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

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 ∆ ÷∆ ÷ ∆ ÷ … 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

» 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 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 | −1ÿ −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 … 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 Ω 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. 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 + 2y − z = 0 2y + 3z = 0 x + 4y + 2z = 0

≈ 1 2 −1 | 0 ’ ≈ 1 2 −1 | 0 ’ ∆ ÷ ∆ ÷ ∆ 0 2 3 | 0 ÷ ∆ 0 2 3 | 0 ÷ ∆ ÷ ∆ ÷ « 1 4 2 | 0 ◊ −R1 + R3 « 0 2 3 | 0 ◊ repeated row

≈ 1 2 −1 | 0 ’ x + 2y − z = 0 x = z − 2y Ω x = 4t ∆ ÷ −3 ∆ 0 2 3 | 0 ÷ 2y + 3z = 0 y = 2 t ∆ ÷ « 0 0 0 | 0 ◊ z is free let z = t

There are infinitely many solutions, one in particular is (8, -3,2). This is a nontrivial solution found by letting t be 2.

7 In a homogeneous system of equations, if you have more variables than equations, you are guaranteed to have nontrivial solutions

x + 3y − 2z = 0 x + 3y + 4z = 0

≈ 1 3 −2 | 0 ’ ∆ ÷ « 1 3 4 | 0 ◊−R1 + R2 x = −3t ≈ 1 3 −2 | 0 ’ x + 3y − 2z = 0 Ω x = −3y y is free ∆ ÷ let y = t « 0 0 6 | 0 ◊ 6z = 0 Ω z = 0 y = t z = 0

With fewer equations than variables you will always have at least one free parameter, this leads to infinitely many nontrivial solutions