Section 8.2 Solving a System of Equations Using Matrices (Guassian Elimination) 2x + y + 3z = 1 ≈2 1 3’≈x ’ ≈ 1 ’ 2 1 3 | 1 ∆ ÷∆ ÷ ∆ ÷ … Ÿ 3x − 2y + 4z = −1 ∆3 −2 4÷∆y ÷= ∆−1÷ Ω …3 −2 4 | −1Ÿ ∆ ÷∆ ÷ ∆ ÷ 2x − 4y + 2z = −2 «2 −4 2◊«z ◊ «−2◊ …2 −4 2 | −2⁄Ÿ %(&(' k k A x b System of Equations Ax = b Augmented Matrix System in matrix form Not every system has a unique solution. There are three different possible solutions • a unique solution ¤ Œ (exactly one solution) Œthe system is called ‹ Œ consistent • infinitely many solutions Œ ›Œ ¤ the system is called • no solution ‹ › inconsistent 1 Starting with an augmented matrix, you have two options: Use row operations to reduce to: row-echelon form reduced row-echelon form Any row consisting of all zeros must Row echelon form + be on the bottom of the matrix Find all the leading ones. For all nonzero rows, the first nonzero All other entries in the column entry must be a 1. This is called the “leading 1”. containing a leading 1 should be zero Take any 2 consecutive nonzero rows: (above and below the leading 1). The leading 1 for the higher row must be to the left of the leading 1 of the lower row. The leading ones must “staircase down” from left to right. Reduction to reduced row-echelon Reduction to row-echelon form is called : form is called : Gaussian elimination Gauss-Jordan elimination The solution is then found by The solution is then found by back-substitution inspection or by a few simple steps Row-echelon, Reduced row-echelon, or Neither 1. 2. 3. 1 2 5 | 3 1 0 1 | 5 1 0 3 | 1 0 0 1 | 2 0 0 1 | 2 0 1 2 | 4 0 0 0 | 0 0 1 4 | 1 row-echelon neither reduced row-echelon 4. 5. 6. 1 3 0 2 | 0 1 2 0 0 | 9 1 3 −4 | 7 1 0 2 2 | 6 0 0 1 0 | 1 0 1 2 | 2 0 0 0 1 | 0 0 0 0 1 | 7 0 0 1 | 5 0 0 0 0 | 0 0 0 0 0 | 0 neither reduced row-echelon row-echelon Order Matters! row-echelon – for each column (move left to right), first get the appropriate leading 1, then get 0’s underneath it. reduced row-echelon – for each column (move left to right), first get the appropriate leading 1, then get 0’s above and below it. 2 3 Permitted Row Operations : (remember: every row represents an equation) a) Multiply a row by a number 1 3 −9 6 | 15 3 ⋅ R Ω » 1 −3 2 | 5 ÿ … Ÿ 1 … Ÿ …5 −2 4 | −1Ÿ …5 −2 4 | −1Ÿ … Ÿ 2 −4 2 | −2⁄ …2 −4 2 | −2⁄Ÿ b) Switch rows » 1 7 3 | 1 ÿ » 1 7 3 | 1 ÿ … Ÿ … Ÿ … 0 −2 4 | −1 Ÿ R2 ↔ R3 …0 1 2 | −2Ÿ … Ÿ 0 1 2 | −2 ⁄ …0 −2 4 | −1⁄Ÿ c) Add a multiple of o%(ne& ro('w to a%no(th&er( ro'w Row that is Row you want not changing to replace » ÿ 1 −2 1 | −1 −3R −3 6 −3 | 3 » 1 −2 1 | −1ÿ … Ÿ 1 … Ÿ 3 −2 4 | −1 −3R + R = New R Ω Ω … Ÿ 1 2 2 + R2 3 −2 4 | −1 …0 4 1 | 2 Ÿ … Ÿ … Ÿ 2 1 3 | 1 ⁄ NewR2 0 4 1 | 2 2 1 3 | 1 ⁄ To get 1’s : a) Switch Rows if there is a 1 in the same column but below the desired spot. » 1 7 3 | 1 ÿ » 1 7 3 | 1 ÿ … Ÿ … Ÿ … 0 −2 4 | −1 Ÿ R2 ↔ R3 …0 1 2 | −2Ÿ … Ÿ 0 1 2 | −2 …0 −2 4 | −1⁄Ÿ ⁄ 1 b) If k is the entry in the desired spot, multiply the row by k if every other entry in the row is divisible by k. » ÿ 1 3 −9 6 | 15 3 ⋅ R Ω » 1 −3 2 | 5 ÿ … Ÿ 1 … Ÿ …5 −2 4 | −1Ÿ …5 −2 4 | −1Ÿ … Ÿ 2 −4 2 | −2⁄ …2 −4 2 | −2⁄Ÿ c) Do step b) followed by step a) if there is another row where every entry is divisible by k. » 2 1 3 | 1 ÿ » 2 1 3 | 1 ÿ » 1 −2 1 | −1ÿ … Ÿ … Ÿ … Ÿ …3 −2 4 | −1Ÿ …3 −2 4 | −1Ÿ …3 −2 4 | −1Ÿ … Ÿ 1 Ω … Ÿ … Ÿ 2 −4 2 | −2⁄ 2 ⋅ R3 1 −2 1 | −1⁄R3 ↔ R1 2 1 3 | 1 ⁄ These are the “easier” ways to get a 1 3 To get 1’s : (continued) d) Use “elimination” step – add one row to a multiple of another row −R −2 −1 −3 | −1 3 −2 4 | −1 −R + R = New R 2 … Ÿ 2 1 1 …2 1 3 | 1 Ÿ Ω + R1 3 −2 4 | −1 … Ÿ NewR 1 −3 1 | −2 2 −4 7 | −2⁄ 1 » 1 −3 1 | −2ÿ … Ÿ …2 1 3 | 1 Ÿ …2 −4 7 | −2⁄Ÿ 1 e) Last Resort – Introduce fractions by multiplying by k » ÿ » 2 −5 7 | 11ÿ 1 Ω −5 7 11 ⋅ R1 … 1 | Ÿ … Ÿ 2 2 2 2 …4 −9 4 | −1Ÿ … Ÿ … Ÿ … 4 −9 4 | −1Ÿ 6 −4 2 | −2⁄ … Ÿ … 6 −4 2 | −2Ÿ ⁄ These are the “harder” ways to get a 1 To get 0’s : Use "elimination" step : add a multiple of o%(ne& ro('w to a%no(th&er( ro'w Row with the Row you want leading 1 in it to replace The leading 1 is always obtained before getting the zero(s) Multiply the row with the “leading” 1 by the same # but opposite sign of the number you would like to be zero. » 1 −2 1 | −1ÿ … Ÿ …3 −2 4 | −1Ÿ −3R1 + R2 = New R2 … Ÿ 2 1 3 | 1 ⁄ −3R1 −3 6 −3 | 3 » 1 −2 1 | −1ÿ Ω Ω … Ÿ + R2 3 −2 4 | −1 …0 4 1 | 2 Ÿ … Ÿ NewR2 0 4 1 | 2 2 1 3 | 1 ⁄ 4 2x + y + 3z = 1 2 1 3 | 1 … Ÿ 3x − 2y + 4z = −1 Ω …3 −2 4 | −1Ÿ 2x − 4y + 2z = −2 …2 −4 2 | −2⁄Ÿ » 2 1 3 | 1 ÿ » 2 1 3 | 1 ÿ » 1 −2 1 | −1ÿ … Ÿ … Ÿ … Ÿ …3 −2 4 | −1Ÿ …3 −2 4 | −1Ÿ …3 −2 4 | −1Ÿ … Ÿ 1 Ω … Ÿ … Ÿ 2 −4 2 | −2⁄ 2 ⋅ R3 1 −2 1 | −1⁄ R3 ↔ R1 2 1 3 | 1 ⁄ » 1 −2 1 | −1ÿ … Ÿ −3R1 −3 6 −3 | 3 …3 −2 4 | −1Ÿ −3R + R = New R Ω 1 2 2 + R2 3 −2 4 | −1 …2 1 3 | 1 Ÿ ⁄ NewR2 0 4 1 | 2 » 1 −2 1 | −1ÿ … Ÿ Ω …0 4 1 | 2 Ÿ …2 1 3 | 1 ⁄Ÿ » 1 −2 1 | −1ÿ … Ÿ …0 4 1 | 2 Ÿ −2R1 −2 4 −2 | 2 … Ÿ Ω + R3 2 1 3 | 1 2 1 3 | 1 ⁄−2R1 + R3 = New R3 NewR3 0 5 1 | 3 » 1 −2 1 | −1ÿ … Ÿ Ω …0 4 1 | 2 Ÿ …0 5 1 | 3 ⁄Ÿ » 1 −2 1 | −1ÿ −R3 0 −5 −1 | −3 … Ÿ Ω + R 0 4 1 | 2 …0 4 1 | 2 Ÿ−R3 + R2 = New R2 2 … Ÿ 0 5 1 | 3 ⁄ NewR2 0 −1 0 | −1 then » 1 −2 1 | −1ÿ × (−1) … Ÿ Ω …0 1 0 | 1 Ÿ …0 5 1 | 3 ⁄Ÿ » 1 −2 1 | −1ÿ … Ÿ …0 1 0 | 1 Ÿ −5R2 0 −5 0 | −5 … Ÿ 0 5 1 | 3 ⁄ Ω + R 0 5 1 | 3 −5R2 + R3 = New R3 3 NewR3 0 0 1 | −2 » 1 −2 1 | −1ÿΩ x − 2y + z = −1 Ω x − 2(1) + (−2) = −1 Ω x − 4 = −1 Ω x = 3 … Ÿ …0 1 0 | 1 ŸΩ y =1 … Ÿ 0 0 1 | −2⁄Ω z = −2 Solution : (3,1, −2) 5 x − 2y + z = −6 2x − 3y = −7 −x + 3y − 3z = 11 1 −2 1 | −6 1 −2 1 | −6 2 −3 0 | −7 −2R1 + R2 0 1 −2 | 5 −1 3 −3 | 11 R1 + R3 0 1 −2 | 5 −R2 + R3 1 −2 1 | −6 x − 2y + z = −6 x = −6 + 2y − z x = −6 +10 + 4t − t 0 1 −2 | 5 y − 2z = 5 y = 5+ 2t 0 0 0 | 0 z is free let z = t x = 4 + 3t y = 5 + 2t Infinitely many solutions z = t 3x + 6y + 6z = 5 3x − 6y − 3z = 2 3x − 2y = 1 1 5 3 6 6 | 5 3 R1 1 2 2 | 3 3 −6 −3 | 2 3 −6 −3 | 2 −3R1 + R2 3 −2 0 | 1 3 −2 0 | 1 −3R1 + R3 5 5 1 2 2 | 3 1 2 2 | 3 −1 3 1 0 −12 −9 | −3 12 R2 0 1 4 | 4 8R2 + R3 0 −8 −6 | −4 0 −8 −6 | −4 5 1 2 2 | 3 3 1 0 1 4 | 4 0 0 0 | −2 0x + 0y + 0z = −2 0 = −2 FALSE No solution Inconsistent System 6 A system of linear equations is said to be homogeneous is each of its equations has a constant term of 0. a11x1 + a12 x2 + ? + a1n xn = 0 a21x1 + a22 x2 + ? + a2n xn = 0 @ @ @ = 0 am1x1 + am2 x2 + ? + amn xn − 0 x1 = 0, x2 = 0,>, xn = 0 is always a solution of a homogeneous system. This solution is called the trivial solution.