Section 8.2 Solving a System of Equations Using Matrices (Guassian Elimination)

Transcription

1 Section 8. Solving a System of Equations Using Matrices (Guassian Elimination) x + y + z = x y + 4z = x 4y + z = System of Equations x 4 y = 4 z A System in matrix form x A x = b b 4 4 Augmented Matrix Not every system has a unique solution. There are three different possible solutions a unique solution (exactly one solution) infinitely many solutions the system is called consistent no solution the system is called inconsistent

2 Starting with an augmented matrix, you have two options: Use row operations to reduce to: row-echelon form Any row consisting of all zeros must be on the bottom of the matrix For all nonzero rows, the first nonzero entry must be a. This is called the leading. Take any consecutive nonzero rows: The leading for the higher row must be to the left of the leading of the lower row. The leading ones must staircase down from left to right. Reduction to row-echelon form is called : Gaussian elimination The solution is then found by back-substitution reduced row-echelon form Row echelon form + Find all the leading ones. All other entries in the column containing a leading should be zero (above and below the leading ). Reduction to reduced row-echelon form is called : Gauss-Jordan elimination The solution is then found by inspection or by a few simple steps Row-echelon, Reduced row-echelon, or Neither row-echelon neither reduced row-echelon neither reduced row-echelon row-echelon Order Matters! row-echelon for each column (move left to right), first get the appropriate leading, then get s underneath it. reduced row-echelon for each column (move left to right), first get the appropriate leading, then get s above and below it.

3 Permitted Row Operations : (remember: every row represents an equation) a) Multiply a row by a number R b) Switch rows 7 4 R c) Add a multiple of one row to another row Row that is not changing 7 R 4 Row you want to replace 4 R + R = New R R 6 + R 4 NewR 4 4 To get s : a) Switch Rows if there is a in the same column but below the desired spot. 7 4 R b) If k is the entry in the desired spot, multiply the row by if every other entry in the row is divisible by k R c) Do step b) followed by step a) if there is another row where every entry is divisible by k R R R R These are the easier ways to get a k

4 To get s : (continued) d) Use elimination step add one row to a multiple of another row 4 R + R = New R R e) Last Resort Introduce fractions by multiplying by k + R 4 NewR R These are the harder ways to get a To get s : Use "elimination" step : add a multiple of one row to another row Row with the leading in it The leading is always obtained before getting the zero(s) Row you want to replace Multiply the row with the leading by the same # but opposite sign of the number you would like to be zero. 4 R + R = New R R 6 + R 4 NewR 4 4 4

5 x + y + z = x y + 4z = 4 x 4y + z = R R R R 6 4 R + R = New R + R 4 NewR R 4 R + R = New R + R NewR R + R = New R R + R = New R then ( ) + + R 5 R NewR 4 5 5R 5 5 R NewR 5 x y + z = y = z = ( ) ( ) x + = x 4 = x = Solution : (,, ) 5

6 x y + z = 6 x y = 7 x + y z = 6 7 R + R R + R R + R 6 x y + z = 6 x = 6 + y z x = t t 5 y z = 5 y = 5 + t z is free let z = t Infinitely many solutions x = 4 + t y = 5 + t z = t x + 6y + 6z = 5 x 6y z = x y = R 5 6 R + R R + R R R R No solution x + y + z = = FALSE Inconsistent System 6

7 A system of linear equations is said to be homogeneous is each of its equations has a constant term of. a x + a x + + a x = n n a x + a x + + a x = n n a x + a x + + a x m m mn n x x x n,,, is always a solution of a homogeneous system. This solution is called the trivial solution. This means that a homogeneous system is always consistent. For a homogeneous system, there are only two different possible solutions : a unique solution (the trivial solution) infinitely many solutions What is more interesting is when there is a solution that has one or more of the variables not zero. A solution of this type is called a non-trivial solution. = x + y z y + z x + 4y + z 4 R + R repeated row x + y z y + z z is free x = z y x = 4t y = t let z = t There are infinitely many solutions, one in particular is (8, -,). This is a nontrivial solution found by letting t be. 7

8 In a homogeneous system of equations, if you have more variables than equations, you are guaranteed to have nontrivial solutions x + y z x + y + 4z 4 R + R x + y z x = y y is free 6 6z z let y = t x = t y = t z With fewer equations than variables you will always have at least one free parameter, this leads to infinitely many nontrivial solutions 8