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IBMathsDPAnalysis & Approaches (AA)HLRevision Notes1. Number & AlgebraSystems of Linear EquationsAlgebraic Solutions

Algebraic Solutions (DP IB Analysis & Approaches (AA)): Revision Note

Dan Finlay

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Row Reduction

How can I write a system of linear equations?

  • To save space we can just write the coefficients without the variables

  • For 2 variables: a subscript 1 x plus b subscript 1 y equals c subscript 1 a subscript 2 x plus b subscript 2 y equals c subscript 2 can be written shorthand as open square brackets table row cell a subscript 1 end cell cell b subscript 1 end cell row cell a subscript 2 end cell cell b subscript 2 end cell end table stretchy vertical line space table row cell c subscript 1 end cell row cell c subscript 2 end cell end table close square brackets

  • For 3 variables: a subscript 1 x plus b subscript 1 y plus c subscript 1 z equals d subscript 1 a subscript 2 x plus b subscript 2 y plus c subscript 2 z equals d subscript 2 a subscript 3 x plus b subscript 3 y plus c subscript 3 z equals d subscript 3 can be written shorthand as open square brackets table row cell a subscript 1 end cell cell b subscript 1 end cell cell c subscript 1 end cell row cell a subscript 2 end cell cell b subscript 2 end cell cell c subscript 2 end cell row cell a subscript 3 end cell cell b subscript 3 end cell cell c subscript 3 end cell end table stretchy vertical line space table row cell d subscript 1 end cell row cell d subscript 2 end cell row cell d subscript 3 end cell end table close square brackets

What is a row reduced system of linear equations?

  • A system of linear equations is in row reduced form if it is written as:

    • open square brackets table row cell A subscript 1 end cell cell B subscript 1 end cell cell C subscript 1 end cell row 0 cell B subscript 2 end cell cell C subscript 2 end cell row 0 0 cell C subscript 3 end cell end table stretchy vertical line space table row cell D subscript 1 end cell row cell D subscript 2 end cell row cell D subscript 3 end cell end table close square brackets which corresponds to table attributes columnalign right center left columnspacing 0px end attributes row cell A subscript 1 x plus B subscript 1 y plus C subscript 1 z end cell equals cell D subscript 1 end cell row cell B subscript 2 y plus C subscript 2 z end cell equals cell D subscript 2 end cell row cell C subscript 3 z end cell equals cell D subscript 3 end cell end table

      • It is very helpful if the values of A1, B2, C3 are equal to 1

What are row operations?

  • Row operations are used to make the linear equations simpler to solve

    • They do not affect the solution

  • You can switch any two rows

    • begin mathsize 14px style stretchy left square bracket table row cell a subscript 1 end cell cell b subscript 1 end cell cell c subscript 1 end cell row cell a subscript 2 end cell cell b subscript 2 end cell cell c subscript 2 end cell row cell a subscript 3 end cell cell b subscript 3 end cell cell c subscript 3 end cell end table stretchy vertical line space table row cell d subscript 1 end cell row cell d subscript 2 end cell row cell d subscript 3 end cell end table stretchy right square bracket end style  can be written as begin mathsize 14px style stretchy left square bracket table row cell a subscript 3 end cell cell b subscript 3 end cell cell c subscript 3 end cell row cell a subscript 2 end cell cell b subscript 2 end cell cell c subscript 2 end cell row cell a subscript 1 end cell cell b subscript 1 end cell cell c subscript 1 end cell end table stretchy vertical line space table row cell d subscript 3 end cell row cell d subscript 2 end cell row cell d subscript 1 end cell end table stretchy right square bracket end style using r1 ↔ r3

      • This is useful for getting zeros to the bottom

      • Or getting a one to the top

  • You can multiply any row by a (non-zero) constant

    • begin mathsize 14px style stretchy left square bracket table row cell a subscript 1 end cell cell b subscript 1 end cell cell c subscript 1 end cell row cell a subscript 2 end cell cell b subscript 2 end cell cell c subscript 2 end cell row cell a subscript 3 end cell cell b subscript 3 end cell cell c subscript 3 end cell end table stretchy vertical line space table row cell d subscript 1 end cell row cell d subscript 2 end cell row cell d subscript 3 end cell end table stretchy right square bracket end style can be written as begin mathsize 14px style stretchy left square bracket table row cell a subscript 1 end cell cell b subscript 1 end cell cell c subscript 1 end cell row cell k a subscript 2 end cell cell k b subscript 2 end cell cell k c subscript 2 end cell row cell a subscript 3 end cell cell b subscript 3 end cell cell c subscript 3 end cell end table stretchy vertical line space table row cell d subscript 1 end cell row cell k d subscript 2 end cell row cell d subscript 3 end cell end table stretchy right square bracket end style using k×r2 → r2

      • This is useful for getting a 1 as the first non-zero value in a row

  • You can add multiples of a row to another row

    • begin mathsize 14px style stretchy left square bracket table row cell a subscript 1 end cell cell b subscript 1 end cell cell c subscript 1 end cell row cell a subscript 2 end cell cell b subscript 2 end cell cell c subscript 2 end cell row cell a subscript 3 end cell cell b subscript 3 end cell cell c subscript 3 end cell end table stretchy vertical line space table row cell d subscript 1 end cell row cell d subscript 2 end cell row cell d subscript 3 end cell end table stretchy right square bracket end style can be written as begin mathsize 14px style stretchy left square bracket table row cell a subscript 1 end cell cell b subscript 1 end cell cell c subscript 1 end cell row cell a subscript 2 plus 5 a subscript 3 end cell cell b subscript 2 plus 5 b subscript 3 end cell cell c subscript 2 plus 5 c subscript 3 end cell row cell a subscript 3 end cell cell b subscript 3 end cell cell c subscript 3 end cell end table stretchy vertical line space table row cell d subscript 1 end cell row cell d subscript 2 plus 5 d subscript 3 end cell row cell d subscript 3 end cell end table stretchy right square bracket end style using r2 + 5r3 → r2

      • This is useful for creating zeros under a 1

How can I row reduce a system of linear equations?

  • STEP 1: Get a 1 in the top left corner

    • You can do this by dividing the row by the current value in its place

    • If the current value is 0 or an awkward number then you can swap rows first

      • begin mathsize 14px style stretchy left square bracket table row 1 cell B subscript 1 end cell cell C subscript 1 end cell row asterisk times asterisk times asterisk times row asterisk times asterisk times asterisk times end table stretchy vertical line space table row cell D subscript 1 end cell row asterisk times row asterisk times end table stretchy right square bracket end style

  • STEP 2: Get 0’s in the entries below the 1

    • You can do this by adding/subtracting a multiple of the first row to each row

      • begin mathsize 14px style stretchy left square bracket table row 1 cell B subscript 1 end cell cell C subscript 1 end cell row 0 asterisk times asterisk times row 0 asterisk times asterisk times end table stretchy vertical line space table row cell D subscript 1 end cell row asterisk times row asterisk times end table stretchy right square bracket end style

  • STEP 3: Ignore the first row and column as they are now complete

    • Repeat STEPS 1 - 2 to the remaining section

      • Get a 1: begin mathsize 14px style stretchy left square bracket table row 1 cell B subscript 1 end cell cell C subscript 1 end cell row 0 1 cell C subscript 2 end cell row 0 asterisk times asterisk times end table stretchy vertical line space table row cell D subscript 1 end cell row cell D subscript 2 end cell row asterisk times end table stretchy right square bracket end style

      • Then 0 underneath: begin mathsize 14px style stretchy left square bracket table row 1 cell B subscript 1 end cell cell C subscript 1 end cell row 0 1 cell C subscript 2 end cell row 0 0 asterisk times end table stretchy vertical line space table row cell D subscript 1 end cell row cell D subscript 2 end cell row asterisk times end table stretchy right square bracket end style

  • STEP 4: Get a 1 in the third row

    • Using the same idea as STEP 1

      • begin mathsize 14px style stretchy left square bracket table row 1 cell B subscript 1 end cell cell C subscript 1 end cell row 0 1 cell C subscript 2 end cell row 0 0 1 end table stretchy vertical line space table row cell D subscript 1 end cell row cell D subscript 2 end cell row cell D subscript 3 end cell end table stretchy right square bracket end style

How do I solve a system of linear equations once it is in row reduced form?

  • Once you row reduced the equations you can then convert back to the variables

    • open square brackets table row 1 cell B subscript 1 end cell cell C subscript 1 end cell row 0 1 cell C subscript 2 end cell row 0 0 1 end table stretchy vertical line space table row cell D subscript 1 end cell row cell D subscript 2 end cell row cell D subscript 3 end cell end table close square brackets corresponds to table attributes columnalign right center left columnspacing 0px end attributes row cell x plus B subscript 1 y plus C subscript 1 z end cell equals cell D subscript 1 end cell row cell y plus C subscript 2 z end cell equals cell D subscript 2 end cell row z equals cell D subscript 3 end cell end table

  • Solve the equations starting at the bottom

    • You have the value for z from the third equation

    • Substitute z into the second equation and solve for y

    • Substitute z and y into the first each and solve for x

Examiner Tips and Tricks

  • To reduce the number of operations you do whilst solving a system of operations, you can do a couple of things:

    • You can set up your original matrix with the equations in any order, so if one of the equations already has a 1 in the top left corner, put that one first

    • You do not need to make every equation so that it only has a single variable with a value of 1, you just need to do that for 1 of the equations and use that result to work out the others

Worked Example

Solve the following system of linear equations using algebra.

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x minus 3 y plus 4 z end cell equals 14 row cell x plus 2 y minus 2 z end cell equals cell negative 2 end cell row cell 3 x minus y minus 2 z end cell equals 10 end table

1-10-2-ib-aa-hl-row-reduction-we-solution

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Number of Solutions to a System

How many solutions can a system of linear equations have?

  • There could be

    • 1 unique solution

    • No solutions

    • An infinite number of solutions

  • You can determine the case by looking at the row reduced form

How do I know if the system of linear equations has no solutions?

  • Systems with no solutions are called inconsistent

  • When trying to solve the system after using the row reduction method you will end up with a mathematical statement which is never true:

    • Such as: 0 = 1

  • The row reduced system will contain:

    • At least one row where the entries to the left of the line are zero and the entry on the right of the line is non-zero

      • Such a row is called inconsistent

    • For example:

      • Row 2 is inconsistent stretchy left square bracket table row 1 cell B subscript 1 end cell cell C subscript 1 end cell row 0 0 0 row 0 0 1 end table stretchy vertical line space table row cell D subscript 1 end cell row cell D subscript 2 end cell row cell D subscript 3 end cell end table stretchy right square bracket if D2 is non-zero

How do I know if the system of linear equations has an infinite number of solutions?

  • Systems with at least one solution are called consistent

    • The solution could be unique or there could be an infinite number of solutions

  • When trying to solve the system after using the row reduction method you will end up with a mathematical statement which is always true

    • Such as: 0 = 0

  • The row reduced system will contain:

    • At least one row where all the entries are zero

    • No inconsistent rows

    • For example:

      • open square brackets table row 1 cell B subscript 1 end cell cell C subscript 1 end cell row 0 1 cell C subscript 2 end cell row 0 0 0 end table stretchy vertical line space table row cell D subscript 1 end cell row cell D subscript 2 end cell row 0 end table close square brackets

How do I find the general solution of a dependent system?

  • A dependent system of linear equations is one where there are infinite number of solutions

  • The general solution will depend on one or two parameters

  • In the case where two rows are zero

    • Let the variables corresponding to the zero rows be equal to the parameters λ & μ

      • For example: If the first and second rows are zero rows then let x = λ & y = μ

    • Find the third variable in terms of the two parameters using the equation from the third row

      • For example: z = 4λ – 5μ + 6

  • In the case where only one row is zero

    • Let the variable corresponding to the zero row be equal to the parameter λ

      • For example: If the first row is a zero row then let x = λ

    • Find the remaining two variables in terms of the parameter using the equations formed by the other two rows

      • For example: y = 3λ – 5 & z = 7 – 2λ

Examiner Tips and Tricks

  • Common questions that pop up in an IB exam include questions with equations of lines

  • Being able to recognise whether there are no solutions, 1 solution or infinite solutions is really useful for identifying if lines are coincident, skew or intersect!

Worked Example

table attributes columnalign right center left columnspacing 0px end attributes row cell x plus 2 y minus z end cell equals 3 row cell 3 x plus 7 y plus z end cell equals 4 row cell x minus 9 z end cell equals k end table

a) Given that the system of linear equations has an infinite number of equations, find the value of k.

1-10-2-ib-aa-hl-general-solution-a-we-solution

b) Find a general solution to the system.

1-10-2-ib-aa-hl-general-solution-b-we-solution

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    Dan Finlay

    Author: Dan Finlay

    Expertise: Maths Lead

    Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.