 2 R +R 3 3 R +R 2 3 2 4 4 R2 R3 2 2 3 2 2 2 4 4 2 R 2+R 3 3 5 5 2 2 3 R2+R 2 2.. Note, x, x 2 are leading variables and x 3 is a free variable. Thus, x = 5 + 5t, x 2 = 2 2t, x 3 = t, for all t R.

Class Example Use Gaussian elimination to solve the system by putting the augmented matrix into RREF: 2x + 3y + 3z = 9 3x 4y + z = 5 5x + 7y + 2z = 4 2 3 3 9 4 4 Solution: 3 4 5 R2+R 3 4 5 5 7 2 4 5 7 2 4 3 R +R 2 5 R +R 3 4 4 4 4 7 2 R2+R3 7. 2 22 24 Thus, the system has no solution.

Definition A system of linear equations is said to be homogeneous if the right hand side of each equation is zero, i.e., each equation in the system has the form ( ) a x + a 2 x 2 + + a n x n =. Note that x = x 2 = = x n = is always a solution to a homogeneous system of equations, called the trivial solution. Any other solution is a non-trivial solution. Two Important Properties.. Sums of solutions are solutions. Suppose (s,..., s n ) and (s,..., s n) are solutions to (*). Then a s + + a n s n = a s + + a n s n = Adding, we get: a (s + s ) + + a n(s n + s n) =, so that (s + s,..., s n + s n) is also a solution.

2. A scalar multiple of a solution to (*) is a solution. Suppose (s,..., s n ) is a solution, so that ( ) a s + + a n s n =. Let λ R. By a scalar multiple of a solution, we mean λ (s,..., s n ) = (λ s,..., λs n ). If we multiply (**) above by λ we get a (λs ) + + a n (λs n ) =, which shows that (λs,..., λs n ) is a solution to (*). Important Consequence: Sums and scalar multiples of solutions to a homogenous system of linear equations are again solutions to the same system of equations.

Definition Let v = (s,..., s n), v 2 = (s 2,..., s2 n),..., v k = (s k,..., sk n ) be solutions to a homogeneous system of m equations in n unknowns. A linear combination of v,..., v k is any expression of the form with each λ i R. If we write this out, we see that λ v + + λ k v k, λ v + λ k v k = (λ s + + λ k s k,..., λ s n + + λ k s k n ). For example, if v = (, 2, 3), v 2 = (,, ), v 3 = (,, 6) are solutions to a homogeneous systems of equations in three variables, we calculate a linear combination as follows: 2 v + 4 v 2 + 8 v 3 = 2 (, 2, 3) + 4 (,, ) + 8 (,, 6) = (2, 4, 6) + (, 4, 4) + (8, 8, 48) = (, 8, 5).

Comment: Expressions of the form v = (s, s 2,..., s n ) are called vectors and live in R n, Euclidean n-space. For various reasons, we use often the s s 2 alternate notation v =, but both expressions mean the same thing.. s n Thus a linear combination of vectors in R n might look like s s s k λ s λ 2. + + λ s k + λ 2s 2 + + λ ks k k 2. = λ s2 + λ 2s2 2 + + λ ks k 2.. sn λ sn + λ 2 sn 2 + + λ k sn k s k n In particular, 2 8 2 2 + 4 + 8 = 4 + 4 + 8 = 8. 3 6 6 4 48 5

Theorem Suppose v,..., v k R n are solutions to a homogeneous system of m linear equations in n unknowns. Then, any linear combination λ v + + λ k v k is also a solution. Moreover, given any homogenous system of m linear equations in n unknowns, there exist solutions, i.e., vectors v,..., v k, in R n such that every solution to the system is a linear combination of v,..., v k. Comments.. The first part of the theorem follows by combining the two Important Points from above. 2. By taking each λ j = above, one gets the zero solution, which is always a solution to any homogeneous system of linear equations. 3. It may be that the zero solution is the only solution, which is still consistent with the statement of the theorem.

Example In this example we illustrate how to find the so-called basic solutions to a homogeneous system of linear equations. Suppose a given system led to the following RREF of the augmented matrix [ ] 2. 4 Thus, x and x 3 are the leading variables and x 2, x 4 are the free variables. The solutions are: for all s, t R. x = 2s + t, x 2 = s, x 3 = 4t, x 4 = t,

Example continued As a solution set, the solutions are{( 2s + t, s, 4t, t) s, t R}. 2s + t Alternately, the solutions are all exprssions of the form s 4t, for all t s, t R.. We can write this last expression as a linear combination: 2s + t 2 ( ) s 4t = s + t 4, t where s, t are arbitrary real numbers.

Example continued Thus, every solution to the original system of equations is a linear combination of the vectors 2 v = and v 2 = 4. We call the vectors (solutions) v, v 2 basic solutions to the system of equations. Note that the basic solutions are in fact solutions, since v is obtained by taking s =, t = in ( ) and v 2 is obtained by taking s =, t = in ( ).

Class Example Find two basic solutions to the homogeneous system of equations having augmented matrix [ ] 3. 4 Solution: Note that x, x 2 are the leading variables and x 3, x 4 are the free 3s + t variables. Thus, in vector form the solutions are all vectors 4t s, t with s, t R arbitrary. Separating this expression into two terms, we get 3s + t 4t s t = s 3 + t 4.

Class Example continued 3 Thus, v = and v 2 = 4 the original system of equations. are basic vectors for the solutions to We will see later in the semester, that basic vectors for the solution space of a homogeneous system of linear equations need not be unique. For example, there are infinitely many pairs of basic vectors for the solution space of the system above.

Class Example Express the solution to the homogenous system with the following augmented matrix in terms of basic solutions: 2 2 2 2. 3 6 3 3 2 2 2 2 Solution: 2 2 R+R2 3 3 3 R 3 6 3 3 +r 3 3 3 3 R2 2 2 3 R2+R3 2. R 3 3 2+R Thus, x, x 3 are the leading variables and x 2, x 4, x 5 are the free variables. The solutions are: x = 2s t u, x 2 = s, x 3 = t, x 4 = t, x 5 = u, for all real numbers s, t, u.

Class Example continued In terms of the basic solutions we have: x x 2 x 3 x 4 x 5 = 2s t u s t t u = s 2 + t + u.